quadratic equations - inyatrust · adfected quadratic equations. solving pure quadratic equations....
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This unit facilitates you in,
defining quadratic equation, pure and
adfected quadratic equations.
solving pure quadratic equations.
solving quadratic equations by
factorisation method.
solving quadratic equations by
completing the square method.
deriving the formula to find the roots of
quadratic equation.
using formula to solve quadratic
equations.
drawing graphs for quadratic expressions.
solving quadratic equations graphically.
establising relation between roots and
coefficients of quadratic equation.
framing quadratic equations.
finding the discriminant and interpret
the nature of roots of quadratic equation.
Quadratic Equations
Quadratic expression andquadratic equation
Pure and adfected quadraticequations
Solution of quadratic equationby
* Factorisation method
* Completing the square method
* Formula method
* Graphical method
Relation between roots andcoefficients
Forming quadratic equations
9
Brahmagupta
(A.D 598-665, India)
Solving of quadratic equationsin general form is often credited
to ancient Indian mathe-maticians. Brahmagupta gavean expl icit formula tosolve a quadratic equationof the form ax2 + bx = c. LaterSridharacharya (A.D. 1025)
derived a formula, now knownas the quadratic formula, forsolving a quadratic equation bythe method of completing thesquare.
An equation means nothing to me, unless it
expresses a thought of God.
- Srinivas Ramanujan
194 UNIT-9
We are familiar with playing number games. Let us consider two such examples.
Example 1 Example 2
* Take a non-zero whole number. * Take a non-zero whole number
* Add 7 to it. * Add 7 to it.
* Equate it to 12. * Multiply the sum by the same
whole number.
* What is the number? * Equate it to 12.
* What is the number?
How to find the numbers? Let the non-zero whole number be 'x'
If we follow the steps, we get
in example 1, x x + 7 and x + 7 = 12 ........(i)
in example 2, x x + 7 andx(x + 7) = 12 ........(ii)
Take equation (i), x + 7 = 12. Here, x is the variable, whose degree is 1.
It is a linear equation. It has only one root.
i.e., x + 7 = 12, x = 12 – 7, x = 5
Take equation (ii), x (x + 7) = 12, x2 + 7x = 12
Here, x is the variable, whose degree is 2. It is a quadratic equation.
How to solve it? How many roots does a quadratic equation have?
It is very essential to learn this, because quadratic equations have wide applications
in other branches of mathematics, in other subjects and also in real life situations.
For instance, suppose an old age home trust decides to build a prayer hall having
floor area of 300 sq.m., with its length one meter more than twice its breadth . What
should be the length and breadth of the hall?
Let, the breadth be x m. Then, its length will be (2x + 1)m
Its area = x(2x + 1)sq.m x(2x + 1) = 300 ( Given)
This information can be diagrammatically represented as follows.
We have, Area = x(2x + 1) = (2x2 + x)m2
So, 2x2 + x = 300. This is a quadratic equation.
Below are given some more illustrations in verbal statement form which when
converted into equation form result in quadratic equation form.
Study the statements. Try to express each statement in equation form.
1. An express train takes one hour less than the passenger train to travel 132 km
between Bangalore and Mysore. If the average speed of the express train is
11 km/hr more than that of the passenger train, what is the average speed of the
two trains?
22 1 300 m mx x x
(2x + 1) m
Quadratic Equations 195
2. A cottage industry produces a certain number of wooden toys in a day. The cost of
production of each toy (in rupees) was found to be 55 minus the number of toys
produced in a day. On a particular day, the total cost of production was ̀ 750. What
is the number of toys produced on that day?
3. A motor boat whose speed is 18 km/hr in still water takes 1 hour more to go 24 km
upstream than to return downstream to the same spot. What is the speed of the
stream?
4. Two water taps together can fill a tank in 93
8 hours. The tap of larger diameter
takes 10 hours less than the smaller one to fill the tank separately. Find the time
in which each tap can separately fill the tank.
How to solve these problems?
Know this!
It is believed that Babylonians were the first to solve quadratic equations. Greek
mathematician Euclid developed a geometrical approach for finding lengths, which
are nothing but solutions of quadratic equations.
Solving of quadratic equations, in general form, is often credited to ancient Indian
mathematicians like Brahmagupta (A.D. 598-665) and Sridharacharya (A.D. 1025).
An Arab mathematician Al-khwarizni (about A.D. 800) also studied quadratic equations
of different types.
Abraham bar Hiyya Ha-Nasi, in his book "Liber embardorum" published in Europe in
A.D. 1145 gave complete solutions of different quadratic equations.
In this unit, let us study quadratic equations, various methods of finding their roots
and also applications of quadratic equations.
Quadratic equation
Recall that we have studied about quadratic polynomials in unit 8.
A polynomial of the form ax2 + bx + c, where a 0 is a quadratic polynomial or
expression in the variable x of degree 2. If a quadratic expression ax2 + bx + c is equated
to zero, it becomes a quadratic equation.
Below are given some verbal statements, when converted to quadratic expression
form and further equated to zero become quadratic equations. Study the examples.
196 UNIT-9
Know this!
The word quadratic is derived from
the Latin word "quadratum" which
means "A square figure".
So it can be stated that, if p(x) is a quadratic polynomial, then p(x) = 0 is a quadratic
equation.
In fact any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2 is a
quadratic equation whose standard form is ax2 + bx + c = 0, a 0.
A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0
where a, b, c, are real numbers and a 0.
Characteristics of a quadratic equation are,
• it is an equation in one variable.
• it is an equation whose single variable is of
degree 2.
• the standard form of quadratic equation is ax2 + bx + c = 0.
Here, a is the coefficient of x2,
b is the coefficient of x,
c is the constant term,
a, b, c are real numbers and a 0.
• the terms are written in descending order of the power of the variable.
In a quadratic equation why is a 0?
What happens to the quadratic equation, if a = 0?
Discuss in the class.
We have discussed that in a quadratic equation, a 0.
What happens to standard form of quadratic equation when b or c or both b and c are
equal to zero?
Sl. Verbal statement Quadratic QuadraticNo. expression equation
1. The sum of a number and five 5x2 + x 5x2 + x =0
times its squares.
2. A wire is bent to form the legsof a right angled triangle.If one of them is 2cm morethan the other, what will beits area?
3. The runs scored by a cricket x2 – x – 6 x2 – x – 6 =0team are 6 less than the differenceof the runs scored by the firstbatsman and the square of runsscored by him
1( 2)
2x x 21
( 2 ) 02
x x
21( 2 )
2x x 2 2 0x x
Quadratic Equations 197
Observe the table given below.
Sl. Value of b value of c Result
1. b = 0 c 0 ax2 + c = 0
2. b 0 c = 0 ax2 + bx = 0
3 b = 0 c = 0 ax2 = 0
4. b 0 c 0 ax2 + bx + c = 0
Observe that, in all the above cases the equation remains as a quadratic equation.
ILLUSTRATIVE EXAMPLES
Example 1: Check whether the following are quadratic equations:
(i) 2x + x2 + 1 = 0 (ii) 6x3 + x2 = 2
(iii)3
( 8) 10 04
x x x (iv) x(x + 1) + 8 = (x + 2) (x – 2)
Sol. (i) 2x + x2 + 1 = 0
Arrange the terms in descending order of their powers. x2 + 2x + 1 = 0
It is in the standard form ax2 + bx +c = 0.
The given equation is a quadratic equation.
(ii) 6x3 + x2 = 2
By rearranging the terms, we get 6x3 + x2 – 2 = 0
The highest degree of the variable is 3.
The given equation is not a quadratic equation.
(iii)3
( 8) 10 04
x x x
By simplifying we get
2 23 248 10 0
4 4x x x x
4x2 – 32x + 3x2 – 24x + 40 = 0
x2 – 56x + 40 = 0. It is of the form ax2 + bx + c = 0.
it is a quadratic equation.
(iv) x (x + 1) + 8 = (x + 2)(x – 2)
By simplifying we get
x2 + x + 8 = x2 – 4 2x 2x + x + 8 + 4 = 0, x + 12 = 0
It is not of the form ax2 + bx + c = 0.
The variable x is only in the first degree.
it is not a quadratic equation.
