quadatrical equations

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139

[ ] [ ]

48 x 2 = ( x + 4) x

96 = x 2 + 4 x

x 2 + 4 x – 96 = 0

x 2 + 12 x – 8 x – 96 = 0

x ( x + 1 2 ) – 8 ( x + 1 2 ) = 0

( x + 12) ( x – 8) = 0

Either ( x + 12) = 0 or ( x – 8) = 0

x = –12 or x = 8

∴ Altitude = x = 8 cms.

Base = x + 4

= 8 + 4

= 12 cms

Example 4 : Rashmi bought some books for Rs. 60. Had she bought 5 more books

to the same amount each book would have cost her 1 rupee less. Find

the number of books bought by Rashmi and price of each book.

Solution : Let the number of books = x

Total cost of the books = Rs. 60

Cost of each book = Rs. x

60

If number of books is ( x + 5)

Then the cost of each book = Rs.)50(

60

+ x

Difference in cost = 1 Re.

Cost of each book cost of each book if number of books – if number of books = Difference amount

is ( x ) is ( x +5)

x

60 –

5

60

+ x

= 1

)5(

60)5(60

+−+

x x

x x

= 1

x = –12 cannot be

considered, because

the length is always

positive



140

[ ] [ ]

x x

x x

5

60300602 +

−+ = 1

1

1

5

3002

=+ x x

∴ x

2 + 5 x = 300

∴ x 2 + 5 x – 300 = 0

x 2 + 20 x – 15 x – 3 0 0 = 0

x ( x + 2 0 ) – 1 5 ( x + 2 0 ) = 0

( x + 20) ( x – 1 5 ) = 0

Either ( x + 20) = 0 or ( x – 1 5 ) = 0

∴ x

= – 20 or x

= 15∴ Number of books = x = 15

Cost of each book =x

60 =

15

60 = Rs. 4

Example 5 : The speed of a motor boat in still water is 15 km/hr. If it goes down

the stream 30 kms and again returns to the starting point in total time

of 4 hrs and 30 minutes, find the speed of the stream.

Solution : Speed in Still water is = 15 km/hr

Total distance travelled = 30 km

Let the speed of the stream = x km/hr

Speed up the stream = (15 – x ) km/hr

Speed down the stream = (15 + x ) km/hr

Total time taken = 4hrs and 30 minutes

Time taken to row down the stream = x +15

30

Time taken to row up the stream = x −15

30

Time taken to row + time taken to row = 4 hours 30 minutes

Down the stream up the stream

x +15

30+

x −15

30=

2

14

x = –20 cannot be

considered because

number of books is always

positive



141

2

9

)15()15(

)15(30)15(30=

−+++− x x

x x

2

9

225

30450304502

=−

++− x

x x

2

9

225

9002 =

− x

∴ 2225

9

)2(900 x −=

200 = 225 – x 2

∴ x 2 = 225 – 200

x 2 = 25

x = ± 5

x = + 5 or x = – 5

∴ Speed of the stream = x = 5km/hr

Exercise : 5.5

1) The sum of a number and twice its square is 105. Find the number.

2) Product of two consecutive integers is 182. Find the integers.

3) The sum of the squares of three consecutive natural numbers is 194. Find the integers.

4) The length of rectangular field is 3 times its breadth. If the area of the field is

147 sq mts. Find the length of the field.

5) Hypotenuse of a right-angled triangle is 20 mts. If the difference between the lengths

of other two sides is 4 mts. Find the measures of the sides.

6) An Aero-plane takes 1 hr. less for a journey of 1200 km. If its speed is increasedby 60 km/hr from its initial speed find the initial speed of the plane.

7) Some students planned a picnic. The budget for the food was Rs. 480. As eight

of them failed to join the party the cost of the food for each member increased

by Rs. 10. How many students participated in the picnic?

8) Sailor Raju covered a distance of 8 km in 1 hr 40 minutes down stream and returns

to the starting point. If the speed of the stream is 2 km/hr, find the speed of the

boat in still water.



142

9) A dealer sells an article for Rs. 24 and gains as much percent as the cost price

of the article. Find the Cost price of the article.

10) Sowmya takes 6 days less than the number of days taken by Bhagya to complete

a piece of work. If both Sowmya and Bhagya together can complete the same

work in 4 days. In how many days will Bhagya complete the work?

