Quadrtic Equations Exercises

Exercise 1

1. Which of the following is the quadratic equation?

(i) 11 = - 4x2 – x3 (ii) 3

4 yy

(iii) 7

z 4z 54

(iv) 23 y 7 3 y

(v) 2

2

q 43

q

Ans:- (i) Given equation is 11 = -4x2 – x3

i.e. x3 + 4x2 + 11 = 0 . Highest index is 3

∴ This is not a quadratic equation.

(ii) Equation is

2

2

34 y

y

on multiplying both sides by y

3 4y y

i.e.y 4y 3 0

This is a quadratic equation.

{ It is of the type ay2 + by + c = 0, a ≠ 0 }

(iii) Given equation is 7

z 4z 5z

Multiply both sides by z to get

z2 – 7 = 4z2 + 5z

i.e. 3z2 + 5z + 7 = 0

This is of the type az2 + bz + c = 0 ( a ≠ 0 )

∴ it is a quadratic equation.

(iv) Given equation is

2

2

3y 7 3 y

i.e. 3y 3 y 7 0

This is of the type ay2 + by + c = 0 ( a ≠ 0 )

∴ it is a quadratic equation.

(v) Given equation is 2

2

q 43

q

i.e. q2 – 4 = - 3q2 i.e. 4q2 – 4 = 0

This is of the type aq2 + bq + c = 0

Where a = 4 ≠ 0, b = 0 , c = -4

∴ It is a quadratic equation.

2. Write the following quadratic equations in standard form

ax2 + bx + c = 0

(i) p (p-6) = 0 (ii) 7

n 4n

(iii) 5

2z z 6z

(iv) 2

2

m 53

m

Ans:- (i) Given equation is p (p-6) = 0

i.e. p2 – 6p = 0 i.e. p2 – 6p + 0 = 0

This is a quadratic equation in p in standard form

(ii) Given equation is 7

n 4n

i.e. n2 – 7 = 4n i.e. n2 – 4n – 7 = 0

This a quadratic equation in n in standard form.

(iii) 5

2z z 6z

Multiply both sides by 2

∴ 2z2 – 5 = z2 – 6z

i.e. z2 + 6z – 5 = 0

This is a quadratic equation in z in standard form.

(iv) Given equation is 2

2

m 53

m

i.e. m2 + 5 = -3m2

i.e. 3m2 + m2 + 5 = 0 i.e. 4m2 + 0m + 5 = 0

This is a quadratic equation in m in standard form.

Exercise 2

1. Explain whether 1

p 1, , 32

are the roots of quadratic equation

2p2 + 5p – 3 = 0.

Ans:- The values of p which satisfy the given quadratic equation

2p2 + 5p – 3 = 0 are the roots of quadratic equation.

i) P = 1, LHS = 2 (1)2 + 5 (1) – 3 = 4 ≠ 0

∴ LHS ≠ RHS

∴ P = 1 is not a root.

ii)

21 1 1

p , LHS 2 5 32 2 2

1 5 1 52 3 3 0

4 2 2 2

RHS

As equation is satisfied , 1

p2

is a root of equation

iii) P = -3, LHS = 2 (-3 )2 + 5 ( -3 ) -3

= 2 × 9 – 15 – 3 = 0 = RHS

As equation is satisfied , p = -3 is a root of equation.

2. If one root of the quadratic equation x2 - 7x + k = 0 is 4, then find the

value of k.

Ans:- As 4 is root of quadratic equation x2 - 7x + k = 0, it will satisfy the

equation .

∴ Put x = 4 in the equation to get

42 – 7 (4) + k = 0 i.e. 16 – 28 + k = 0

∴ k = 12

Ans = K = 12

3. State whether k is a root of the equation y2 – (k – 4) y – 4 k = 0

Ans:- Quadratic equation is y2 – (k – 4) y – 4 k = 0

Put x = k in above equation

LHS = k2 – (k – 4) k – 4k = k2 – k2 + 4k – 4k = 0

= RHS

As x = k satisfies the equation , it is a root of the equation.