198 UNIT-9
EXERCISE 9.1
1. Check whether the following are quadratic equations:
(i) x2 – x = 0 (ii) x2 = 8 (iii)2 1
02
x x (iv) 3x – 10 = 0
(v)2 29
5 04
x x (vi)22
5 65
x x (vii) 22 3 0x x (viii)22
313
x
(ix) x3 – 10x + 74 = 0 (x) x2 – y 2 = 0
2. Simplify the following equations and check whether they are quadratic equations.
(i) x(x + 6) = 0 (ii) (x - 4)(2x – 3) = 0
(iii) (x + 9)(x – 9) = 0 (iv) (x + 2)(x – 7) = 5
(v) 3x + (2x – 1)(x – 9) = 0 (vi) (x + 1)2 = 2(x – 3)
(vii) (2x – 1)(x – 3) = (x + 5)(x – 1) (viii) x2 + 3x + 1 = (x – 2)2
(ix) (x + 2)3 = 2x(x2 – 1) (x) x3 – 4x2 – x + 1 = (x – 2)3
3. Represent the following in the form of quadratic equations.
(i) The product of two consecutive integers is 306.
(ii) The length of a rectangular park (in metres) is one more than twice its breadth
and its area is 528m2.
(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8
km/hr less, then it would have taken 3 hours more to cover the same distance.
Observe the examples given in the table.
A B
x2 = 144 x2 – 7x + 10 = 0
x2 – 81 = 0 17y – y2 + 30 = 0
2y2 = 0 p2 – 20 = – 8p
We can observe, that,
• All the equations in A and B are quadratic equations.
• All the quadratic equations in column A have the variable in second degree only.
They are called pure quadratic equations.
• All the quadratic equations in column B have the variable in both second degree
and first degree. They are called adfected quadratic equations.
Solution of quadratic equations:
Consider the pure quadratic equation x2 – 25 = 0.
Let us take real values for 'x' and substitute in the equation.
Let x = 1 Let x = 5
LHS = x2 – 25 = 12 – 25 = – 24 LHS = x2 – 25 = 52 – 25 = 0
LHS RHS LHS = RHS
Compare the equations in
columns A and B.
Quadratic Equations 199
Let x = – 2 Let x = – 5
LHS = x2 – 25 =(–2)2 – 25 = – 21 LHS = x2 – 25 = (–5)2 – 25 = 0
LHS RHS LHS = RHS
Try this for different values of x.
We observe that LHS = RHS only for two values of x, i.e., x = 5 and x = – 5.
We say, +5 and –5 satisfy the equation x2 – 25 = 0.
+5 and –5 are called roots of the quadratic equation
+5 and –5 are also called zeroes of the polynomial x2 – 25.
The roots of the quadratic equation satisfy the equation, which means LHS = RHS.
The above discussion holds good for adfected quadratic equations also.
Consider the adfected quadratic equation x2 – 3x – 10 = 0
Let x = 1 Let x = –3
LHS = x2 – 3x – 10 = 12 – 3(1) – 10 LHS = x2 – 3x – 10 = (–3)2 – 3(–3) – 10
= – 12 = 8
LHS RHS LHS RHS
Let x = 5 Let x = – 2
LHS = x2 – 3x – 10 = 52 – 3(5) – 10 LHS = x2 – 3x – 10 = (–2)2 – 3(–2) – 10
= 0 = 0
LHS = RHS LHS = RHS
We observe that the quadratic equation x2 – 3x – 10 = 0 is satisfied only for the
values x = 5 and x = – 2.
5 and –2 are the roots of the quadratic equation x2 – 3x – 10 = 0
5 and –2 are the zeroes of the quadratic polynomial x2 – 3x – 10.
In general, a real number 'k' is called a root of the quadratic equation
ax2 + bx + c = 0, a 0 if ak2 + bk + c = 0.
We also say that x = k is a solution of the quadratic equation, or that k satisfies the
quadratic equation.
Note that the zeroes of the quadratic polynomial ax2 + bx + c and the roots of the
quadratic equation ax2 + bx + c = 0 are the same.
We know that a quadratic polynomial has at most two zeroes.
Any quadratic equation has at most two roots.
Solving a quadratic equation means, finding the roots of quadratic equation. The
roots can be verified by substituting the values in the quadratic equation and checking
whether they satisfy the equation. The roots also form the solution set of the quadratic
equation.
200 UNIT-9
Solution of a pure quadratic equation
1. Solve x2 – 225 = 0
Sol. x2 – 225 = 0, x2 = 225 x = 225
x = 15 x = +15 or x = – 15
2. Solve 143 = t2 – 1
Sol. 143 = t2 – 1
Rewrite the equation in standard form
t2 = 143 + 1, t2 = 144
t = + 144 , t = +12 t = +12 or t = – 12
3. If V = r2h, then solve for 'r' and find the value of 'r' when V = 176 and h = 14.
Sol. Given V = r2h
r2h = V, r2 = V
h
r = V
h
If V = 176 and h = 14, we get
r = 176
14 (by substituting the values) r =
176 7 176
22 14
84
7
22 142
4 = 2
r = +2 or r = – 2
5. Shweta owns a 625 m2 land in square shape. She wants to fence the land with
barbed wire. Calculate the length of the wire required for 4 rounds.
Sol. Let length of the square be x m., Area of the square = x2
x2 = 625, x = 625 = 25 x =+25 or x = – 25
Since the length cannot be negative, we take x = +25.
length of one side of the square = +25m
Perimeter of square = 4x m
wire required for four rounds = 4 × 4x = 16x m
required length of the wire = 16 × 25 = 400 m
EXERCISE 9.2
1. Classify the following equations into pure and adfected quadratic equations.
(i) x2 = 100 (ii) x2 + 6 = 6 (iii) p(p – 3) = 1 (iv) x2 + 3 = 2x
(v)(x + 9)(x – 9) = 0 (vi) 2x2 = 72 (vii) x2 – x = 0 (viii) 7x = 35
x
(ix)1
5xx
(x)81
4xx
(xi) (2x – 5) 2 = 81 (xii)2( 4) 2
18 9
x
Recall:
Positive real numbers will
have two roots. One is +ve
& the other root is –ve
1 1 & 0 0
Quadratic Equations 201
2. Solve the quadratic equations:
(i) x2 – 196 = 0 (ii) 5x2 = 625 (iii) x2 + 1 = 101 (iv) 7x = 64
7x
(v) (x + 8)2 – 5 = 31 (vi)2 3 1
72 4 4
x(vii) –4x2 + 324 = 0 (viii) –37.5x2 = –37.5
3. In each of the following, determine whether the given values of 'x' is a solution of
the quadratic equation or not.
(i) x2 + 14x + 13 = 0 ; x = –1, x = – 13 (ii) 7x2 – 12x = 0 ; x = 1
3
(iii) 2m2 – 6m + 3 = 0 ; m = 1
2(iv) 2 2 4 0y y ; y = 2 2
(v)2 1
3x
xx
; x = 1 and x = – 1 (vi) (3k + 8)(2k + 5) = 0 ; k = 2
23
and k = 1
22
(vii)1
2 2
x
x ; x = 2 and x = 1 (viii) 6x2 – x – 2 = 0 ; x =
1
2 and x =
2
3
4. (i) If A = r2, solve for r, and find the value of 'r' if A = 77 and =22
7.
(ii) If r2 = l2 + d2 solve for d, and find the value of 'd' if r = 5 and l = 4.
(iii) If c2 = a2 + b2 solve for b and find the value of b, if a = 8 and c = 17.
(iv) If A = 23
4
a solve for a and find the value of a, if A = 16 3 .
(v) If k = 1
2mv2 solve for 'v' and find the value of v, if k = 100 and m = 2.
(vi) If v2 = u2 + 2as solve for 'v' and find the value of 'v', if u = 0, a = 2, s = 100.
Solution of adfected quadratic equations
We know that the general form of an adfected quadratic equation is ax2 + bx + c = 0, a 0.
This equation can also occur in different forms such as ax2 + bx = 0, ax2 + c = 0 and ax2 = 0.
ax2 = 0 and ax2 + c = 0 are pure quadratic equations and we have learnt in the
previous section the method of solving them.
How to solve an adfected quadratic equation?
There are several methods of solving the adfected quadratic equations depending
on their forms, ie., ax2 + bx + c = 0 or ax2 + bx = 0. These methods apply for pure quadratic
equations also.
Let us learn them.
(A) Solution of a quadratic equation by factorisation method.
We have learnt to factorise quadratic polynomials by splitting their middle terms.