6. Nature of the roots of a quadratic equation.1) Consider the equation x

2 – 2 x + 1 = 0

This is in the form a x 2 + b x + c = 0

The coefficients are a = 1, b = –2, c = 1

x =a2

ac4bb 2 −±−

x =1x2

1x1.4)2()2( 2 −−+−−

x =2

442 −±

x =2

02+

x =2

02+ or x =2

02−

x = 1 or x = 1 → roots are equal

2) Consider the equation x 2 – 2 x – 3 = 0


the coefficients are a = 1, b = –2, c = –3

x =a2

ac4bb 2 −±−

x =1x2

16)2( ±−−

x =2

42±+



144

Discriminant (b2 – 4ac) Nature of the roots

∆ = 0 Roots are real and equal

∆ > 0 (Positive) Roots are real and distinct

∆ < 0 (negative) Roots are imaginary

Example 1 : Determine the nature of the roots of the equation 2 x 2 – 5 x – 1 = 0.

Consider the equation 2 x 2 – 5 x – 1 = 0

This is in form of a x 2 + b x + c = 0

The co-efficient are a = 2, b = –5, c = –1

∆ = b2 – 4ac

∆ = (–5)2 –4(2) (–1)

∆ = 25 + 8

∆ = 33

∴ ∆ > 0Roots are real and distinct

Example 2 : Determine the nature of the roots of the equation 4 x 2 – 4 x + 1 = 0

Consider the equation 4 x 2 – 4 x + 1 = 0

This is in the form of a x 2 + b x + c = 0

The co-efficient are a = 4, b = –4, c = 1

∆ = b2 – 4ac

∆ = (–4)2 –4 (4) (1)

∆ = 16 – 16

∴ ∆ = 0

Roots are real and equal

Example 3 : For what values of ‘m’ roots of the equation x 2 + m x + 4 = 0 are

(i) equal (ii) distinct

Consider the equation x 2 + m x + 4 = 0


the co-efficients are a = 1, b = m, c = 4

∆ = b2 – 4ac

∆ = m2 – 4(1) (4)

∆ = m2 – 16

1) If roots are equal ∆ = 0 ∴ m2 – 16 = 0

m2 = 16

∴ m = 16 ∴ m = ± 4



145

2) If roots are distinct ∆ > 0 ∴ m2 – 16 > 0

∴ m2 > 16

m2 > 16

m > ± 4

Example 4 : Determine the value of ‘k’ for which the equation k x 2 + 6 x + 1 = 0 has

equal roots.

Consider the equation k x 2 + 6 x + 1 = 0


the co-efficients are a = k, b = 6, c = 1

∆ = b2 – 4ac

since the roots are equal, b2

– 4ac = 0 (

∴

∆ = 0)(6)2 – 4(k)(1) = 0

3 6 – 4 k = 0

4k = 36

k =4

36 = 9

∴ k = 9

Example 5 : Find the value of ‘p’ for which the equation x 2 – ( p + 2 ) x + 4 = 0 hasequal roots.

Consider the equation x 2 – ( p + 2 ) x + 4 = 0


Coefficients are a = 1, b = –(p + 2), c = 4

since the roots are equal ∆ = 0

b2 – 4 a c = 0

[–(p + 2)]2 – 4(1)(4) = 0

(p + 2)2 – 16 = 0

p + 2 =± 16

p + 2 = ± 4

p + 2 = + 4 or p + 2 = –4

∴ p = 4 – 2 or p = –4 – 2

∴ p = 2 or p = –6



146

If m and n are the roots of the

quadratic equation

a x 2 + b x + c = 0

Sum of the rootsa

b−=

Product of rootsa

c+=

Exercise : 5.6

A. Discuss the nature of roots of the following equations

1) y2 – 7y + 2 = 0 2) x 2 – 2 x + 3 = 0 3) 2n2 + 5n – 1 = 0

4) a2 + 4a + 4 = 0 5) x 2 + 3 x – 4 = 0 6) 3d2 – 2d + 1 = 0

B. For what positive values of ‘m’ roots of the following equations are

1) equal 2) distinct 3) imaginary

1) a2 – ma + 1 = 0 2) x 2 – m x + 9 = 0

3) r2 – (m + 1) r + 4 = 0 4) mk 2 – 3k + 1 = 0

C. Find the value of ‘p’ for which the quadratic equations have equal roots.

1) x 2 – p x + 9 = 0 2) 2a2 + 3a + p = 0 3) pk 2 – 1 2 k + 9 = 0

4) 2y2 – py + 1 = 0 5) (p + 1) n2 + 2 ( p + 3 ) n + ( p + 8 ) = 0

6) (3p + 1)c2

+ 2 (p + 1) c + p = 0

7. Relationship between the roots and co-efficient of the terms of the quadratic

equation.