Exercise 3

1. Solve the quadratic equation 20

x 12 0x

by factorization method

Ans: Quadratic equation is 20

x 12 0x

2. Solve the quadratic equation 10x2 + 3x – 4 = 0 by factorization method.

Ans:- Quadratic Equation is 10x2 + 3x – 4 = 0

Split middle term 3x as 8x – 5x

3. Solve the quadratic equation 2x 3 3 x 6 0 using factorization

method

Ans:- Quadratic equation is 2x 3 3 x 6 0

Exercise 4

1. Solve quadratic equation x ( x – 1 ) = 1 by completing square

Ans:- Quadratic equation is x ( x – 1 ) = 1

i.e. x2 – x = 1

2

2

2

2

1Third term coefficient of x

2

1 11

2 4

1add in g both sides

4

1 1 1 5x x 1 i.e. x

4 4 2 4

taking square roots

1 5 5x

2 4 2

5 1 51x

2 2 2

∴ roots are 1 5 1 5

and2 2

2. Solve the equation 6m2 + m = 2 by completing square

Ans:- Quadratic equation is 6m2 + m = 2

Divide both sides by 6

2 1 1m m

6 3

2

2

2

2

1Third term coefficient of m

2

1 1 1

2 6 144

1add to both sides

144

1 1 1 1m m

6 144 3 144

48 11 49i.e. m

12 144 144

Taking roots on both sides

1 7m

12 12

1 7 1 7m or m

12 12 12 12

1 7 1 7i.e. m or m

12 12 12 12

6 8i.e. m or m

12 12

1 2i.e. m or m

2 3

1 2Thus roots are and

2 3

Exercise 6

1. Find discriminant of the quadratic equation 23 x 2 2 x 2 3 0

Ans:- Quadratic equation is 23 x 2 2 x 2 3 0 ……… (1)

Compare it with ax2 + bx + c = 0 to get

a 3 , b 2 2 and c 2 3

Now discriminant = ∆ = b2 – 4ac

22 2 4 3 2 3

4 2 8 3 8 24 32

2. Determine the nature of the roots of the equation 3y2 +9y + 4 = 0 ,

using discriminant.

Ans:- Quadratic equation is 3y2 + 9y + 4 = 0

Compare it with ay2 + by + c = 0 to get, a = 3, b = 9, c = 4

Discriminant ∆ = b2 – 4ac = 92 – 4 (3) (4)

= 81 – 48 = 33

As ∆ > 0 , the roots of quadratic equation are real and unequal.

3. Determine the nature of the roots of quadratic equation 22x 5 3 x 16 0 using discriminant.

Ans:- Quadratic Equation is 22x 5 3 x 16 0

Compare it with ax2 + bx + c = 0 to get, a = 2, b = 5 3 , c = 16

The discriminant is ∆ = b2 – 4ac

2

5 3 4 2 16

25 3 128 75 128

53

As ∆ < 0, roots are not real.

4. Find the value of k, for which the equation

( k – 12 ) x2 + 2 ( k – 12 ) x + 2 = 0 has real and equal roots.

Ans:- Quadratic equation is ( k – 12 ) x2 + 2 ( k – 12 ) x + 2 = 0

Compare it with ax2 + bx + c = 0 to get,

a = k – 12, b = 2 ( k – 12 ) , c = 2

As roots of equation are real and equal, discriminant = 0

i.e. b2 – 4ac = 0

i.e. [ 2 ( k – 12 ) ]2 – 4 ( k – 12 ) . 2 = 0

4 ( k – 12 )2 – 8 ( k – 12 ) = 0

Divide by 4 ( k – 12 )

{ ( k – 12 ) is coefficient of x2 ≠ 0 }

In a quadratic equation,

∴ k – 12 – 2 = 0 i.e. k = 14

5. Find the value of k for which equation k 2 x 2 – 2 ( k – 1 ) x + 4 = 0 has

real and equal roots.

Ans:-

Quadratic equation is

k 2 x 2 – 2 ( k – 1 ) x + 4 = 0

Compare it with ax 2 + bx + c = 0 to get,

a = k 2, b = -2 ( k – 1 ) and c = 4

As roots are real and equal,

discriminant = 0

i e. b2 – 4ac = 0

[ -2 ( k – 1 ) ]2 – 4 k2 ( 4 ) = 0

i.e. 4 ( k – 1 )2 – 16 k2 = 0

divide by 4

∴ ( k – 1 )2 – 4k2 = 0

i.e. k2 – 2k + 1 – 4k2 = 0

i.e. – 3k2 - 2k + 1 = 0

multiply by (-1)

∴ 3k2 + 2k – 1 = 0

i.e.3k2 + 3k – k – 1 = 0

3 k ( k + 1 ) – ( k + 1 ) = 0

i.e. ( k + 1 ) ( 3k – 1 ) = 0

i.e. k + 1 = 0 or 3k – 1 = 0

i.e. k = -1 or 1

k3

Exercise 5

1. Solve the equation 2 x 1x 0

3, using formula method.