We shall use this knowledge for finding the roots of a quadratic equation.
202 UNIT-9
Factorisation method is used when the quadratic equation can be factorised into
two linear factors. After factorisation, the quadratic equation is expressed as the product
of its two linear factors and this is equated to zero.
That is, ax2 + bx + c = 0 and (x m) (x n) = 0
Then, we apply zero product rule and equate each factor to zero and solve for the
unknown.
i.e., x m = 0 x = m
x n = 0 x = n
So, m and n are the roots of the quadratic equation ax2 + bx + c = 0.
Consider the quadratic equation x2 + 5x + 6 = 0.
The middle term is 5x and coefficient of x is 5.
Let us split 5 such that m + n = 5 and mn = 6
x2 + 3x + 2x + 6 = 0, x(x + 3) + 2(x + 3)= 0 (x + 3)(x + 2) = 0
Now, x2 + 5x + 6 = 0 is split into two linear factors (x + 3) and (x + 2).
By using zero product rule, we get (x + 3)(x + 2) = 0
x + 3= 0 or x + 2 = 0, If x + 3= 0, x = – 3, If x + 2 = 0, x = – 2
the solution set is {–3, –2}.
Thus, –3 and –2 are the two roots of the quadratic equation x2 + 5x + 6 = 0.
This can also be diagrammatically represented using algebraic tiles as follows.
x
x
x2
+ x x x x x x x x x x +
1 1 1 1 1
11 11
11 11
11 11
11 11
11 11
11 11
x x2+5 +6
On rearranging all the tiles we get the figure
Observe that the length = x + 3 and breadth = x + 2
and total area is x2 + 5x + 6.
The area of rectangle, A = l × b
x2 + 5x + 6 = (x + 3)(x + 2)
From this, we can conclude that
If the quadratic polynomial ax2 + bx + c represents
the area of a square or rectangle, then the length and
breadth represent the two factors of it.
Study the examples on solving quadratic equation by factorisation method.
Zero product rule:
Let a and b be any two realnumbers or factors. If a × b = 0then, either a = 0 or b = 0 botha and b are equal to zero
x x x x x
x
x2
x x 1 1 1
1 1 11 1 1
11
( + 2)x
( + 3)x
Quadratic Equations 203
ILLUSTRATIVE EXAMPLES
Example 1: Solve x2 – 3x + 2 = 0
Sol. Given x2 – 3x + 2 = 0
x2 – 2x – 1x + 2 = 0 (Resolving the expression)
x(x – 2) –1(x – 2) = 0 (Taking common factors)
(x – 2)(x – 1) = 0 (Taking common factors)
x – 2 = 0 or x – 1= 0 (Equating each factor to zero)
x = + 2 or x = 1
–2 and + 1 are the roots of x2 – 3x + 2 = 0
Example 2: Solve 6x2 – x – 2 = 0
Sol. Given: 6x2 – x – 2 = 0
6x2 + 3x – 4x – 2 = 0, 3x(2x + 1) – 2(2x + 1) = 0, (2x + 1)(3x – 2) = 0
2x + 1 = 0 or 3x – 2 = 0, 2x = – 1 or 3x = 2 x = 1
2 or x =
2
3
1
2 and
2
3 are the roots of 6x2 – x – 2 = 0
Example 3: Find the roots of 3x2 – 2 6x + 2 = 0.
Sol. Given: 23 2 6 2x x = 0
23 6 6 2x x x = 0, 3 3 2 2 3 2x x x = 0
3 2 3 2x x = 0 3 2 0x or 3 2x = 0
3x = 2 or 3x = 2 , x = 2
3 or x =
2
3
So, the root is repeated twice, one for each repeated factor 3 2x .
The two equal roots of 23 2 6 2 0x x are 2 2
, 3 3
.
Example 4: Solve 24 10x = 3 – 4x
Sol. Given: 24 10x = 3 – 4 x
Squaring on both sides, we get
2
24 10x = (3 – 4x)224 – 10x = 9 – 24x + 16x2, 16x2 – 24x + 10x + 9 – 24 = 0
16x2 – 24x + 10x – 15 = 0, 8x(2x – 3) + 5(2x – 3) = 0
(2x – 3)(8x + 5) = 0,2x – 3 = 0 or 8x + 5 = 0
2x = 3 or 8x = – 5, x = 3
2 or x =
5
8
3
2 and
5
8 are the roots of 24 10x = 3 – 4x.
204 UNIT-9
Steps of finding roots of quadratic equation by factorisation method.
1. Write the given equation in standard form of quadratic equation, ax2 + bx + c = 0.
2. Resolve the quadratic expression (LHS) by splitting its middle term.
3. Take the common factor and obtain the two linear factors.
4. Equate each factor to zero.
5. Simplify each linear equation and find the value of unknown.
EXERCISE 9.3
Solve the quadratic equations by factorisation method:
1. x2 + 15x + 50 = 0 2. x2 – 3x – 10 = 0 3. 6 – p2 = p
4. 2x2 + 5x – 12 = 0 5. 13m = 6(m2 + 1) 6. 100x2 – 20x + 1 = 0
7. 22 7 5 2 0x x 8. x2 + 4kx + 4k2 = 0 9.7
6mm
10.1
xx
= 2.5 11. 21y2 = 62y + 3 12. 0.2t2 – 0.4t = 0.03
13. 4x2 + 32x + 64 = 0 14. 25 2 3 5x x 15.1 34
1 15
x x
x x
16.1 3 1
32 4 3
x x
x x17. a2b2x2–(a2+b2)x +1= 0 18. 2(x + 1)2 – 5(x + 1) = 12
19. (x – 4)2 + 122 = 152 20. (2x – 3) = 22 2 21x x
Completing the square method
Let us consider the quadratic equation. x2 + 5x + 5 = 0
Here, the middle term 5x cannot be split into terms such that m + n = 5 and
mn = 5. This means we cannot resolve the equation as product of two factors and therefore,
it cannot be solved by factorisation method. This is the limitation of factorisation method
of solving quadratic equations. This method can be used only when it is possible to split
the middle term and factorise the given quadratic polynomial.
Then, how to solve quadratic equations where the quadratic polynomial cannot be
factorised?
We can also diagrammatically represent the
given quadratic equation x2 + 5x + 5= 0 to understand
why it cannot be resolved into factors.
Observe the following figure.
We see that the figure representing x2 + 5x + 5
is not a complete square or rectangle. It means that,
if the figure is a complete square or a rectangle then
we can solve it by factorization method.
Such in complete squares or rectangles can be converted into complete square or
rectangles by adding some quantities to it. This method of adding quantities to make it a
perfect square/rectangle is called completing the square method.
x2 + 5x + 5 x x x
x2
x x 1 1 1
1 1
Quadratic Equations 205
B. Solving a quadratic equation by completing the square method.
Consider the equation x2 + 5x + 5 discussed above. We see from the diagram that
(x2 + 5x + 5) is 12 or 1 less to form a complete square or rectangle.
In other words, {(x2 + 5x + 5) + 1} will form a rectangle and it
can be factorised.
i.e., x2 + 5x + 5 + 1= 0 x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0
x = –3 or x = –2
Consider another example of solving x2 + 4x + 5 = 0
The middle term is 4x and it cannot be split such that the sum of the terms is 4 and
the product is +5.
Hence, x2 + 4x + 5 do not represent a complete square.
What is the minimum quantity to be added to given equation or subtracted from it,
to make it a complete square?
Recall the identity that represents a complete square.
we know, a2 + 2ab + b2 = (a + b)(a + b)
This indicates that any quadratic
polynomial can be resolved into linear factors, if
it is in the form (a2 + 2ab + b2).
In, a2 + 2ab + b2
the middle term is 2ab, where
2 is the constant
a is square root of first term.
b is square root of last term.
In the given equation x2 + 4x + 5 = 0, the middle term is 4x.
Compare 4x with 2ab.
2ab = 4x 2 × x × b = 4x [a = x] b =4
2
x
x = 2
Observe that, b = 2 which is half of the coefficient of x.
By squaring b, we get b2 = 22 = 4.
The LHS of the quadratic equation x2 + 4x = –5 can be converted to a complete
square by adding 4 to the LHS. If 4 is added to LHS, the value of the equation changes. In
order to maintain the original value we have to add 4 to LHS and RHS or we can add and
subtract 4 to the LHS.