If ‘m’ and ‘n’ are the roots of the quadratic equation a x 2 + b x + c = 0 then

m =a2

ac4bb 2 −+−, n =

a2

ac4bb 2 −−−

∴ m + n =a2

ac4bb 2 −+− +a2

ac4bb 2 −−−

m + n =a2

ac4bbac4bb 22 −−−−+−

∴ m + n =a2

b2−

m + n =a

b-

mn =

−+−a2

ac4bb2

−−−a2

ac4bb2

mn =

( )2

222

a4

ac4b)b( −−−



147

mn =( )

2

22

a4

ac4bb −−

mn = 2

22

a4

ac4bb +−

∴ mn = 2a4ac4 =

ac ∴ mn =

ac

Example 1 : Find the sum and product of the roots of equation x 2 + 2 x + 1 = 0

x 2 + 2 x + 1 = 0


The coefficients are a = 1, b = 2, c = 1

Let the roots be m and n

i) Sum of the roots m + n =a

b− =

1

2−

∴ m + n = –2

ii) Product of the roots mn =a

c =

1

1

∴ mn = 1

Example 2 : Find the sum and product of the roots of equation 3 x 2 + 5 = 0

3 x 2 + 0 x + 5 = 0


The coefficients are a = 3, b = 0, c = 5

Let the roots are p and q

i) Sum of the roots p + q =a

b− =

3

0

∴ p + q = 0

ii) Product of the roots pq =a

c =

3

5∴ pq =

3

5



148

Example 3 : Find the sum and product of the roots of equation 2m2 –8m=0

2m2 – 8m + 0 =0


Let the roots be α and β

i) Sum of the roots a

b−

=β+α 2

)8(−−

= = 4

ii) Product of the rootsa

c=αβ

2

0= = 0

Example 4 : Find the sum and product of the roots of equation x 2 – (p+q) x + pq = 0

x 2 – ( p + q ) x + pq = 0

The coefficients are a = 1, b = –(p + q), c = pq


b−

m + n =( )[ ]1

qp +−−

∴ m + n = (p + q)

ii) Product of the roots mn = a

c = 1

pq

∴ mn = pq

Exercise : 5.7

Find the sum and product of the roots of the quadratic equation :

1) x 2 + 5 x + 8 = 0 2) 3a2 – 10a – 5 = 0 3) 8m2 – m = 2

4) 6k 2 – 3 = 0 5) pr2 = r – 5 6) x 2 + (ab) x + ( a + b ) = 0

8. To form an equation for the given roots

Let ‘m’ and ‘n’ are the roots of the equation

∴ x = ‘m’ or x = ‘n’

i.e., x – m = 0 , x – n = 0

( x – m) ( x – n) = 0

∴ x 2 – m x – n x + mn = 0

x

2

– ( m + n ) x + mn = 0



149

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

x2 – (m + n) x + mn = 0

Example 1 : Form the quadratic equation whose roots are 2 and 3

Let ‘m’ and ‘n’ are 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 x 2 – ( m + n ) x + mn = 0

x 2 – (5) x + ( 6 ) = 0

∴ x 2 – 5 x + 6 = 0

Example 2 : Form the quadratic equation whose roots are5

2 and

2

5

Let ‘m’ and ‘n’ are the roots

∴ m =5

2 and n =

2

5

∴ Sum of the roots = m + n =5

2 +

2

5 =

10

254+

∴ m + n =10

29

Product of the roots = mn =2

5x

5

2∴ mn = 1

Standard form x 2 – (m + n) x + mn = 0

∴ x 2 –

10

29 x + 1 = 0

∴ 10 x 2 – 29 x + 10 = 0



150

Example 3 : Form the quadratic equation whose roots are 3 + 2 5 and 3 – 2 5


∴ m = 3 + 2 5 a n d n = 3 – 2 5

Sum of the roots = m + n

= 3 + 2 5 + 3 – 2 5

∴ m + n = 6Product of the roots = mn

= (3 + 2 5 ) (3 – 2 5 )