Ans:- Quadratic equation is 2 x 1x 0

3

i.e. 3x2 + x – 1 = 0

Comparing it with ax2 + bx + c = 0

To get, a = 3, b = 1, c = -1

Now b2 – 4ac = 12 – 4 ( 3 ) ( -1) = 1 + 12 = 13

The roots are 2b b 4ac

x2a

1 13

6

∴ The roots of the given equation are 1 13 1 13

and6 6

2. Solve the quadratic equation 3q2 = 2q + 8 using formula method.

Ans:- Quadratic equation is 3q2 = 2q + 8

i.e. 3q2 – 2q – 8 = 0

Compare it with aq2 + bq + c = 0

To get, a=3, b= -2, c = -8

Now, b2 = 4ac = (-2)2 – 4 (3) (-8) = 4 + 96 = 100

The roots are

2b b 4acq

2a

2 100 2 10

2 3 6

∴ roots of quadratic equations are

2 10 2 10 12 8and i.e. and

6 6 6 6

4i.e. 2 and

3

Exercise 7

1. If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the

other. Find k

Ans:- Given Q.E. is kx2 – 5x+ 2 = 0

Let α, β be the roots of given Q. E.

We are given α = 4 β

Now,

2

2

2

2

2

2

5coefficient of x 5

k kcoefficient of x

5 5i.e. 4 i.e. 5

k k

1i.e. ....... 1

k

cons tant kAlso product of roots .

coefficient of x

2 2 1i.e. 4 . i.e.

k 4k 2k

1Put from 1

k

1 1 1 1i.e.

k 2k 2kk

k 2k

divide by k 0

k 2

2. Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are in the

ratio 2 : 5

Ans:- Quadratic equation is x2 + kx + 40 = 0

The roots are in the ration 2 : 5

Let roots be

α = 2m, β = 5 m

2

2

2

2

2 2

2

coefficientof xsum of roots

coefficient of x

k2m 5m k

1

k7m k m

7

cons tantAlso, product of roots .

coefficientof x

402m 5 m

1

10m 40

m 4

kPut m

7

k k4 i.e. 4

7 49

k 49 4 196

k 14

3. Find k, if one of the roots of quadratic equation kx2 – 7x + 12 = 0 is 3

Ans:- The given quadratic equation is kx2 – 7x + 12 = 0

As 3 is a root of above equation,

It will satisfy the equation.

∴ put x = 3 in above equation to get

k (3)2 – 7 (3) + 12 = 0

i.e. 9k – 21 + 12 = 0 i.e. 9k = 9

k = 1

4. If the roots of the equation x2 + px + q = 0 differ by 1, prove that

p2 = 1 + 4q

Ans:- Quadratic equation is x2 + px + q = 0

Roots differ by 1

Let α = β + 1

2

2

coefficient of xnow,

coefficient of x

pi.e. 1 i.e. 2 1 p

1

1 p2 1 p

2

Cons tantAlso,

Coefficient of x

qi.e. 1 .

1

1 pput

2

1 p 1 p1 . q

2 2

1 p 2 1 pq

2 2

1 p 1 pq

2 2

1 2

2

2

pq

4

1 p 4q

i.e. p 1 4q

5 Find k, if the sum of the roots of the quadratic equation

4x2 + 8kx + k + 9 = 0 is equal to their product.

Ans:- Quadratic Equation is

4x2 + 8kx + k + 9 = 0

We are given that

Sum of roots = product of roots

2 2

Coefficient of x Cons tan ti.e.

Coefficient of x Coefficient of x

i.e. – (coefficient of x) = constant

i.e. – 8 k = k + 9

i.e. -8 K – k = 9

i.e. –9k =9

i.e. k = -1

6. If α and β are the roots of the equation x2 – 5x + 6 = 0, find

(i) α2 + β2 (ii)

Ans:- Quadratic equation is x2 – 5x + 6 = 0

As α and β are the roots of above equation,

5 6

5 and . 61 1

(i) α2 + β2 = ( α + β )2 - 2α β = 52 – 2 × 6 = 25 – 12 = 13

(ii) 2 2

13from i

6

7. If one root of the quadratic equation kx2 – 20x + 34 = 0 is 5 2 2 find k

Ans:- Quadratic Equation is kx2 – 20 x + 34 = 0

As α = 5 2 2 is a root of above equation,

β = 5 2 2 must be the other root

2

coeffic i ent of xNow, sum of roots

coefficient of x

205 2 2 5 2 2

k

20 20i.e. 10 k 2

k 10

Exercise - 8

1. Form a quadratic equation with roots – 2 and 11

2.

Ans:- The given roots are α = -2 and β = 11

2

11 4 11 72 and

2 2 2

112 11

2

Quadratic Equation is

x2 – (sum of roots) x + product of roots = 0

i.e. x2 – (α + β) x + α . β = 0

on putting the values

2 7x x 11 0

2

Multiplying both sides by 2

∴ 2x2 – 7x – 22 = 0

This is required Q.E.