Again, take the equation x2 + 4x + 5 = 0
x2 + 4x = –5
x2 + 4x + 4 = –5 + 4, x2 + 4x + 4 = –1
(x + 2)2 = –1
a2 + 2ab + b2 a
2 ab
ab
a
b
b
b
a
a
a
( +
)
ab
( + )a b
x x x x2
x x 1 1 1
1 1 1
206 UNIT-9
So, solving x2 + 4x + 5 = 0 is equivalent to solving (x + 2)2 + 1 = 0 or (x + 2)2 = –1.
This means, we can convert a quadratic equation to the form (x + a)2 – b2 = 0 and then find
its roots.
In the figure given below, observe how (x2 + 4) is being converted to (x + 2)2 – 4 or
(x + 2)2 – 22.
4 4 x
x
x x x
x
x
+ = =
2
2
+ 4 =x x2
= + 2 + 2x x x( +2) + 2
x x x2x x
2 + 4
2
2
2
2
2
2
2
2
x
x x=
+ 2x
+2
x
( + 2) + 2 + 2 – 2x x x 2 2
( + 2) – x2
22
Now, let us solve x2 + 4x + 5 = 0.
Find, half of the coefficient of x. 1
42
= 2 b = 2
By squaring it, we get b2 = 22 = 4
By adding and subtracting 4,
2 4 4 4 5x x = 0 (x + 2)2 + 1 = 0 (x + 2)2 = –1
Taking square root on both the sides, We get, x + 2 = + 1
If, x + 2 = + 1 , x = – 2 + 1 or x + 2 = – 1 , x = – 2 – 1
– 2 + 1 and – 2 – 1 are the roots of x2 + 4x + 5 = 0
So, we have solved the quadratic equation x2 + 4x + 5 = 0 by the process of completing
the square. This known as the method of completing the square.
Consider the square of the binomial 2
bx
2 2
2
2 2
b bx x bx
Note that the last term
2
2
b is the square of half the coefficient of x.
Quadratic Equations 207
Hence, (x2 + bx) lacks only the term
2
2
bof being the square of
2
bx . Thus, if the
square of half the coefficient of x is added to the expression of the form x2 + bx, the resultis the square of a binomial. Such an addition is known as completing the square.
ILLUSTRATIVE EXAMPLES
Example 1: Solve the quadratic equation x2 + 6x – 7 = 0 by completing the square.
Sol. Given: x2 + 6x – 7 = 0 2 × a × b = 6x
x2 + 6x + 9 – 9 – 7 = 0 2 × x × b = 6x ( a = x)
x2 + 3x + 3x + 9 = +16 b = 6
2
x
x
x(x + 3) + 3(x + 3) = 16 b = 3
(x + 3)(x + 3) = 16 b2 = 9
(x + 3)2 = (4)2 Taking square root on both the sides,
x + 3 = 4 (Half of coefficient of x = 6
2 b = 3)
If, x + 3 = 4 x + 3 = – 4
x = 4 – 3 x = – 4 – 3
x = 1 x = – 7
1 and –7 are the roots of x2 + 6x – 7 = 0
Example 2: Solve 3x2 – 5x + 2 = 0 by completing the square method.
Sol. Given 3x2 – 5x + 2 = 0
Here, the coefficient of x2 is 3 and it is not a perfect square.
Such quadratic equations are solved in two ways. Let us do both of them.
(i) Multiply the equation through out by 3.
(3x2 – 5x + 2 = 0) × 3 9x2 – 15x + 6 = 0
Now, half of the coefficient of x is 5
2. b =
5
2 and b2 =
25
2
So, 9x2 – 15x +
2 25 5
62 2
= 0(3x)2 – 15x +
25
2 =
25
2 – 6
25
32
x = 25
64
= 1
4
(Taking square root on both the sides)
3x – 5
2 =
1
2
3x – 5
2 = +
1
2 or 3x –
5
2 = –
1
2
208 UNIT-9
3x = 1
2 +
5
2 3x =
5
2 –
1
2
3x = 6
2 = 3 3x =
4
2 = 2
x = 1 x = 2
3
(ii) Dividing the equation through out by 3.
(3x2 – 5x + 2 = 0) 3
x2 – 5 2
3 3x = 0
Now, let us proceed as earlier.
2
21 5 5 5
2 3 6 6b b
So,
2 2
2 5 5 5 2
3 6 6 3x x = 0
25
6x =
25 2
6 3
25
6x =
25 2 1
36 3 36
By taking square root on both the sides,
5
6x =
1
6
5
6x = +
1
6 , x =
5 1
6 6 =
61
6 x = 1 or
x – 5
6 = –
1
6 , x =
5 1 4 2
6 6 6 3 x =
2
3
1 and 2
3 are the roots of the quadratic equation 3x2 – 5x + 2 = 0.
Steps of finding the roots of a quadratic equation by completing the square method
Step 1: Write the equation in standard form.
Step 2: If the coefficient of x2 is 1, go to step 3
If not, multiply or divide both the sides of the equation by the coefficient of x2.
Step 3: Find half the coefficient of x and square it. Add this number to both the sides of
the equation or add and subtract on LHS of the equation.
Step 4: Solve the equation, using the square root property.
If x2 = p, then x = + p or x = p where, p is non - negative number.
Quadratic Equations 209
EXERCISE 9.4
Solve the following quadratic equations by completing the square.
(i) 4x2 – 20x + 9 = 0 (ii) 4x2 + x – 5 = 0 (iii) 2x2 + 5x – 3 = 0
(iv) x2 +16x – 9 = 0 (v) x2 – 3x + 1 =0 (vi) t2 + 3t = 7
(vii) 3x (x – 5) = 2x(x + 7) (viii)5 7
1
x
x = 3x + 2
(ix) a2x2 – 3abx + 2b2 = 0 (x) 4x2 + 4bx – (a2 – b2) = 0
We know that, in mathematics calculations and solving problems are made easier
by using formulae. In the same way, quadratic equations can be easily solved using a
formula. The quadratic formula, which is very useful for finding its roots can be derived
using the method of completing the square. Let us derive the quadratic formula and
learn how to use it for finding roots of the quadratic equations.
c) Solution of a quadratic equation by formula method
Consider the quadratic equation ax2 + bx + c = 0, a 0.
Divide the equation by a (i.e., coefficient of x2) b c
x xa a
= 0
Find half the coefficient of x and square it 1
2
b
a =
2
b
a
2
2
b
a =
2
24
b
a
Transpose the constant c
a to RHS
2 bx x
a = –
c
a
Add
2
2
b
ato both sides of the equation
2 2
2
2 2
b b b cx x
a a a a
Factorise LHS and simplify RHS
2 2 2
2 2
4
2 4 4
b b c b acx
a a a a
Take square root on both sides of the equation 2 2
2
4 4
2 24
b b ac b acx
a aa
2 4
2 2
b b acx
a a
2 4
2
b b acx
a
The roots of the quadratic equation ax2 + bx + c = 0 are
2 4
2
b b ac
a and
2 4
2
b b ac
a
x = 2 4
2
b b ac
a is known as quadratic formula.
210 UNIT-9
In the above derivation, we have eliminated the coefficient of x2, which is not a
perfect square by dividing the equation by a. However, it can also be solved by multiplying
the equation by 4a. This method is also called Sridharacharya's method. Sridharacharya
(1025 A.D) is credited with deriving the formula for solving quadratic equations by the
method of completing the square. Study the derivation of quadratic formula as evolved by
the ancient Indian mathematician, Sridharacharya.
Consider the general form of quadratic equation. ax2 + bx + c = 0, where a 0.
Multiply both the sides by 4a {(ax2 + bx + c) = 0} × 4a,
4a2x2 + 4abx + 4ac = 0, 4a2x2 + 4abx = –4ac
Add b2 to both the sides, we get 4a2x2 + 4abx + b2 = b2 – 4ac,
(2ax)2 + 4abx + b2 = b2 – 4ac, (2ax + b)2 = b2 – 4ac
Taking square not on both side 2ax + b = 2 4b ac ,
2ax = 2 4b b ac , x =
2 4b b ac ,
x = 2 4
2
b b ac
a and x =
2 4
2
b b ac
a
ILLUSTRATIVE EXAMPLES
Example 1: Solve x2 – 7x + 12 = 0 by formula method.