= (3)2 –(2 5 )2

= 9 – 20

∴ mn = – 11

x 2 – ( m + n ) x + mn = 0

∴ x 2 – 6 x – 11 = 0

Example 4 : If ‘m’ and ‘n’ are the roots of equation x 2 – 3 x + 1 = 0 find the value

of (i) m2n + m n2 (ii)n

1

m

1+

Consider the equation x 2 – 3 x + 1 = 0


The coefficients are a = 1, b = –3, c = 1Let ‘m’ and ‘n’ are the roots


b− =

1

)3(−− = 3

∴ m + n = 3


c

mn =1

1∴ mn = 1

∴ (i) m2n+mn2 = mn (m + n)

= 1(3) = 3

(ii)m

1 +

n

1=

mn

mn + =

mn

nm + =

1

3

∴ m

1 + n

1 = 3



151

Example 5 : If ‘m’ and ‘n’ are the roots of equation x 2 – 3 x + 4 = 0 form the

equation whose roots are m2 and n2.

Consider the equation x 2 – 3 x + 4 = 0



i) Sum of the roots = m + n =a

b− =

1

)3(−−

∴ m + n = 3

ii) Product of the roots = mn =a

c =

1

4

∴ 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

x 2 – (m2 + n2) x + m2n2 = 0

∴ x 2 – (1) x + (16) = 0

∴ x 2 – x + 16 = 0

Example 6 : If one root of the equation x 2 – 6 x + q = 0 is twice the other, find the

value of ‘q’

Consider the equation x 2 – 6 x + q = 0


The coefficients are a = 1, b = –6, c = q

Let the ‘m’ and ‘n’ are the roots


b− =

1

)6(−−

∴ m + n = 6



152


c =

1

q

∴ 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 =3

6∴ m = 2

We know that q = mn

q = m(2m)

q = 2m2

q = 2(2)2

q = 8

∴ q = 8

Example 7 : Find the value of k so that the equation x 2 – 2 x + (k + 3) = 0 has one

root equal to zero.

Consider the equation x

2

– 2 x + (k + 3) = 0The coefficients are a = 1, b = –2, c = k + 3


Product of the roots = mn

∴ mn =a

c

mn = 1

3k +

∴ 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



153

Exercise : 5.8

A. Form the equation whose roots are

1) 3 and 5 2) 6 and –5 3) –2 and2

34)

3

2 and

2

3

5) 2 + 3 and 2 – 3 6) –3 + 2 5 and –3 – 2 5

B.

1) If ‘m’ and ‘n’ are the roots of the equation x 2 – 6 x + 2 = 0 find the value of

i) (m + n) mn ii)m

1 +

n

1

2) If ‘a’ and ‘b’ are the roots of the equation 3m2 = 6m + 5 find the value of

i)ab

ba + 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 areq

p and

p

q

5) Find the value of ‘k’ so that the equation x 2 + 4 x + (k + 2) = 0 has one root equal

to zero.

6) Find the value of ‘q’ so that the equation 2 x 2 – 3q x + 5q = 0 has one root which

is twice the other.

7) Find the value of ‘p’ so that the equation 4 x 2 – 8p x + 9 = 0 has roots whose

difference is 4.

8) If one root of the equation x 2 + p x + q = 0 is 3 times the other prove that 3p2 = 16q

Graphical method of solving a Quadratic Equation

Let us solve the equation x 2 – 4 = 0 graphically,

x 2 – 4 = 0

∴ x 2 = 4

let y = x 2 = 4

∴ y = x 2

and y = 4



154

y = x 2

x = 0 y = 02 y = 0

x = 1 y = 12 y = 1

x = 2 y = 22 y = 4

x = –1 y = (–1)2

y = 1x = –2 y = (–2)2 y = 4

x 0 1 –1 2 –2 3

y 0 2 2 8 8 6

( x, y) (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) ( 3 ,6)

Step 1: Form table of

corresponding values

of x and y

Satisfying the equation

y = x 2

Step 2: Choose the scale on x axis, 1 cm = 1 unit

y axis, 1 cm = 1 unit.

Step 3: Plot the points (0, 0);

(1, 1); (–1, 1); (2, 4)

and (–2, 4) on graph

sheet.

Step 4: Join the points by asmooth curve.

Step 5: Draw the straight line

y = 4 Parallel to x -axis

Step 6: From the intersecting

points of the curve and

the line y = 4, draw

perpendiculars to the x axis

Step 7: Roots of the equations are x = +2 or x = –2

The graph of a quadratic polynomial is a curve called ‘parabola’

Example 1 : Draw a graph of y = 2 x 2 and find the value of 3 , using the graph.