2. Form a quadratic equation if its one of the roots is 5 3

Ans:- If one of the root of a quadratic equation is 5 3 , then the other root

is 5 3

2 2

Let 5 3 , 5 3

5 3 5 3 2 5

. 5 3 . 5 3 5 3 5 3 2

∴ Required quadratic equation is

x2 – ( α + β ) x + α . β = 0 2i.e. x 2 5 x 2 0

3. If the sum of the roots of the quadratic equation is 3 and sum of their

cubes is 63. Find the quadratic equation.

Ans:- Suppose α, β are the roots of a quadratic equation. Then we are given

α + β = 3 and α 3 + β3 = 63

We have

(α + β) 3 = α 3 + 3 α 2β + 3 αβ2 + β3

i.e. (α + β) 3 = α 3 + 3 α β (α + β) + β3

(α + β) 3 = α 3 + β3 + 3 α β . ( α + β )

∴ 33 = 63 + 3 α β . 3

27 = 63 + 9 α β

∴ 9 α β = 27 – 63 = - 36

α β = -4

now, quadratic equation is

x2 – ( sum of roots) x + (product of roots) = 0

i.e. x2 – ( α + β ) x + α . β = 0

Put α + β = 3 and α β = - 4

∴ x2 – 3x – 4 = 0

This is the required Q.E.

4. If the difference of the roots of the quadratic equation is 5 and the

difference of the cubes is 215 , find the quadratic equation

Ans:- Suppose α, β are the roots of a quadratic equation.

We are given α – β = 5 and α3 – β3 = 215

Now, (α – β)3 = α3 - 3α2β + 3 α β2 – β3 = α3 – 3 α β ( α – β) – β3

On rearranging terms

3 α β . ( α – β) = (α3 – β3) – (α – β)3

Put α – β = 5 and α3 – β3 = 215

∴ 3 α β × 5 = 215 - 53 = 215 – 125 = 90

15 α β = 90

α β = 6

Now, ( α + β)2 = ( α – β)2 + 4 α β = 52 + 4 (6) = 25 + 24 = 49

∴ α + β = ± 7

i.e. α + β = 7 or α + β = - 7

now, Quadratic equation is

x2 – ( α + β) x + α β = 0

# put α + β= 7 and α β = 6

∴ Q.E. is x2 – 7x + 6 = 0

# put α + β = -7 and α β = 6

∴ Q.E. is x2 – (-7) x + 6 = 0

i.e. x2 + 7x + 6 = 0

Ans: There are two such quadratic equations . They are

x2 – 7x+ 6 = 0 and x2 + 7x+ 6 = 0

Exercise 9

1. Solve: (x2 + 2x) (x2 + 2x – 11) + 24 = 0

Ans:- Equation is (x2 + 2x) (x2 + 2x – 11) + 24 = 0

Take x2 + 2x = m

∴ equation becomes

m (m – 11) + 24 = 0

m2 – 11m + 24 = 0

m – 8m – 3m + 24 = 0

m (m – 8) – 3 (m – 8) = 0

(m – 8) (m – 3) = 0

∴ m – 8 = 0 or m – 3 = 0

i.e. m = 8 or m = 3

But m = x2 + 2x

# m = 8 gives x2 + 2x = 8 i.e. x2 + 2x – 8 = 0

( x + 4 ) ( x – 2 ) = 0

i.e. x + 4 = 0 or x – 2 = 0

x = - 4 or x = 2

# m = 3 gives

x2 + 2x = 3 i.e. x2 + 2x – 3 = 0

i.e. ( x + 3 ) ( x – 1 ) = 0

i.e. x + 3 = 0 or x – 1 = 0

i.e. x = -3 or x = 1

Ans: x = -4, x = 2, x = -3 or x = 1 or S.S = {-4, -3, 1, 2}

2. Solve: 2

2

1 12 x 9 x 14 0

xx

Ans:- Equation is 2

2

1 12 x 9 x 14 0

xx

We know the identity

2

2

2

1 1x x 2

xx

Note:

2

2 2

2 2

1 1 1 1x x 2x. x 2

x x x x

2

2

2

1 1x x 2

xx

∴ given equation becomes

2

2

1 12 x 2 9 x 14 0

x x

1Take x m

x

2 m 2 9m 14 0

i.e. 2m2 – 4 – 9m + 14 = 0

2m2 – 9m + 10 = 0

2m2 – 4m – 5m + 10 = 0

2m (m – 2) – 5 (m – 2) = 0

(m – 2) . (2m – 5) = 0

i.e. m – 2 = 0 or 2m – 5 = 0

i.e. m = 2 or 5

m2

As 1

m xx

# m = 2 , gives 1

x 2x

i.e. 2x 1

2x

i.e. x2 + 1 = 2x

i.e. x2 – 2x + 1 = 0 i.e. ( x – 1)2 = 0

i.e. x – 1 = 0 i.e. x = 1

#

22

5 1 5m gives x

2 x 2

x 1 5i.e. i.e. 2 x 1 5x

x 2

i.e. 2x2 + 2 – 5x = 0

i.e. 2x2 – 5x + 2 = 0

i.e. 2x2 – x – 4x + 2 = 0

i.e. x (2x – 1) – 2 (2x – 1) = 0

i.e. (2x – 1) = 0 or (x – 2) = 0

i.e. 2x – 1 = 0 or x – 2 = 0

1i.e. x or x 2

2

Ans: 1

x 1, x , x 22

Or solution set = 1

{1, , 2}2

3. Solve : 2

2

1235y 44

y

Ans:- Equation is 2

2

1235y 44

y

Take y2 = m ∴ equation becomes

12

35m 44m

On multiplying both sides by m, we get

35m2 + 12 = 44m i.e. 35m2 – 44m + 12 = 0

i.e. 35m2 – 14m – 30m + 12 = 0

i.e. 7m ( 5m – 2 ) – 6 ( 5m – 2 ) = 0

i.e. (5m – 2) (7m – 6) = 0

i.e. 5m – 2 = 0 or 7m – 6 = 0

i.e. 2 6

m or m5 7

As m = y2

# 22 2 2m gives y y

5 5 5

# 26 6 6m gives y y

7 7 7

Ans: 2 6

y ,5 7

i.e. solution set = 2 6

,5 7

4. Solve : 3x4 – 13x2 + 10 = 0

Ans:- Equation is 3x4 – 13x2 + 10 = 0

Now x4 = x2 × x2 = ( x2) 2

∴ equation becomes

3 (x2)2 – 13x2 + 10 = 0

Take x2 = m

∴ 3m2 – 13 m + 10 = 0

i.e. 3m2 – 3m – 10m + 10 = 0

i.e. 3m ( m – 1 ) – 10 ( m – 1 ) = 0

i.e. ( m – 1 ) ( 3m -10 ) = 0

i.e. m – 1 = 0 or 3m – 10 = 0

i.e. m = 1 or m = 10

3

As m = x2

# m = 1 gives x2 = 1 ∴ x = ± 1

# 210 10 10m gives x x

3 3 3

Ans: 10

x 1,3

5. Solve : 2

2

152y 12

y

Ans:- Equation is 2

2

152y 12

y

Take y2 = m

∴ equation becomes 15

2m 12m

multiplying both sides by m

2m2 + 15 = 12m

i.e. 2m2 – 12m + 15 = 0

This can not be solved by factorization method

Compare it with am2 + bm + c = 0 , to get

a = 2, b = -12, c = 15

Now, b2 – 4ac = (-12)2 – 4 (2) (15)

= 144 – 120 = 24

Roots are 2b b 4ac

m2a

12 24 12 2 6

i.e. m2 2 4

2

6 6m

2

6 6 6 6m or m

2 2

As m y

#

26 6 6 6m gives y

2 2

6 6y

2

#

26 6 6 6m gives y

2 2

6 6y

2

Ans: 6 6 6 6

y ,2 2

Exercise 10

1. Tinu is younger than Pinky by three years. The product of their ages is

180. Find their ages.

Ans:- Suppose Tinu’s age is x yrs.

∴ Pinky’s age = (x + 3) yrs.

The product of their ages is given to be 180 yrs.

∴ x. (x + 3) = 180

i.e. x2 + 3x – 180 = 0

∴ (x + 15) ( x – 12) = 0

∴ x = -15 or x = 12

As age can not be negative, x ≠ -15

∴ x = 12

Tinu’s age = 12 yrs.

Pinky’s age = 12 + 3 = 15 yrs.

2. A natural number is greater than twice its square root by 3. Find the

number.

Ans:- Let n be the natural number. Its square root is n . It is given that the

natural number is greater than twice its square root by 3.

i.e. n = 2 n + 3

i.e. n – 3 = 2 n (note this step)

On squaring both sides

(n – 3)2 = 4n

i.e. n2 – 6n + 9 = 4n

n2 – 6n – 4n + 9 = 0 i.e. n2 – 10n + 9 = 0

(n- 1) (n – 9) = 0

i.e. n – 1 = 0 or n – 9 = 0

i.e. n = 1 or n = 9

But n = 1 does not satisfy the condition. Hence it is rejected

∴ n = 9

Required natural number is 9.

Note: If we consider 1 1 , then only condition is satisfied.

Here, 1 = 2 (-1) + 3

3. A natural number is greater than the other by 5. The sum of their squares

is 73. Find those numbers.

Ans:- Let the natural numbers be x and x + 5.