Sol. Given x2 – 7x + 12 = 0
This is in the form ax2 + bx + c = 0 where, a = 1, b = –7 and c = 12
Quadratic formula is, x = 2 4
2
b b ac
a
By substituting the values, we get
x = 2( 7) ( 7) 4 1 12
2 1 =
7 49 48
2=
7 1
2 =
7 1
2
If x = 7 1 8
2 2 = 4 x =
7 1 6
2 2 = 3
4 and 3 are the roots of x2 – 7x + 12 = 0.
Example 2: Solve m2 = 2 + 2m
Sol. Given m2 = 2 + 2m
Rewrite the equation in standard form i.e., m2 – 2m – 2 = 0
This is of the form ax2 + bx + c = 0 where, a = 1, b = –2, c = –2
x = 2 4
2
b b ac
a(Quadratic formula) x =
2( 2) ( 2) 4(1)( 2)
2 1
Quadratic Equations 211
x = 2 4 8
2 =
2 12
2 =
2 2 3 2 1 31 3
2 2
x = 1 3 and 1 3 are the roots of the quadratic equation m2 = 2 + 2m.
Example 3: Solve 1 2 4
1 2 4x x x
Sol. Given 1 2 4
1 2 4x x x Simplify the equation.
1
1x=
4 2
4 2x x
1
1x=
2 12
4 2x x
1
1x=
2( 2) 1( 4)2
( 4)( 2)
x x
x x
1
1x=
2 4 42
( 4)( 2)
x x
x x
1
1x=
2
( 4)( 2)
x
x x, (x + 4)(x + 2)= 2x(x + 1), x2 + 6x + 8= 2x2 + 2x
x2 – 4x – 8 = 0. It is a quadratic equation, Here, a = 1, b = –4, c = –8
x = 2 4
2
b b ac
a =
2( 4) ( 4) 4 (1)( 8)
2(1) =
4 48
2
x = 4 4 3
2 =
4 1 32 1 3
2
x = 2 1 3 and x = 2 1 3
2 1 3 and 2 1 3 are the roots of the given quadratic equation.
Example 4: Solve x2 + x – (a + 2)(a + 1) = 0 by using quadratic formula.
Sol. Given x2 + x – (a + 2)(a + 1) = 0, Here, a = 1, b = 1, c = –(a + 2)(a + 1)
x = 2 4
2
b b ac
a =
21 1 4 ( 2)( 1)
2(1)
a a =
21 1 4 ( 3 2)
2
a a
x = 21 1 4 12 8
2
a a =
21 4 12 9
2
a a =
21 (2 3)
2
a =
1 (2 3)
2
a
x = 1 2 3
2
a and x =
1 2 3
2
a x =
2 2
2
a and x =
2 4
2
a
x = a + 1 x = –(a + 2)
The roots of x2 + x – (a + 2)(a + 1)= 0 are (a + 1) and –(a + 2)
212 UNIT-9
Steps for solving a quadratic equation using the quadratic formula
Step1 : Write the equation in standard form, ax2 + bx + c = 0
Step 2: Compare the equation with standard form and identify the values of a, b, c.
Step 3: Write the quadratic formula 2 4
2
b b acx
a
Step 4: Substitute the values of a, b and c in the formula.
Step 5: Simplify and get the two roots.
EXERCISE 9.5
Solve the following quadratic equations by using the formula method.
1. x2 – 4x + 2 = 0 2. x2 – 2x + 4 = 0
3. x2 – 7x + 12 = 0 4. 2y2 + 6y = 3
5. 15m2 – 11m + 2 = 0 6. 8r2 = r + 2
7. p = 5 – 2p2 8. (2x + 3)(3x – 2) + 2 = 0
9. 4x2 – 4ax + (a2 – b2) = 0 10. 2 9 13x x
11. a(x2 + 1) = x(a2 + 1) 12. 36x2 – 12ax + (a2 – b2) = 0
13.1 1 1
02 3 4x x x
14.3 2 8
5 4 2b b b
So far we have learnt to find the roots of given quadratic equations by different
methods. We see that the roots are all real numbers. What is the nature of these roots?
What determines the nature of the roots?
Is it possible to determine the nature of roots of a given quadratic equation, without
actually finding them? Discuss in class and try to answer these questions.
Now, let us learn about this.
Nature of the roots of quadratic equations
Study the following examples
1. Consider the equation x2 – 2x + 1 = 0
This is in the form of ax2 + bx + c = 0, a = 1, b = –2, c = 1
x = 2 4
2
b b ac
a, =
2( 2) ( 2) 4 1 1
2 1
= 2 4 4
2, =
2 0
2
x = 2 0
2 or x =
2 0
2 x = 1 or x = 1 roots are equal
Quadratic Equations 213
2. Consider t he equat ion x2 – 2x – 3 = 0
This is in the form ax2 + bx + c = 0, where a = 1, b = –2, c = –3
x =2 4
2
b b ac
a =
2( 2) ( 2) 4 1 ( 3)
2 1 =
2 4
2
x = 2 4
2 or x =
2 4
2, x =
6
2 or x =
2
2
x = 3 or x = -1 roots are distinct
3. Consider the equation x2 – 2x + 3 = 0
This is in the form ax2 + bx + c = 0, where a = 1, b = –2, c = 3
x = 2 4
2
b b ac
a =
2( 2) ( 2) 4(1)(3)
2 1
= 2 4 12
2 =
2 8
2
x = 2 2 2
2 =
2 1 21 2
2
1 2x or 1 2 roots are imaginary
From the above examples, it is evident that the roots of a quadratic equation can be
real and equal, real and distinct or imaginary.
Also, observe that the value of b2 – 4ac determines the nature of the roots. We say
the nature of roots depends on the values of b2 – 4ac.
The value of the expression b2 – 4ac discriminates the nature of the roots of
ax2 + bx + c = 0 and so it is called the discriminant of the quadratic equation. It is denoted
by the symbol and read as 'delta'.
In general, the roots of the quadratic equation ax2 + bx + c = 0 are x = 2 4
2
b b ac
a
• If, b2 – 4ac = 0 then, the equation has two equal roots which are real. Thus, x = 2
b
a
• If, b2 – 4ac > 0 then, the equation has two distinct roots which are real.
Thus, x = 2 4
2
b b ac
a, x =
2 4
2
b b ac
a
• If, b2 – 4ac < 0 then, the equation has no real roots.
Since, 2( 4 )b ac cannot be found and we say it is imaginary.
214 UNIT-9
The above results are presented in the table given below.
Discriminant Nature of roots
= 0 Real and equal
> 0 Real and distinct
< 0 No real roots (imaginary roots)
ILLUSTRATIVE EXAMPLES
Example 1: Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0
Sol. This is in form of ax2 + bx + c = 0. The co-efficient are a = 2, b = –5, c = –1
= b2 – 4ac, = (–5)2 –4(2) (–1) = 25 + 8, = 33
> 0. Therefore, roots are real and distinct
Example 2: Determine the nature of the roots of the equation 4x2 – 4x + 1 = 0
Sol. Consider the equation 4x2 – 4x + 1 = 0
This is in the form of ax2 + bx + c = 0. The co-efficient are a = 4, b = –4, c = 1
= b2 – 4ac,= (–4)2 – 4(4) (1), = 16 – 16
= 0. Therefore, roots are real and equal
Example 3: For what positive values of 'm' roots of the equation x2 + mx + 4 = 0 are
(i) equal (ii) distinct
Sol. Consider the equation x2 + mx + 4 = 0
This is in the form ax2 + bx + c = 0. The co-efficients are a = 1, b = m, c = 4
= b2 – 4ac, = m2 – 4(1)(4), = m2 – 16
(i) If roots are equal, then = 0
m2 – 16 = 0, m2 = 16, m = 16 m = 4
(ii) If roots are distinct, then 0
m2– 16 0, m2 16, m 16 m
EXERCISE 9.6
A. Discuss the nature of roots of the following equations
(i) y2 - 7y + 2 = 0 (ii) x2 - 2x + 3 = 0 (iii) 2n2 + 5n - 1 = 0
(iv) a2 + 4a + 4 = 0 (v)) x2 + 3x - 4 = 0 (vi) 3d2 - 2d + 1 = 0
B. For what positive values of ‘m’ roots of following equations are
1) equal 2) distinct 3) imaginary
i) a2-ma+1=0 ii) x2-mx+9=0 iii) r2-(m+1) r+4=0 iv) mk2-3k+1=0
C. Find the value of ‘P’ for which the quadratic equations have equal roots.
i) x2-px+9 = 0 ii) 2a2+3a+p = 0 iii) pk2-12k+9 = 0 iv) 2y2-py+1 = 0
v) (p+1) n2+2 (p+3)n+(p+8)=0 vi) (3p+1)c2+2(p+1)c+p=0
Quadratic Equations 215
Relationship between the roots and co-efficients of the terms of the quadratic
equation.
