quadratic equations - · illustration 6: find the values of p for which the quadratic equation...

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26513942.

4

QUADRATIC EQUATIONS

QUADRATIC POLYNOMIAL

A polynomial of degree two is called a quadratic polynomial. E.g. x2 + 4, x

2 – 5x + 6, x

2 + 3 x,

2 x2 + 2x – 6. A quadratic polynomial can have at most three terms namely, terms containing x

2, x

and constant.

The general term of a quadratic polynomial in x is ax2 + bx + c, where a, b, c are real numbers and a

0. Quadratic polynomials are generally denoted by f(x), g(x), h(x) etc.

In the quadratic polynomial f(x) = ax2 + bx + c; a, b, c are called coefficients.

If for x = , where is a real number, the value of quadratic expression becomes zero, then is

called zero of ax2 + bx + c.

There will be two zeros for a quadratic polynomial. They are denoted by and .

Note: f(x) = ax2 + bx + c is also called quadratic expression.

Illustration 1: Find whether

(i). x = 2 is a zero of x2 – 5x+ 6.

(ii). x = -4, x = -1 are zeros of x2 + 5x + 4

(iii). x = -1

3, x = -

1

4 are zeros of 6x

2 + 5x + 1

Solution: (i). x = 2

x2 – 5x + 6 = 2

2 – 5(2) + 6 = 4 – 10 + 6 = 0

x = 2 is a zero of x2 – 5x + 6.

(ii). x = -4,

x2 + 5x + 4 = (-4)

2 + 5(-4) + 4

= 16 – 20 + 4 = 0

x = -4 is a zero of x2 + 5x + 4

x = - 1

x2 + 5x + 4 = (-1)

2 + 5(-1) + 4 = 1 – 5 + 4 = 0


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Quadratic Equations- 83

ASCENT Program & School

( I N T E G R A T E D P R E P A R A T I O N )

A S C E N T P l u s

x = -1 is zero of x2 + 5x + 4

x = - 4 and x = -1 are zeros of x2 + 5x + 4.

(iii). x = -1

3, 6x

2 + 5x + 1 = 6

21

3

+ 51

3

+ 1 = 6 1

9 -

5

3 + 1

x = -1

4, 6x

2 + 5x + 1 =

5 50

3 3

= 6

21

4

+ 51

4

+ 1 = 2

16 0 .

x = - 1

3 is a zero and x = -

1

4 is not a zero of 6x

2 + 5x + 1.

QUADRATIC EQUATION

Zeros of quadratic polynomial ax2 + bx + c = 0 where a, b, c are real numbers and a 0 is found by

solving the corresponding equation ax2 + bx + c = 0 called a quadratic equation.

If the real number and are two zeros of the quadratic polynomial ax2 + bx + c, then and are

roots of the quadratic equation ax2 + bx + c = 0.

The general form of a quadratic equation is ax2 + bx + c = 0 where a, b, c are real numbers and a 0.

There will be two roots for a quadratic equation and can be found by solving the equation ax2 + bx +

c = 0.

Roots are also called solutions of ax2 + bx + c = 0.

Exercise 1: (i) Check whether the following equations are quadratic or not:

(a) 4x2 + 5x – 3 = 0 (b) x

2 = 4

(c) 6x = 7x2

(d) (2x – 1) (3x + 5) = 0

(e) (x + 1) (x + 2) = 7(3x + 1) (x + 3) (f) x3 + 3x

2 – 4x + 5 = 0

(g) x - 1

x = 2x

2 (x 0) (h) 8x

2 – 5 = (4x + 3) (2x + 1)

(ii) Determine whether the given value of x is a solution of the given equation or not

(a) x =3

2, 2x

2 – 7x + 6 = 0 (b) x = -

5

2, x =

2

3, 6x

2 + 11x – 10 = 0

(c) x = 2, 5x2 – 4x – 1 = 0 (d) x = - 2 , x = 2 2 , x

2 - 2 2 x – 4 = 0

(e) (3x + 8 (2x + 5) = 0 , x =8

3, x =

5

2.

(iii) Find whether x = a + b, x = a – b are roots of x2 – 2ax = b

2 – a

2.

SOLVING A QUADRATIC EQUATION BY FACTORIZATION

Illustration 2: Solve 81x

2 – 64 = 0.

Solution: 81x

2 – 64 = 0.