According to given condition,

x2 + (x+5)2 = 73

i.e. x2 + x2 + 10x + 25 = 73

2x2 + 10x + 25 – 73 = 0

2x2 + 10x – 48 = 0

Dividing both sides by 2

x2 + 5x -24 = 0

i.e. (x +8) (x -3) = 0

x+ 8 = 0 or x – 3 = 0

i.e. x = -8 or x = 3

As x is a natural number, x ≠ -8

∴ x = 3

Two natural numbers are 3 and 3 +5 = 8

4. The sum of the first n natural numbers is given by n n 1

S2

Find n if the sum (S) is 276.

Ans:- We are given n (n 1)

S 2762

∴ n (n+1) = 276 × 2 = 552

i.e. n2 + n = 552

i.e. n2 + n -552 = 0

i.e. n2 +24n -23n -552 = 0

i.e. n(n + 24)-23(n + 24) = 0

i.e. (n + 24) (n -23) = 0

∴ n + 24 = 0 or n -23 = 0

i.e. n = -24 or n = 23

As n is a natural number , n ≠ -24

∴ n = 23

Note: We have n n 1

2762

i.e. n(n+1) = 2 × 276 = 552

LHS is a product of two consecutive natural numbers. We express RHS

also as product of two consecutive natural numbers.

Now, 552 = 23 × 24

∴ n (n+1) = 23 × 24

∴ n = 23 (i.e. n +1 = 24)

5. A rectangular playground is 420 sq. m .If its length is increased by 7 m

and breadth is decreased by 5 m, the area remains the same .Find the

length and breadth of the playground .

Ans:- Let x m be the length of playground whose area is 420 sq. m. 420

breadth mx

Now length is increased by 7m and breadth is decreased by 5m.

∴ New length = (x + 7) m

420New breadth 5 m

x

It is given that area remains the same.

420x 7 5 420

x

x 7 420 5xi.e. 420

x

(x + 7 ) (420 – 5x) = 420 x

420x – 5x2 + 2940 – 35x – 420x = 0

i.e. -5x2 -35x + 2940 = 0

divide both sides by -5

∴ x2 + 7x – 588 = 0

i.e. x2 + 28x -21x -588 = 0

i.e. x(x +28) -21 (x + 28) = 0

i.e. (x + 28) (x – 21) = 0

i.e. x = -28 or x = 21

As length cannot be negative, x ≠ -28

∴ x = 21

Length = 21m 420

Breadth 20m21

6. If the cost of bananas is increased by Re 1 per dozen , one can get 2

dozen less for Rs. 840. Find the original cost of one dozen of banana.

Ans:- Suppose cost of one dozen banana = Rs. x

∴ Number of dozens for Rs. 840

840x

New cost of banana after increase by 1 Re is Rs. (x + 1) per dozen

∴ Number of dozens for Rs. 840

840x 1

As per given condition 840 840

2x x 1

840 x 1 840x2

x x 1

On cross multiplying

i.e. 2x.(x+1) = 840x + 840 – 840x

i.e. 2x2 + 2x = 840

i.e. 2x2 + 2x -840 = 0

divide by 2

i.e. x2 + x -420 =0

i.e. x2 + 21x -20x -420 = 0

i.e. x(x + 21) – 20 (x + 21) = 0

i.e. (x +21) (x -20)=0

i.e. x + 21 = 0 or x – 20 = 0

i.e. x = -21 or x = 20

But number of dozens cannot be negative

∴ x ≠ -21 ∴ x = 20

Cost of one dozen of banana = Rs. 20 /-

Miscellaneous Sums

1. Form an equation for the following

The ten’s place digit in a two digit number is greater than the square of

digit at unit’s place (x) by 5 and the number formed is 61.

Ans:-

As unit’s place digit is x, as per given condition ten’s place digit = x 2 + 5

As weightage of that place is 10, the number formed is 10 ( x 2 + 5 ) + x

But it is given to be 61

∴ equation is 10 ( x 2 + 5 ) + x = 61

2. Find m if roots of a quadratic equation ( m + 1 ) x 2 – 2 ( m – 1 ) x + 1 =

0 are real and equal

Ans:-


( m + 1 ) x 2 – 2 ( m – 1 ) x + 1 = 0

Comparing with the standard equation

ax 2 + bx + c = 0 , we get

a = m + 1, b = – 2 ( m – 1 ), c = 1

we have

∆ = b 2 – 4ac

= [ - 2 ( m – 1 ) ] 2 – 4 ( m + 1 ) (1)

= 4 ( m 2 – 2m + 1 ) – 4m - 4

= 4m 2 – 8m + 4 – 4m – 4

∆ = 4m 2 – 12m ---------- (1)