If ‘m’ and ‘n’ are the roots of the quadratic equation ax2+bx+c=0 then
m = 2 4
2
b b ac
a and n =
2 4
2
b b ac
a
m+n = 2 4
2
b b ac
a +
2 4
2
b b ac
a =
2 24 4
2
b b ac b b ac
a =
2
2
b
a
m + n =b
a
mn =
2 4
2
b b ac
a
2 4
2
b b ac
a =
22 2
2
4
4
b b ac
a
mn =
2 2
2
4
4
b b ac
a =
2 2
2 2
4 4
4 4
b b ac ac
a a
mn = c
a
If m and n are the roots of the quadratic equation ax2 + bx + c = 0
Sum of the roots = (m+n) =-b
aProduct of roots = mn =
+c
a
ILLUSTRATIVE EXAMPLES
Example 1: Find the sum and product of the roots of equation x2 + 2x + 1 = 0
Sol. This is in the form ax2 + bx + c = 0, where a =1, b = 2, c=1
Let the roots be m and n
i) Sum of the roots m+n = b
a =
2
1 2m n
ii) Product of the roots mn = c
a =
1
1 1mn
Example 2: Find the sum and product of the roots of equation 3x2 + 5 = 0
Sol. This is in the form ax2 + bx + c = 0, where a = 3, b = 0, c = 5
Let the roots be p and q
i) Sum of the roots p+q = b
a =
0
3 p q 0+
ii) Product of the roots pq = c
a =
5
3
5pq
3
216 UNIT-9
Example 3: Find the sum and product of the roots of equation x2 – (p + q)x + pq = 0.
Sol. The coefficients are a = 1, b = –(p + q), c = pq
i) Sum of the roots m + n = b
a m + n =
( )
1
p q ( )m n p q
ii) Product of the roots mn =1
c pq
a mn pq .
EXERCISE 9.7
Find the sum and product of the roots of the quadratic equation:
1. x2 – 5x + 8 = 0 2. 3a2 – 10a – 5 = 0 3. 8m2 – m = 2
4. 6k2 – 3 = 0 5. pr2 = r – 5 6. x2 + (ab) x + (a + b) = 0
Framing a quadratic equation :
We know how to find the roots, of given quadratic equations and also their sum and
product.
Is it possible to frame a quadratic equation if sum and product of its roots are given?
If 'm' and 'n' are the roots then the standard form of the equation is
x2 – (sum of the roots) x + product of the roots = 0, i.e. x2 – (m + n) x + mn = 0.
ILLUSTRATIVE EXAMPLES
Example 1: Form the quadratic equation whose roots are 2 and 3.
Sol. Let 'm' and 'n' be the roots m = 2, n = 3
Sum of the roots = m + n = 2 + 3 m + n = 5
Product of the roots = mn = (2) (3) mn = 6
Standard form is x2 – (m + n)x + mn = 0 x2 – (5)x + (6) = 0
2 5 6 0x x
Example 2: Form the quadratic equation whose roots are 3 2 5 and 3 2 5
Sol. Let 'm' and 'n' be the roots m = 3 2 5 and n = 3 2 5
Sum of the roots = m + n = 3 2 5 3 2 5 6m n
Product of the roots= mn = 3 2 5 3 2 5 = (3)2 – 2
2 5 = 9 – 20 11mn
x2 – (m + n) x + mn = 0 x2 – 6x – 11 = 0
Example 3: If 'm' and 'n' are the roots of equation x2 – 3x + 1 = 0. Find the value of
(i)m2n + mn2 (ii) 1 1
m n.
Sol. Consider the equationx2 – 3x + 1 = 0
This is in the form ax2 + bx + c = 0, where a = 1, b = –3, c = 1
Quadratic Equations 217
(i) Sum of the roots m + n = ( 3)
31
b
a 3m n
(ii) Product of the roots mn =c
a=
1
1 1mn
Therefore,
(i) m2n + mn2 = mn (m + n) = 1 × 3 = 3
(ii) 1 1
m n =
3
1
n m m n
mn mn= 3
1 13
m n
Example 4: If 'm' and 'n' are the roots of equation x2 – 3x + 4 = 0 form the equation
whose roots are m2 and n2.
Sol. Consider the equation x2 – 3x + 4 =0. Here a = 1, b = –3, c = 4
(i) Sum of the roots = m + n = ( 3)
1
b
a 3m n
(ii) Product of the roots = mn = c
a=
4
1 mn = 4
If the roots are m2 and n2
Sum of the roots m2 + n2 = (m + n)2 – 2mn = (3)2 – 2(4) = 9 – 8 m2 + n2 = 1
Product of the roots m2n2 = (mn)2 = 42 m2n2 = 16
x2 – (m2 + n2)x + m2n2 = 0 x2 – (1)x + (16) = 0 2 16 0x x
Example 5: If one root of the equation x2 – 6x + q = 0 is twice the other, find the
value of 'q'.
Sol. Consider the equationx2 – 6x + q = 0, where a = 1, b = –6, c = q
(i) Sum of the roots m + n = ( 6)
1
b
a 6m n
(ii) Product of the roots mn = 1
c q
a mn = q
If one root is m then twice the root is 2m . m = m and n = 2m
m + n = 6 m + 2m = 6, 3m = 6 m = 6
3= 2
We know that q = mn q = m(2m) = 2m2 = 2(2)2 = 8 q = 8
Example 6: Find the value of k so that the equation x2 – 2x + (k + 3) = 0 has one root
equal to zero.
Sol. Consider the equation x2 – 2x + (k + 3) = 0. Here, a = 1, b = –2, c = k + 3
Product of the roots = mn = c
a mn =
3
1
k mn = k + 3
Since 'm' and 'n' are the roots, and one root is zero then m = m and n = 0, mn = k + 3
m(0) = k + 3 0 = k + 3 k = –3.
218 UNIT-9
EXERCISE 9.8
A. Form t he equat ion whose root s are
i) 3, 5 ii) 6, –5 iii) –3, 3
2iv)
2
3,
3
2
v) ( 2 3 ), ( 2 3 ) (vi) ( 3 2 5 ), ( 3 2 5 )
B.1. If 'm' and 'n' are the roots of the equation x2 – 6x + 2 = 0 find the value of
(i) (m + n) mn (ii) 1 1
m n(iii) 3 2 3 2m n n m (iv)
1 1
n m
2. If 'a' and 'b' are the roots of the equation 3m2 = 6m + 5, find the value of
(i) a b
b a(ii) (a + 2b)(2a + b)
3. If 'p' and 'q' are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of
(i) (p + q)2 + 4pq (ii) p3 + q3
4. Form a quadratic equation whose roots are p
q and
q
p
5. Find the value of 'k' so that the equation x2 + 4x + (k + 2) = 0 has one root equal to
zero.
6. Find the value of 'q' so that the equation 2x2 – 3qx + 5q = 0 has one root which is
twice the other.
7. Find the value of 'p' so that the equation 4x2 – 8px + 9 = 0 has roots whose difference
is 4.
8. If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.
D. Solution of a quadratic equation by graphical method.
You are familiar with drawing graphs for linear equations and also solving
simultaneous linear equations graphically.
Now let us draw graphs for quadratic expressions and learn to solve guadratic
equations graphically.
Example 1: Consider the quadratic equation
x2 = 0. We need co-ordiantes of points to plot
them and draw the graph.
Let y = x2 Prepare the table for values of 'x'
and 'y' for the equation y = x2.
2 1 0 1 2
4 1 0 1 4
x
y
Plot the points A(–2, 4), B(–1, 1), C(0, 0),
D(1, 1) and E(2, 4).
Join the points by a smooth curve.
The graph of y = x2 is a curved line.