(9x)2 – (8)

2 = 0

(9x + 8) (9 x – 8) = 0

x = - 8

9 or x =

8

9


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x = - 8 8

,9 9

are solutions of 81x2 – 64 = 0

Illustration 3: Solve 5x2 – 7x – 6 = 0.

Solution: 5x2 – 7x – 6 = 0

5x2 – 10x + 3x – 6 = 0

5x (x – 2) + 3(x – 2) = 0

(x – 2 ) (5x + 3) = 0

x – 2 = 0 or 5x + 3 = 0

x = 2 or x = - 3

5

x = 2, - 3

5 are solutions of 5x

2 – 7x - 6 = 0

Illustration 4: Solve

x 1 x 23

x 1 x 2 (x 1, - 2).

Solution: x 1 x 2

3x 1 x 2

x 1 x 23 0

x 1 x 2

x 1 x 2 x 1 x 2 3 x 1 x 20

x 1 x 2

(x + 1) (x + 2) + (x – 1) (x – 2) – 3(x – 1) (x + 2) = 0

x2 + 3x + 2 + x

2 – 3x + 2 – 3(x

2 + x – 2) = 0

- x2 – 3x + 10 = 0 - (x

2 + 3x – 10) = 0

x2 + 3x – 10 = 0

(x + 5) (x – 2) = 0 x = - 5, or x = 2.

x = -5, 2 are the solutions of the given equation.

Exercise 2: (i) Solve: x2 + 2 3 x + 3 = 0

(ii) Solve: x 1 x 3 10

x 2, 4x 2 x 4 3

(iii) Solve: 6

x 5x

(iv) Solve: abx2 + (b

2 – ac)x – bc = 0.

SOLVING A QUADRATIC EQUATION BY THE METHOD OF COMPLETION OF

A SQUARE: QUADRATIC FORMULA

ax2 + bx + c = 0

Dividing throughout by a (a 0), we get

x2 +

b

ax +

c

a = 0

By adding and subtracting the square of

21

coefficient of x2

,

We get

2 2

2 b b b cx 2. x 0

2a 2a 2a a


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2 2

2

b b 4acx

2a 4a

If b2 – 4ac 0, then 2b 4ac is a real number

2

2

b b 4acx

2a 4a

2b b 4acx

2a 2a

2b b 4ac

x2a

Note: Here D= b2 – 4ac is called discriminant of the quadratic equation ax

2 + bx + c = 0

Case –I : If b2 – 4ac > 0, then the equation has two distinct real roots , given by

=2b b 4ac

2a

, =

2b b 4ac

2a

Case -II: If b2 – 4ac = 0 i.e. b

2 = 4ac then the equation has real and equal roots , given by

= -b

2a, = -

b

2a. In this case equation will have repeated root .

Case -III: If b2 – 4ac < 0 then the equation does not have real roots.

Note: The quadratic equation ax2 + bx + c = 0 will have real roots if D = b

2 – 4ac 0.

Illustration 5: Determine whether the following quadratic equation have real roots and if they have,

find them

(i). 2x2 + 11x – 6 = 0

(ii). x2 - 6x + 9 = 0

(iii). x2 + x + 1 = 0

(iv). x2 – 4x – 9 = 0

Solution: (i). 2x2 + 11x – 6 = 0

Comparing this equation with ax2 + bx + c = 0

We have a = 2, b = 11, c = -6

Discriminant D = b2 – 4ac = 11

2 – 4(2) (-6) = 121 + 48 = 169

D > 0, given equation will have two distinct real roots say ,

given by = 2b b 4ac

2a

=

11 169 11 13 1

4 4 2

= 2b b 4ac

2a

=

11 169 11 136

4 4

the two roots are 1

and 62

.

(ii). x2 – 6x + 9 = 0

a = 1, b = - 6, c = 9

D = b2 – 4ac = (-6)

2 – 4(1) (9) = 36 – 36 = 0.


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Equation has a repeated root given by = b

2a = =

63

2 1

(iii). x2 + x + 1 = 0

a = 1, b = 1, c = 1

D = b2 – 4ac = 1

2 – 4(1) (1) = - 3 < 0

The equation does not have real roots.

(iv). x2 – 4x – 9 = 0

a = 1, b = -4, c = -9

D = b2 – 4ac = (-4)

2 – 4(1) (-9) = 16 + 36 = 42 > 0

The equation has two roots given by

4 42 4 42

,2 2

x = 4 42 4 42

,2 2

are the required solutions .