We are given that the given equation has real and equal roots

∴ ∆ = 0

∴ 4m 2 – 12m = 0

∴ 4m ( m – 3 ) = 0

∴ 4m = 0 or m – 3 = 0

∴ m = 0 or m = 3

3. If α and β are the roots of the equation ax 2 + bx + c = 0 find the value

of

Ans:-


ax 2 + bx + c = 0

As α and β are the roots of the equation

2 2

2 2 2

2

2 2

2

2 2

2

2

b cand

a a

and

We have, ( 2

( 2

b 2cb c2

aaa a

cc

aa

b 2ac a b 2ac

c aca

b 2ac

ac

4. If one root of the quadratic equation x 2 + 6x + k = 0 is h 2 6 , find k.

Ans:-


x 2 + 6x + k = 0

Given that h 2 6 is one root

∴ the second root is h 2 6

∴ α = h 2 6 and β = h 2 6

∴ α + β = h 2 6 + h 2 6 = 2h ----------- (1)

Comparing with ax 2 + bx + c = 0, we get

a = 1, b = 6, c = k

b 6

6 (2)a 1

∴ from (1) and (2)

2h = - 6 ∴ h = - 3

2

2

cAgain

a

kh 2 6 h 2 6

1

h 24 k

( 3) 24 k

k 9 24 15

∴ h = – 3 and k = – 15

5. Solve the quadratic equation 2

1 1

x 2 x by factorization method.

Ans:- Quadratic equation is

2

1 1

x 2 x

On cross multiplying we get

x 2 = x + 2 i.e. x 2 – x – 2 = 0

i.e. x 2 – 2x + x – 2 = 0

i.e. x ( x – 2 ) + 1 ( x – 2 ) = 0

i.e. ( x – 2 ) ( x + 1 ) = 0

i.e. x = 2 or x = – 1

Thus, roots of the quadratic equation are 2 and – 1

6. Solve the quadratic equation ( 2y + 3 ) 2 = 81 by factorization method


( 2y + 3 ) 2 = 81

i.e. ( 2y + 3 ) 2 – 81 = 0

i.e. ( 2y + 3 ) 2 – 9 2 = 0

i.e. ( 2y + 3 – 9 ) ( 2y + 3 + 9 ) = 0

i.e. ( 2y – 6 ) ( 2y + 12 ) = 0

i.e. 2y – 6 = 0 or 2y + 12 = 0

i.e. y = 3 or y = – 6

Roots are 3 and – 6

7. Form the quadratic equation whose roots are – 3 and 5

2

Ans:- Let α = – 3 and β = 5

2 be the given roots of a quadratic equation in x.

15

2 2

155and 3

2 2

Now, quadratic equation is

x 2 – ( sum of roots ) x + ( product of roots ) = 0

2 1 15i.e. x x 0

2 2

Multiplying both sides by 2

∴ 2x 2 + x – 15 = 0 is required quadratic equation

8. Form a quadratic equation if one of the roots is 3 2 3

Ans:- Let α = 3 2 3 be the given root. Then other root must be 3 2 3

∴ α + β = 3 2 3 + 3 2 3 = 6

αβ = (3 2 3 ) (3 2 3 ) = 3 2 – (2 3 ) 2 = 9 – 12 = – 3

∴ quadratic equation is


i.e. x 2 – 6x – 3 = 0

9. If α, β are the roots of the equation 4x 2 – 5x + 2 = 0, find the equation

whose roots are

i) α + 3β and 3α + β

ii) and

iii) 2 2

and

iv) 1 1

and

Ans:-

As roots of 4x 2 – 5x + 2 = 0 are α and β

( 5) 5

4 4

2 1

4 2

i) Required quadratic equation has roots α + 3β and β + 3α

sum of roots = α + 3β + β + 3α = 4α + 4β = 4 ( α + β )

= 4 × 5

4 = 5

2 2

2 2

2

2

product of roots ( ) 3 9

3[ ( 2 ( 6

5 1( 4 3 4

4 2

75 3225 1073 2

16 16 16

Now, required quadratic equation is


on putting values, we get

2 2107x 5x 0 i.e. 16x 80x 107 0

16

ii) Required quadratic equation has roots and

2 2 2

2

( ) 2sum of roots

5 1 252 19 94 2 16 2

1 1 16 8

2 2

product of roots 1

Required quadratic equation is


2

2

9i.e. x x 1 0

8

i.e. 8x 9x 8 0

iii) Required quadratic equation has roots 2 2

and

3 3 32 2

3 3 3

3 3 3

3

( 3sum of roots

( 3

( 3

5 1 5 125 120125 1535 54 2 4 64 8 64sum of roots 2

1 1 1 64 32

2 2 2

Also,

2 2 1

product of roots2



2

2

5 1i.e. x x 0

32 2

i.e. 32x 5x 16 0

iv) Roots of required quadratic equation are 1 1

and

1 1 1 1sum of roots (

(

5 / 45 5 5 5 10 152

4 1/ 2 4 4 4 4 4

2 2 2

2

1 1 1product of roots

( ) 21 1

2511(5 / 4) 2

1 1 1 1 9162 2 2 22 1/ 2 1/ 2 2 1/ 2 2 16

1 9 292

2 8 8



2

2

15 29i.e. x x 0

4 8

i.e. 8x 30x 29 0

10. If the difference between the roots of the quadratic equation is 3 and

difference of their cubes is 189, find the quadratic equation

Ans:- Let α and β be the roots of the equation

∴ α – β = 3 and α 3 – β 3 = 189

We have

α 3 – β 3 = ( α – β ) 3 + 3 α β ( α – β )