X O
X
Y
6
5
4
3
2
1
1 2 3–1–2–3
Y
E(2, 4)A(–2, 4)
B(—1, 1) D(1, 1)C(0, 0)
x - axis : 1cm = 1unit y - axis : 1 cm =1unit
Quadratic Equations 219
Observe the graph of two more quadratic equations.
Example 2: 2x2 = 0 Example 3: 21
02
x
Let y =2x2
2 1 0 1 2
8 2 0 2 8
x
y Let y = 21
2x
2 1 0 1 2
2 ½ 0 ½ 2
x
y
Thus, we observe that the graphs of quadratic equations x2 = 0, 2x2 = 0 and 21
02
x
are curved lines. This curved line representing quadratic equations is called "Parabola".
From the graph of these quadratic equations we observe that:
• The parabola of the graphs are symmetrical with respect to the y-axis.
• The point where the curvature is greatest is called the vertex.
The parabola takes a turn at this point.
Now let us learn more about parabola.
Consider expressions (x2 – 4x) and (–x2 + 2x + 5).
1. x2 – 4x 2. –x2 + 2x + 5
Here a is greater than zero Here a is less than zero
Let y = x2 – 4x Let y = – x2 + 2x + 5
• Prepare a table of values. • Let us prepare a table of values.
1 0 1 2 3 4 5
5 0 3 4 3 0 5
x
y2 1 0 1 2 3
3 2 5 6 5 2
x
y
• Plot the points of each ordered pair (x, y). • Plot the points of each ordered pair (x, y).
• Draw a smooth curve through the plotted • Draw a smooth curve through the
points. plotted points.
Y
X
O X
Y
5
4
3
2
1
–1
–2
–3
–4
–5
1 2 3 4 5–1–2–3–4–5
D(3,–3)B(1, 3)
A(–1, 5)F(5, 5)
E(4,0)
C(2 –4)
x - axis : 1cm = 1unit y - axis : 1 cm =1unit
Y
X O X
Y
6
5
4
3
2
1
–1
–2
–3
–4
–5
1 2 3 4 5–1–2–3–4–5
D(1,6)
C(0, 5) E(2, 5)
F(3,2)B(–1,2)
A(–2,–3)
x - axis : 1cm = 1unit y - axis : 1 cm =1unit
X–2–3–4–5
9
8
7
6
5
4
3
2
1
X O–1
Y
Y
E(2,8)A(-2,8)
B(–1,2) D(1,2)
C(0
,0)
x - axis : 1cm = 1unit y - axis : 1 cm =1unit
X I
2.5
2
1.5
1
0.5
X O
Y
Y
I
P(-2,2)
R(0
,0)
T(2,2)
S(1, 0.5)Q(-1,0.5)
x - axis : 1cm = 1unit
y - axis : 1 cm =0.5unit
220 UNIT-9
From the above two graphs we observe the following:
EXERCISE 9.9
I. Draw the graphs of the following quadratic equations:
i) y = –x2 ii) y = 3x2 iii) y = x2 + 6x iv) y = x2 – 2x
v) y = x2 – 8x + 7 vi) y = (x + 2)(2 – x) vii) y = x2 + x – 6 viii) y = x2 – 2x + 5
Graphical solutions of quadratic equations:
In this section let us study the graphical method of solving quadratic equations.
ILLUSTRATIVE EXAMPLES
Example 1: Draw the graph of y = x2 – x – 2 and
find its roots.
Sol. Let us draw the graph of this equation and
find the roots graphically.
Step 1: Prepare a table of values to
y = x2 – x – 2.
2 1 0 1 2
4 0 2 2 0
x
y
Step 2:Plot all the points on the graph.
Step 3:Draw a smooth curve through the
plotted points.
Step 4:Mark the intersecting points of the
curve with the x-axis.
The coordinates where the parabola intersects
the x-axis are the roots of the equation.
The coordinates of the intercepts are B(–1, 0)
and E(2, 0).
The roots of the equation are
1 and 2x x
–x2 + 2x + 5
• The parabola opens downwards.
• As x increases y also increases until
the miximum point is reached.
• The highest value of y is 6.
x2 – 4x
• The parabola opens upwards.
• As x increases y decreases until the
minimum value of y is reached.
• The smallest value of y is –4.
Verification by factorisation
method:
x2 – x – 2 = 0
x2 + 1x – 2x – 2 = 0
x(x + 1) – 2(x + 1) = 0
(x + 1)(x – 2) = 0
x + 1= 0 or x – 2 = 0
x = –1 or x = 2
Y
X O X
Y
5
4
3
2
1
–1
–2
–3
–4
–5
1 2 3 4 5–1–2–3–4–5
D(1,–2)
B(–1, 0)
A(–2, 4)
E(2,0)
C(0 –2) - interceptx - interceptx
x - axis : 1cm = 1unit y - axis : 1 cm =1unit
Quadratic Equations 221
Example 2: Solve the quadratic equation x2 – 10x + 25 = 0 graphically.
Sol. Prepare the table of values for the equation
y = x2 – 10x + 25.
1 2 3 4 5 6
16 9 4 1 0 1
x
y
Mark the point at which the curve touches the
x-axis.
The parabola intersects the x-axis at only one
point, E (5,0)
The roots of the equation are 5 and 5.
Example 3: Draw the graph and find the roots of y = x2 + 2.
Sol. Prepare a table of values for the equation
y = x2 + 2.
2 1 0 1 2
6 3 2 3 6
x
y
The parabola does not intersect the x-axis.
There is no real value of x for x2 + 2 = 0.
Hence, there are no real roots.
Let us now record the details of the above three examples in the table. Study them:
x2 – x – 2 x2 – 10x + 25 x2 + 2
It has 2 points of It has 1 point of It has no point of
intersection intersection Intersection
X X
2 solutions
X X
No solutions
ax2 + bx + c = 0 ax2 + bx + c = 0 ax2 + bx + c = 0
has 2 real and has 2 real and has no real
unequal roots repeated roots roots
b2 – 4ac > 0 b2 – 4ac = 0 b2 – 4ac < 0
Y
X O
X
Y
25
20
15
10
5
1 2 3 4 5 6–1–2–3–4
A(1,16)
= –10 +25y x x2
x - intercept
B(2,9)
C(3,4) D(4
,1)
E(5
,0)
F(6
,0)
Y
X O
X
Y
8
7
6
5
4
3
2
1
1 2 3 4 5–1–2–3–4–5
A(–2, 6)
B(–1, 3) D(1, 3)
E(2, 6)
C(0, 2)
X X
1 solution
222 UNIT-9
Y
X O
X
11
10
9
8
7
6
5
4
3
2
1
1 2 3 4 5 –1–2–3–4–5
=10y
D(–2, 8) C(2, 8)
A(1, 2)B(–1, 2)
= –2.3x = 2.3x
Y
Example 4 : Solve the equation y = (2 – x)(4 + x) graphically.
Sol. Step 1: The given equation isy = (2 – x)(4 + x).
Simplify the RHS and bring it to thestandard form ax2 + bx + c = 0
y = (2 – x)(4 + x) = –x2 – 2x + 8
y = –x2 – 2x + 8
4 3 2 1 0 1 2
0 5 8 9 8 5 0
x
y
Mark the intersecting points of the
curve with the x-axis. The
interesecting points are (–4,0) and (2,0).
Hence the roots are (2 and –4)
Example 5: Draw the graph of y = 2x2 and find the value of 5 using the graph.
Sol. Step 1: Prepare the table of values for y = 2x2
0 1 1 2 2 5
0 2 2 8 8 10
x
y
Step 2: Plot all the points on the graph
sheet and draw the parabola
Step 3: When 5x , 2
2 5 10y
Draw the straight line y = 10 parallel
to x-axis.
Step 4: From the intersecting points of
the parabola and the straight line, draw
perpendiculars to the x-axis.
The point on x-axis at which the
perpendiculars meets are the values
of 5 .
x = 2.3
5 = 2.3
Y
X O X
Y
9
8
7
6
5
4
3
2
1
1 2 3 4 5–1–2–3–4–5
C(–2, 8) E(0, 8)
F(1, 5)B(–3, 5)
D(–1, 9)
A(–4, 0)G
(2, 0)
5 = ± 2.3
Exact ��solution Exact ��
solution Approximate
solutionApproximate
solution
Quadratic Equations 223
EXERCISE 9.10
I. Draw the graph of the following equations.
i) y= x2 ii) y= 3x2 iii) y = x2 – 4x iv) y = –x2 + 8x – 16
v) y = 1
2x2 – 2 vi) y =
1
2x2 – 4
II. 1. Draw the graph of y = 2x2 and find the value of 7 .
2.Draw the graph of y = 1
2x2 and find the value of 10 .