Illustration 6: Find the values of p for which the quadratic equation 6x2 + px + 6 = 0 has real

roots.

Solution: D = b2 – 4ac = p

2 – 4 6 6 = p

2 – 144

As the equation has real roots D 0

p2 – 12

2 0 (p + 12) (p – 12) 0 …..(1)

(1) holds good if

(i). p + 12 0 and p – 12 0

p - 12 , p 12

p 12.

Or

(ii). p + 12 0 and p – 12 0

p - 12 and p 12

p - 12

Required valued of p are p - 12 or p 12.

Exercise 3: (i) Examine which of the following quadratic equations has real root(s). If so, find the

root(s).

(a) 6x2 – x – 2 = 0 (b) x

2 – x + 5 = 0

(c) x2 – 4x + 4 = 0 (d) 2x

2 + 3x – 4 = 0

(e) 3x2 + 2 5 x – 5 = 0 (f)

1 2 4

x 1 x 2 x 4

(x -1, -2, -4)

(g) 3y2 + (6 + 4a)y + 8a = 0

(ii) Find the values of p for which the following equations have real roots

(a) px2 + 2x + 1 = 0 (b) 3x

2 + 4x – p = 0

(c) 5x2 + px + 5 = 0 (d) 4x

2 + px + 4 = 0

Illustration 7: The sum of squares of two consecutive positive integers is 221. Find the integers.

Solution: Let x be one of the positive integers. Then the other is x + 1

sum of squares of the integers = x2 + (x + 1)

2 = 221


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x2 + x

2 + 2x + 1 – 221 = 0

2x2 + 2x – 220 = 0

2(x2 + x – 110) = 0

x2 + x – 110 = 0

(x – 10) (x – 11) = 0 x = 10 or x = 11.

consecutive positive integers are 10 and 11.

Illustration 8: One side of a rectangle exceeds its other side by 2 cm. If its area is 195 cm2,

determine the sides of the rectangle.

Solution: Let one side be x cm

Then other side will be (x + 2) cm

Area of rectangle = x (x + 2)

x (x + 2) = 195

x2 + 2x – 195 = 0

x2 + 15x – 13x – 195 = 0

x(x + 15) – 13(x + 15) = 0

(x – 13) (x + 15) = 0

x = 13 or x = -15.

Since, side of the rectangle cannot be negative

x = 13.

sides of rectangle are x, x + 2 = 13 cm, 15 cm.

Illustration 9: The hypotenuse of a right angled triangle is 25 cm. The difference between the

lengths of the other two sides of the triangle is 17 cm. Find the lengths of these sides.

Solution: Let the length of the shorter side be x cm.

Then, the length of the longer side = (x + 17) cm

AB = x, BC = x + 17, CA = 25

A

B C

x

x+17

25

By Pythagoras theorem AB

2 + BC

2 = 25

2 x

2 + (x + 17)

2 = 25

2

2x2 + 34x – 336 = 0

x2 + 17x – 168 = 0

(x – 7) (x + 24) = 0 x = 7, x = - 24

But side of triangle cannot be negative

x = 7

Length of the shorter side x = 7 cm

Let of the longer side = x + 17 = 7 + 17 = 24 cm.

Exercise 4: (i) Find two consecutive positive even integers the sum of whose square is 340.

(ii) A two digit number is such that product of its digits is 30. If the digits are

interchanged the resulting number will exceed the previous by 9. What is the

number?


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A S C E N T P l u s

(iii) The sum of the ages of a father and his son is 45 years. Five years ago the

product of their ages was four times the father’s age at that time. Find their

present ages.

Relation between roots and coefficients of a Quadratic equation:

If , are roots of ax2 + bx + c = 0 then

=2b b 4ac

2a

, =

2b b 4ac

2a

+ = 2b b 4ac

2a

+

2b b 4ac

2a

= sum of roots

= - 2b b

2a a

Sum of roots = + =

2

Coeffic ient of xb

a Coeffic ient of x

= 2b b 4ac

2a

2b b 4ac

2a

=

2 2 2 2

2 2 2

b b 4ac b b 4ac 4ac c

a4a 4a 4a

Product of roots = = 2

c cons tan t term

a coefficient of x

Some of Important Formulae:

1. 2 +

2 = ( + )

2 - 2

2. (3 +

3) = ( + )

3 - 3 ( + )

3. 4 +

4 = (

2 +

2)

2 - 2

2

2

4. ( - )2 = ( + )

2 - 4

Take ax2 + bx + c = 0

Divide the whole equation by a ( since a 0)

2 b cx x 0

a a 2 b c

x x 0a a

x2 – ( + )x + = 0 i.e. x

2 – (sum of the roots) x + Product of the roots = 0

The quadratic equation whose roots are , is given by x2 – ( + )x + = 0.