Step 1: Form the table of

corresponding values of x and y satisfying the

equation y = 2 x 2

Step 2: Choose the scale on x

axis, 1 cm = 1 unit and

y axis, 1 cm = 1 unit


(1, 2) (–1, 2); (2, 8) and

(–2, 8) on graph sheet.



155

x 0 1 –1 2 –2

y 0 1 1 4 4

( x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)

x 0 1 –1 2 –2

y 2 1 3 0 4

( x, y) (0, 2) (1, 1) (–1, 3) (2, 0) (–2, 4)

Step 4: Join the points by a

smooth curve

Step 5: Draw the straight line

y = 6 Parallel to x -axis.



the line y = 6, draw

perpendiculars to the

x -axis.

Step 7: Value of 3 = ± 1.7

x = –1.7 or x = + 1.7

Example 2 : Draw a graph of y = x 2 and y = 2- x and hence solve the equation

x 2 + x – 2 = 0


corresponding values of

x and y satisfying the

equation y = x 2


corresponding values of x and y satisfying the

equation y = 2 – x .


axis 1 cm = 1 unit and

y axis, 1 cm = 1 unit.


(1, 1); (–1, 1); (2, 4)and (–2, 4) on the graph

sheet.

Step 5: Join the points by a

smooth curve.

Step 6: Plot the points (0, 2) ;

(1, 1); (–1, 3); (2, 0)

and (–2, 4) on graph

sheet



156

x 0 1 –1 2 –2

y 2 1 1 4 4

( x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)

x

0 1 2 –1 –2y 2 3 4 1 0

( x, y) (0, 2) (1, 3) (2, 4) (–1, 1) (–2, 0)

Step 7: Join the points to get a line.


Curve and the line, draw

perpendiculars to the

x -axis

Step 9: Roots of the equation are ∴ x = 1 or x = –2

Example 3 : Solve the equation

Method I : x 2 – x – 2 = 0

Split the equation

y = x 2 and y = 2 + x


corresponding values x

and y satisfying the

equation y = x 2

Step 2: Form the table of corresponding values x and y satisfying theequation y = 2 + x

Step 3: Choose the scale on

x axis, 1 cm = 1 unity axis, 1 cm = 1 unit

Step 4: Plot the points (0, 0);(1, 1); (–1, 1); (2, 4)and (–2, 4) on the graphsheet.

Step 5: Join the points by asmooth curve

Step 6: Plot the points (0, 2);(1, 3) (2, 4); (–1, 1) and

(–2, 0) on the graphsheet.

Step 7: Join the points to get astraight line

Step 8: From the intersectingpoints of Curve and theline, draw the perpendi-

culars to the x -axis.

Step 9: Roots of the equation are x = –1 or x = 2



157

x 0 1 –1 2 –2

y –2 –2 0 0 4

( x, y) (0, –2) (1, –2) (–1, 0) (2, 0) (–2, 4)

Method II :


corresponding values of

x and y satisfying

equation y = x 2 – x – 2.


axis 1 cm = 1 unit and

y axis 1 cm = 1 unit.

Step 3: Plot the points (0, –2);

(1 –2); (–1, 0); (2, 0)

and (–2, 4) on the graph

sheet.

Step 4: Join the points to form

a smooth curve

Step 5: Mark the intersecting


the x – axis.

Step 6: Roots of the equations are x = –1 or x = 2

Exercise : 5.9

A. 1) Draw the graph of y = x 2 and find the value of 7

2) Draw the graph of y = 2 x 2 and find the value of 3

3) Draw the graph of y =2

1 x

2 and find the value of 10

B. 1) Draw the graph of y = x 2 and y = 2 x + 3 and hence solve the equation

x 2 – 2 x – 3 = 0

2) Draw the graph of y = 2 x 2 and y = 3 – x and hence solve the equation2 x 2 + x – 3 = 0

3) Draw the graph of y = 2 x 2 and y = 3 + x and hence solve the equation

2 x 2 – x – 3 = 0

C. Solve graphically

1) x 2 + x – 12 = 0 2) x

2 – 5 x + 6 = 0 3) x 2 + 2 x – 8 = 0

4) x 2 + x – 6 = 0 5) 2 x 2 – 3 x – 5 = 0 6) 2 x 2 + 3 x – 5 = 0

quadatrical equations

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