Solving problems based on quadratic equations.
We come across many situations in our daily life where we can solve them
by applying the methods of solving quadratic equations. Let us consider some
examples.
ILLUSTRATIVE EXAMPLES
Example 1 : The product of two consecutive positive odd numbers is 195. Find thenumbers.
Sol. Step 1: Framing the equation
Let one of the odd positive number be x . The other odd positive number will be (x + 2)
The product of the numbers is x(x + 2) = 195 x2 + 2x – 195 = 0
Step 2 : Solving the equation
x2 + 2x – 195 = 0
x2 + 15x – 13x – 195 = 0 (x + 15)(x – 13) = 0
x + 15 = 0 or x – 13 = 0 x = – 15 or x = 13
Step 3: Interpreting and finding the solution
Since the numbers must be positive, x = – 15 is not taken.
x = 13 and x + 2 = 13 + 2 = 15
the two consecutive odd positive numbers are 13 and 15.
Example 2: A man travels a distance of 196 km by train and returns in a car whichtravels at a speed of 21 km/hr faster than the train. If the total journey takes 11hours, find the average speed of the train and the car respectively.
Sol. Let the speed of the train be x km/hr.
Then the speed of the car is (x + 21) km/hour
Time taken for journey by train = 196
xhours
Time taken for journey by car = 196
21xhours
224 UNIT-9
Total journey period = Journey period by train + Journey period by car = 11 hrs.
196 196
21x x = 11
196( 21) 196
( 21)
x x
x x = 11
196x + 4116 + 196x = 11x2 + 231x
11x2 – 161x – 4116 = 0
x =
2( 161) ( 161) 4(11)( 4116)
2 11
= 161 455
22
x = 28 or x – 13.36
Since the speed should be positive, the average speed of the train is 28 km/hr and
the average speed of the car is (28 + 21) = 49 km/hr
Example 3: Anirudh bought some books for Rs. 60. Had he bought 5 more books for
the same amount each book would have cost him 1 rupee less. Find the number of
books bought by Anirudh and the price of each book.
Sol. Let the number of books be = x
Total cost of the books = ̀ 60
Cost of each book = ̀ 60
x
If the number of books is (x + 5), Then, the cost of each book = ` 60
5x
Difference in cost
=is one rupee
60 60
5x x = 1
60( 5) 60
( 5)
x x
x x = 1
2
60 300 60
5
x x
x x = 1
2
300
5x x =
1
1
Cost of each book when Cost of each book when number
number of books is ( ) of books is ( + 5)x x
Quadratic Equations 225
x2 + 5x = 300
x2 + 5x – 300 = 0
x2 + 20x – 15x – 300 = 0
x(x + 20) – 15 (x + 20) = 0
(x + 20)(x – 15) = 0
x + 20 = 0 or x – 15 = 0
x = – 20 or x = 15
Number of books = x = 15
Number of books cannot be negative. Hence – 20 is rejected.
Cost of each book = 60 60
15x = ` 4
EXERCISE 9.11
1. Find two consecutive positive odd numbers such that the sum of their squares is
equal to 130.
2. Find the whole number such that four times the number subtracted from three
times the square of the number makes 15.
3. The sum of two natural numbers is 8. Determine the numbers, if the sum of their
reciprocals is 8
.15
4. A two digit number is such that the product of the digits is 12. When 36 is added to
this number the digits interchange their places. Determine the number.
5. Find three consecutive positive integers such that the sum of the square of the
first and the product of other two is 154.
6. The ages of Kavya and Karthik are 11 years and 14 years. In how many years time
will the product of their ages be 304.
7. The age of a man is twice the square of the age of his son. Eight years hence, the
age of the man will be 4 years more than three times the age of his son. Find their
present age.
8. The area of a rectangle is 56 cm. If the measure of its base is represented by x + 5
and the measure of its height by x – 5, find the dimensions of the rectangle.
9. The altitude of a triangle is 6cm greater than its base. If its area is 108 cm2. Find
its base and height.
10. In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = x + 7, and
AB = x + 8, find the lengths of the diagonals AC and BD .
226 UNIT-9
11. If twice the area of smaller square is subtracted from the area of a larger square,
the result is 14 cm2. However, if twice the area of the larger square is added to
three times the area of the smaller square, the result is 203 cm2. Determine the
sides of the two squares.
12. In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.
If DC = x, BD = 2x – 1 and BC = 2x + 1, find the lengths of all three sides of the
triangle.
13. A motor boat whose speed is 15km/hr in still water goes 30 km downstream and
comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.
14. A dealer sells an article for ` 24 and gains as much percent as the cost price of the
article. Find the cost price of the article.
15. Nandana takes 6 days less than the number of days taken by Shobha to complete a
piece of work. If both Nandana and Shobha together can complete the same work in
4 days, in how many days will shobha alone complete the work?
16. A particle is projected from ground level so that its height above the ground after t
second is given by (20t – 5t2)m. After how many seconds is it 15 m above the ground?
Can you explain briefly why there are two possible answers?
Quadratic Equations 227
ANSWERS
EXERCISE 9.2
2] (i) 14 (ii) 5 5 (iii) 10 (iv) 8
7(v) –2 and –14 (vi) 4 (vii) 9 (viii) 1
4] (i) r = 2
7(ii) d = 3 (iii) b = 15 (iv) a = 8 (v) v = 10 (vi) v = 20
EXERCISE 9.3
1] –10, –5 2] –2, 5 3] –3, 2 4] 3
2, –4 5]
2
3,
3
26]
1
10,
1
107] 2,
2
–5–
8] –2k, –2k 9] 7, –1 10] 2
1, 2 11]
21
1, 3 12]
10 2
-3 1, 13] –4, –4 14]
3– 5
5,
15] 5 3
– ,2 2
16] 5
2, 5 17]
1 1,
b2 2a 18]
2
5– , 3 19] 13, –5 20] 6, –1
Quadratic Equations
Pure quadratic
equations
ax2 + c = 0
Adfected quadratic
equations
ax2 + bx + c = 0
Relation between
roots and
coefficients
Nature of roots
Solving quadratic
equations
Sum of roots
m + n = -b
a
Discriminant
= b2 – 4ac
Roots are
real
and equal
Roots are
real
and distinct
Roots are
imaginary
Product of roots
mn = c
a
Framing a Q.E.
2x (m n)x mn 0
Factorisation method
Completing the
square method
Formula method
Graphical method
= 0
> 0
< 0
Finding 2 , 3 .... by
using parabola
Solving Q.E. by
drawing parabola
Drawing parabola
228 UNIT-9
EXERCISE 9.4
(i) 9 1
,2 2
(ii) 1, -5
4(iii) –3,
1
2(iv) –8 ± 73 (v)
3± 5
2(vi)
-3± 37
2
(vii) 29, 0 (viii) 3, –1 (ix) 2b b
a a, (x)
–b a
2
EXERCISE 9.5
1] 2 2 2] 1 -3 3] 4, 3 4] 15–3
25]
2 2
5 6, 6]
651
16
7]41-1
48]
1 -4
2 3, 9]
b
2
a10] 20, 8 11] a,
1
a12]
b
6
a
13]39
314]
58
13, 2
EXERCISE 9.6
C. (i) 6 (ii) 9
8(iii) 4 (iv) 22 (v)
1
3(vi) 1,
1
2
EXERCISE 9.7
1] 5, 8 2] 10 -5
3 3, 3]
1 -1
8 4, 4] 0,
-1
25]
1 5
p p, 6] –ab, (a+b)
EXERCISE 9.8
B. 1] (i) 12 (ii) 3 (iii) 24 (iv) 7
2] (i) -22
5(ii)
19
33] (i) 6 (ii) 5 5] –2 6] 5 7]
5
2
EXERCISE 9.11
1] 7, 9 2] 3 3] 3, 5 or 5, 3 4] 26
5] 8, 9, 10 6] 5 7] 32, 4 8] 14 cm, 4cm
9] 12cm, 18cm 10] 10cm, 24cm 11] 5cm, 8cm
12] 17cm, 17cm, 16cm 13] 5 km/hr
14] `20 15] 12 days 16] 3 or 1
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