Illustration 10: If , are roots of ax2 + bx + c = 0. Find the quadratic equation whose roots are

(a) 2, 2 (b) + 3, + 3

(c) ,4 4

(d)

1 1,

Solution: , are roots of ax2 + bx + c = 0

+ = -b

a, =

c

a

(a) We have to find the equation whose roots are 2, 2


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A S C E N T P l u s

for the required equation

sum of roots = 2 + 2 = 2( + ) = b 2b

2a a

product of roots = (2) (2) = 4 = 4c 4c

a a

The equation whose roots are 2, 2 is

x2 – (sum )x + (product ) = 0

x2 +

2b

ax +

4c

a = 0 i.e. ax

2 + 2bx + 4c = 0

(b) Sum = + 3 + + 3 = + + 6 = b b 6a

6a a

product = ( + 3) ( + 3) = + 3 + 3 + 9

= + 3( + ) + 9 = c 3b c 3b 9a

9a a a

required equation is x2 – (sum) x + product = 0

x2 -

b 6a c 3b 9ax 0

a a

i.e. ax2 – (-b + 6a)x + (c – 3b + 9a) = 0

(c) Sum = b / a b

4 4 4 4 4a

Product = c

4 4 16a

Required equation is

x2

b

4a

x + c

16a = 0 i.e. 16ax

2 + 4bx + c = 0

(d) Sum = 1 1 b / a b

c / a c

product 1 1 1 1 a

c / a c

required equation is 2 b ax x 0

c c

i.e. cx2 + bx + a = 0

Exercise 5: (i) Form the quadratic equation whose roots are

(a) -2, - 5 (b) 2 3 - 5, 2 3 + 5

(ii) If , are roots of ax2 + bx + c = 0 (c 0), find the values of

(a) 2 + 2 (b) 3 + 3

(c) 1 1

(d)

(iii) If , are roots of 2x2 + x + 3 = 0, find the quadratic equation whose roots are

(a) -, - (b) 1 1

,

(c) 3, 3 (d) ,2 2

(e) - 1, - 1


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ANSWERS TO EXERCISES

Exercise 1:

(i). (a) yes (b) yes (c) yes (d) yes

(e) yes (f) no (g) no (h) No

(ii). (a) yes (b) yes (c). no (d) no (e) no

(iii). yes

Exercise 2:

(i). x = - 3 , - 3 (ii). 5, 5

2

(iii). x = 2, 3 (iv). c b

,b a

Exercise 3:

(i). (a) roots are 2 1

,3 2

(b) no roots

(c) repeated root = 2 (d) roots are 3 41 3 41

,2 2

(e) roots are 1

5, 53

(f) 2 1 3 , 2 1 3

(g) 4a

2,3

(ii). (a) p 1 (b) p 4

3

(c) p -10 or p 10 (d) p - 8 or p 8

Exercise 4:

(i) 12, 14

(ii) 56

(iii) Fathers age = 36 years, son’s age = 9 years

Exercise 5:

(i). (a) x2 + 7x + 10 = 0 (b) x

2 - 4 3 x – 13 = 0

(ii). (a)

2

bc

a (b)

2 2

3

b c 2ac

a

(c) b

c (d)

2b 2ac

ac

(iii). (a) 2x2 – x + 3 = 0 (b) 3x

2 + x + 2 = 0

(c) 2x2 + 3x + 27 = 0 (d) 8x

2 + 2x + 3 = 0

(e) 2x2 + 5x + 6 = 0


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SOLVED PROBLEMS

SECTION -A

Problem 1: Solve

1 1 1x 3, 5

x 3 x 5 6

Solution: 1 1 1

x 3 x 5 6

x 5 x 3 1

x 3 x 5 6

8 1

x 3 x 5 6

48 = x2 + 2x – 15

x2 + 2x – 63 = 0

x = -9 or 7

Problem 2: Solve

2x 1 3x 90

x 3 2x 3 x 3 2x 3.

Solution: Obviously, given equation is valid only when x – 3 0 , 2x + 3 0

2x 2x 3 x 3 3x 90

x 3 2x 3

2x2 + 5x + 3 = 0

(2x + 3) (x + 1) = 0

2x + 3 = 0 or x + 1 = 0

but 2x + 3 0 x = -1.

Problem 3: Find the solutions set of x4 – 5x

2 + 6 = 0.

Solution: x4 – 5x

2 + 6 = 0

(x2)

2 – 5x

2 + 6 = 0 Let t = x

2

t2 – 5t + 6 = 0

(t – 2) (t – 3) = 0

t = 2 or t = 3

x2 = 2 or x

2 = 3

x = 2 or x = 3

solution set x = 2, 3

Problem 4: Solve 3x + 3

-x – 2 = 0.

Solution: 3x + 3

-x – 2 = 0

3x +

x

1

3 - 2 = 0 let t = 3

x


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A S C E N T P l u s

t + 1

t - 2 = 0

t2 – 2t + 1 = 0

(t – 1)2 = 0 t = 1

3x = 1 3

x = 3

0 x = 0.

Problem 5: Find two consectutive positive odd integers such that sum of their squares is 290.

Solution: If the first of the two required consecutive positive odd integers is x, then the second

will be x + 2

By the given condition x2 + (x + 2)

2 = 290

2x2 + 4x – 286 = 0 (x + 13) (x – 11) = 0

since the required odd number is positive , we have x = 11

required integers are 11, 13.

Problem 6: Find the number having two digits such that it is 4 times the sum and three times the

product of its two digits.

Solution: When x 0, x = 1 or x = 2 or 3 …, or 9 and y = 0 or 1 or 2 … or 9

Let the required number be 10x + y

By given conditions

10x + y = 4(x + y) …(1)

10x + y = 3xy …(2)

From (1) 6x = 3y 2x = y

Substituting value of y in (2)

10x + 2x = 3x 2x

12x = 6x2

6x2 – 12x = 0

6x (x – 2) = 0

as x 0 , x = 2

Substituting x = 2 in (2)

10 2 + y = 3 2y

20 + y = 6y 5y = 20

y = 4

Required number = 10x + y = 10 2 + 4 = 24.

Problem 7: In a cricket match Anil took one wicket less than twice the number of wickets taken

by Ravi. If the product of the number of wickets taken by them is 15, find the number

of wickets taken by each of them.

Solution: Let the number of wickets taken by Anil be x and that by Ravi be y

By given conditions

x = 2y – 1 …(1)

xy = 15

y =15

x …(2)

Substituting (2) in (1)

x = 15

2 1x

x2 + x – 30 = 0


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A S C E N T P l u s

(x – 5) (x + 6) = 0

x = 5 or - 6

x cannot be negative x = 5

from (2) y = 15

35

.

Anil has taken 5 wickets and Ravi has taken 3 wickets.

Problem 8: Find the values of m for which the equation x2 -2mx + 7m – 12 = 0 has equal roots.

Solution: Comparing the given equation x2 – 2mx + 7m – 12 = 0 with

ax2 + bx + c = 0, a = 1, b = -2m, c = 7m – 12

Discriminant D = b2 – 4ac

= (-2m)2 – 4(1) (7m – 12) = 4m

2 – 28m + 48

If the roots are equal D = 0

4m2 – 28m + 48 = 0 4(m

2 – 7m + 12) = 0

m2 – 7m + 12 = 0 (m – 3) (m – 4) = 0

m = 3 or 4

Problem 9: Determine the equation, sum of whose roots is 1 and sum of their squares is 13.

Solution: Let , be roots of the required quadratic equation

+ = 1, 2 +

2 = 13

(+ )2 =

2 +

2 + 2

1 = 13 + 2

2 = -12

= -6.

required quadratic equation is x2 – ( + )x + = 0

x2 – x – 6 = 0.

Problem 10: If and are roots of ax2 + bx + c = 0, find the equation whose roots are ,

.

Solution: Let , be roots of ax2 + bx + c = 0

+ = -b

a, =

c

a

For the required equation sum =

= 2 2

=

2

2

=

2

2

b c2

b 2aca a

c ac

a

Product = 1

Required equation is x2 – (sum)x + product = 0

x2 - 2b 2ac

x 1 0ac

acx2 - (b

2 – 2ac) x+ ac = 0 .


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SECTION -B

Problem 1: If and are roots of ax2 + bx + c = 0 then prove that

22

a b b

Solution: + = -b

a, =

c

a

2 2

a b

= 3 3 2 2a b

=

3 2a 3 b 2

=

3 2b c b b 2c

bca 3 ba a a a a

a bc c

a a

Problem 2: If , are roots of x2 + px + q = 0 and , are roots of x

2 + rx + s = 0, prove that

( - ) ( - ) ( - ) ( - ) = (q – s)2 + (p – r) (sp – rq).

Solution: + = - p, = q, + = - r, = s

( - ) ( - ) ( - ) ( - ) = [2 - ( + ) + ] [

2 - ( + ) + ]

= [2 + r + s] [

2 + r+ s] =

2

2 + r ( + ) + r

2 + s(

2 +

2) + s

2 + rs( + )

= q2 – rpq + r

2q + s(p

2 – 2q) + s

2 – srp

= (q – s)2 + (p – r) (sp – rq)

Problem 3: If the sum of the roots of the equation ax2 + bx + c = 0 is equal to sum of the squares

of their reciprocals then show that bc2, ca

2, ab

2 are in A.P.

Solution: Let , be roots of the given equation

Given that + = 2 2

1 1

+ = 2 2

2 2

+ =

2

2 2

2

2

2

2

2

b 2cb aa

a c

a

2 2

2 2

b c b 2c

a aa a


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2a2c = b

2a + c

2b

ab2, a

2c, bc

2 are in A.P.

Problem 4: If the roots of the equation ax2 + bx + c = 0 are in the ratio m : n, prove that

mnb2 = ac(m + n)

2

+ =b

a , =

c

a

Solution: : = m : n

m

n

m n

m n

(Applying Componendo and Dividendo)

2 2

2 2

m n

m n

2 2

2 2

m n

4 m n

22 2

2 2

2

m nb / a

b c m n4aa

22

2 2

m nb

b 4ac m n

b2(m – n)

2 = b

2 (m + n)

2 – 4ac( m + n)

2

4ac( m + n)2 = b

2 [(m + n)

2 – (m – n)

2]

4ac(m + n)2 = b

2 [4mn]

mnb2 = ac(m + n)

2

Problem 5: If the roots of the equation a(b – c)x2 + b(c – a)x + c(a – b) = 0 are equal. Show

that1 1 2

a c b .

Solution: Comparing the given equation with Ax2 + Bx + C = 0.

A = a(b – c), B = b(c – a), C = c(a – b)

As the roots are equal D = B2 – 4AC = 0

[b(c – a)]2 – 4[a(b – c)] [c(a – b)] = 0

a2b

2 + b

2c

2 – 2ab

2c – 4a

2bc + 4c

2a

2 – 4abc

2 + 4ab

2c

= 0

a2b

2 + b

2c

2 + 2ab

2c – 4a

2bc + 4c

2a

2 – 4abc

2 = 0

(ab + bc – 2ca)2 = 0

ab + bc = 2ca

dividing both sides by abc

1 1 2

a c b .

Problem 6: Show that the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are real.


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A S C E N T P l u s

Solution: Comparing the given equation with Ax2 + Bx + C = 0.

A = (b – c), B = (c – a), C = (a – b)

Discriminant D = B2 – 4AC

(c – a)2 – 4(b – c) (a – b)

= c2 + a

2 – 2ac – 4ab + 4b

2 + 4ca – 4bc

= a2 + 4b

2 + c

2 – 4ab – 4bc + 2ca

= (a – 2b + c)2

As D is a perfect square it is always positive

D > 0

roots of given equation are real.

Problem 7: Let a, b, c be real numbers with a 0. Let , be the roots of the equation

ax2 + bx + c = 0. Express the roots of a

3x

2 + abcx + c

3 = 0 in terms of , ..


then + = - b

a, =

c

a

Let 1, 1 be roots of a3x

2 + abcx + c

3 = 0

1 + 1 = 3 2

abc bc b c

a aa a

1 + 1 = 2 +

2 ….(1)

1 1 = 3

3 3 3 2 2

3

c

a

11 = (2) (

2) ….(2)

From (1) and (2), 1 = 2, 1 =

2

roots of a3x

2 + abcx + c

3 = 0 are

2 and

2.

Problem 8: Find the condition that one root of ax2 + bx + c = 0 is square of the other root.

Solution: If is one root, then 2 will be another root

+ 2 =

b

a , ….(1)

. 2 =

c

a

3 =

c

a ….(2)

(1)3 ( +

2)

3 =

3

3

b

a

3 +

6 + 3

3( +

2) =

3

3

b

a

2 3

2 3

c c 3c b b

a a aa a

2 3

2 3

ac c 3bc b

a a

a(ac + c2 – 3bc ) = - b

3 .

ac(a + c) = 3abc – b3 .


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Problem 9: If the roots of the equation x2 – px + q = 0 differ by unity then prove that p

2 = 4q + 1.

Solution: Let , be the roots of x2 – px + q = 0

+ = p, = q

given - = 1

( - )2 = 1

( + )2 - 4 = 1

p2 – 4q = 1

p2 = 4q + 1.

Problem 10: If , are the roots of ax2 + bx + c = 0 (a 0) and + k, + k are roots of

Ax2 + Bx + C = 0 (A 0) for some constant k, then prove that

2 2

2 2

b 4ac B 4AC

a A

.


+ = b

a , =

c

a

+ k , + k are the roots of Ax2 + Bx + C = 0

( + k) + ( + k) = B

A ,

( + k) ( + k) = C

A

( + k) - ( + k) = -

Squaring both sides,

[( + k) - ( + k)]2 = ( - )

2

[( + k) + ( + k)]2 – 4( + k) ( + k) = ( + )

2 - 4

2 2B C b c

4 4A A a a

2 2

2 2

B 4AC b 4ac

A a


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ASSIGNMENT PROBLEMS

SUBJECTIVE

SECTION –A

1. Solve 4x2 – 4a

2x+ (a

4 – b

4) = 0 by factorization method.

2. Solve 9x2 – 9(a + b)x + (2a

2 + 5ab + 2b

2) = 0 by factorization method.

3. Solve x a x b a b

x b x a b a

(x a, x b) by factorization method.

4. Using quadratic formula, solve p2x

2 + (p

2 – q

2)x – q

2 = 0.

5. Find the value of k for which

(k + 1)x2 + 2(k + 3)x + (k + 8) = 0 has equal roots.

6. If the roots of (1 + m2)x

2 + 2mcx + (c

2 – a

2) = 0 has equal roots prove that

c2 = a

2(1 + m

2).

7. If the roots of (c2 – ab)x

2 – 2(a

2 – bc) x + (b

2 – ac) = 0 are equal then show that either a = 0 or

a3 + b

3 + c

3 = 3abc.

8. Find the value of k, if the roots of the equation x2 + kx + 64 = 0 and x

2 – 8x + k = 0

(k > 0) are equal.

9. A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the

slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.

10. Two numbers differ by 3, and their products is 504. Find the numbers.


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A S C E N T P l u s

SECTION –B

1. If one root of quadratic equation ax2 + bx + c = 0 is equal to nth power of the other root, show

that 1 1

n nn 1 n 1ac a c b 0 .

2. Show that the equation 2(a2 + b

2)x

2 + 2(a + b)x + 1 = 0 has no real roots.

3. If the ratio of roots of the quadratic equation x2 + px + q = 0 be equal ratio of roots of x

2 + lx

+ m = 0. Prove that p2m = l

2q.

4. If the roots of the equation (x – a) (x – b) – k = 0 are c and d, then prove that the roots of (x –

c) (x – d) + k = 0 are a and b.

5. Let p and q be the roots of x2 – 2x + A = 0 and let r and s be the roots of x

2 – 18x + B = 0. If

p < q < r < s are in arithmetic progression, find the values of A and B.

6. Show that if the roots of (a2 + b

2)x

2 + 2x(ac + bd) + c

2 + d

2 = 0 are real, they will be equal.

7. If , are the roots of the equation x2 – px + q = 0 and 1, 1 are the roots of the equation

x2 – qx + p = 0. Form the quadratic equation whose roots are

1 1

1 1

and

1 1

1 1

.

8. Show that the roots of x2 + mx + n = 0 are real when m = k +

n

k.

9. If one root is equal to the square of the other root of the equation x2 + x – k = 0, what is the

value of k?

10. Let p(x) = x2 + ax + b be a quadratic polynomial in which a and b are integers. Given any

integer n, show that there is an integer m such that p(n) p(n + 1) = p(m).


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A S C E N T P l u s

OBJECTIVE

Multiple Choice Questions:

1. If one root of quadratic equation 2x2 + kx – 6 is 2 then the other root is

(A) -1 (B) 2

(C) 3

2 (D)

3

2

2. The roots of 4x2 – 4ax + (a

2 – b

2) = 0 are

(A) a + b, a – b (B) a b a b

,2 2

(C) a + b , ab (D) a b

, ab2

3. The roots of x2 +

1a x 1 0

a

are (a 0)

(A) a, 1

a (B) -a, -

1

a

(C) a, - 1

a (D) - a,

1

a

4. The value of k if (k + 1)x2 – 2(k – 1) x + 1 = 0 has equal roots

(A) 0, 3 (B) 1, 2

(C) -3, -2 (D) 3, 2

5. The values of ‘k’ for which x2 – 4x + k = 0 has distinct real roots are

(A) k 4 (B) k > 4

(C) k < 4 (D) k 4

6. If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation

p(x2 + x) + k = 0 has equal roots then k =

(A) 7

4 (B) -

7

4

(C) 4

7 (D) -

4

7

7. If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal then

(A) a, b, c are in A.P. (B). a, c, b are in A.P.

(C) b, a, c are in A.P. (D) none of these

8. A plane left 30 minutes later than the scheduled time and in order to reach its destination

1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. It’s

usual speed is

(A) 750 km/hr (B) 1000 km/hr

(C) 500 km/hr (D) 300 km/hr

9. If p and q are roots of x2 + px + q = 0 then

(A) p = 1, q = - 2 (B) p = 0, q = 1

(C) p = - 2, q = 0 (D) p = -2, q = 1


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A S C E N T P l u s

10. If , are roots of ax2 + bx + c = 0 then (1 + ) (1 + ) is equal to

(A) a b c

a

(B)

a b c

a

(C) a b c

a

(D)

b c a

a

11. If , are roots of x2 – 7x + 12 = 0 then the equation whose roots are 3, 3 is

(A) (3x)2 – 7(3x) + 12 = 0 (B)

2x x

7 12 03 3

(C) (x – 3)2 – 7(x – 3) + 12 = 0 (D) (x + 3)

2 – 7(x + 3) + 12 = 0

12. If , are roots of 2x2 – 3x – 6 = 0 then equation whose roots are

2 + 2,

2 + 2 is

(A) 4x2 + 49x + 118 = 0 (B) 4x

2 – 49x + 118 = 0

(C) 4x2 – 49x – 118 = 0 (D) x

2 – 49x + 118 = 0

13. Match the following:

The condition that one root of ax2 + bx + c = 0 is

A) reciprocal of the other root 1) nb2 = ac(1 + n)

2

B) nth power of the other root 2) ac(a + c) + b3 = 2abc

C) square of the other root 3)

1 1n nn 1 n 1a c ac b 0

D) n times the other root 4) c = a

5) ac(a + c) = 2abc + b3

14. Match the following: The roots of the equation

1) x2 – 2x + 4 = 0 are A) real

2) 6x2 – x – 2 = 0 are B) real and equal

3) 2x2 + 3x + 2 = 0 are C) real and distinct

4) (b – c) x2 + (c – a)x + (a – b) = 0 are D) imaginary

Fill in the Blanks:

15. The values of m for which the equation x2 – 2mx + 7m – 12 = 0 has equal roots are _______.

16. The values of x for which x2/3

+ x1/3

– 2 = 0 are __________

Short answer questions:

17. Find the value of k for which one root is square of the other root of x2 – x – k = 0.

18. If , are roots of ax2 + bx + c = 0 then find the value of

2 2

2 2

.

19. Write the condition that the roots of ax2 + bx + c = 0 are real.

20. Find the nature of roots of the equation 9x2 – 30x + 25 = 0.


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A S C E N T P l u s

ANSWERS TO ASSIGNMENT PROBLEMS

SUBJECTIVE

SECTION –A

1. x = 2 2 2 2a b a b

or2 2

2. 2a b a 2b

or3 3

3. x = 0 or a + b

4. x = - 1 or 2

2

q

p

5. m = 1

3

8. k = 16

9. 50 km/hr , 40 km /hr

10. 21, 24 or -21, -24

SECTION –B

5. A = - 3, B = 77 9. k = - 1


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A S C E N T P l u s

OBJECTIVE

1. C 2. B

3. B 4. A

5. C 6. A

7. A 8. A

9. A 10. B

11. B 12. B

13. A 4, B 3 , C 2, D 1 14. 1 B, 2 C, 3 D, 4 A

15. m = 3 or m = 4 16. x = 0

17. k = 2 5 18. 2

2

c

a

19. b2 – 4ac 0 20. real and equal

quadratic equations - · illustration 6: find the values of p for which the quadratic equation...

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