∴ 189 = (3) 3 +3 α β ( 3)

∴ 189 = 27 + 9 α β

∴ 9 α β = 189 – 27

∴ α β = 162

9

∴ α β = 18

Now,

(α + β) 2 = (α - β) 2 + 4 α β

∴ (α + β) 2 = 3 2 + 4 × 18

∴ (α + β) 2 = 9 + 72 = 81

∴ α + β = ± 9

∴ The required quadratic equation is

x 2 – ( α + β ) x + α β = 0

∴ x 2 – ( ± 9x ) + 18 = 0

∴ x 2 – 9x + 18 = 0 and x 2 + 9x + 18 = 0

are the equations.

11. If α + β = 5 and α 3 + β 3 = 35, find the quadratic equation whose roots

are α and β.

Ans:- We are given

α + β = 5 and α 3 + β 3 = 35

α 3 + β 3 = ( α + β ) 3 - 3 α β ( α + β )

∴ 35 = 5 3 – 3 α β (5)

∴ 35 = 125 - 15 α β

∴ 15αβ = 90

∴ α β = 6

As α, β are roots of quadratic equation, the required quadratic equation

is

x 2 – (α + β ) x + α β = 0

i.e. x 2 – 5x + 6 = 0

12. If the difference of the roots of the quadratic equation is 4 and difference

of their cubes is 208, find the quadratic equation.

Ans:- Let α and β be the roots of the equation

∴ α – β = 4 and α 3 – β 3 = 208

We have

α 3 – β 3 = ( α – β ) 3 + 3 α β ( α – β )

∴ 208 = (4) 3 +3 α β (4 )

∴ 208 = 64 + 12 α β

∴ 12 α β = 208 – 64 = 144

∴ α β = 12

Now,

(α + β) 2 = (α - β) 2 + 4 α β

∴ (α + β) 2 = 4 2 + 4 × 12

∴ (α + β) 2 = 16 + 48 = 64

∴ α + β = ± 8

∴ The required quadratic equation is

x 2 – ( α + β ) x + α β = 0

∴ x 2 – ( ± 8x ) + 12 = 0

∴ x 2 – 8x + 12 = 0 and x 2 + 8x + 12= 0

are the required equations.

13. If one root of the quadratic equation ax 2 + bx + c = 0 is the square of

the other, show that b3 + a 2 c +ac 2 = 3 abc

Ans:- Quadratic equation is ax 2 + bx + c = 0

Let first root be α. Then second root must be its square i.e. α 2

2

2

coeff of xsum of roots

coeff of x

b(1)

a

2

2 3

constantproduct of roots

coeff of x

c ci.e. (2)

a a

To get the required result, we have to eliminate α from (1) and (2)

On cubing both sides of (1), we get

33

2

33 2 2 2 3

3

3 3 3

33 3 2 6

3

6 3 2

33 3 2 3 2

3

b

a

bi.e. 3 ( ) ( )

a

using a b a 3ab (a b ) b

bi.e. 3 ( )

a

now , ( )

b3 ( ) ( )

a

Put values from (1) and (2) to get

2 3

3

2 3

2 2 3

b bc c c3

a a a a a

c bc 3bci.e.

a a a a

multiply both sides by a 3 to get

a 2 c – 3abc + ac 2 = - b 3

on rearranging terms we get

b3 + a 2 c +ac 2 = 3 abc

14. If the sum of the roots of the quadratic equation 1 1 1

x p x q r is zero.

Show that the product of the roots is 2 2( p q )

2


2

2

1 1 1

x p x q r

x q x p 1i.e.

x q x p r

i.e. ( 2x p q ) r x q x p

i.e. 2x r pr qr x px qx pq

i.e. x ( p q 2r ) x pq pr qr 0

If α and β are the roots of the above equation then

(p q 2r )

(p q 2r )1

We are given

α + β = 0

∴ - ( p + q – 2r ) = 0

∴ p + q = 2r

p q

r (1)2

Now,

2 2 2

2 2

pq pr qrproduct of roots

1

pq r ( p q )

p qpq (p q) ......... from (1)

2

(p q) 2pq (p q 2pq )pq

2 2

(p q )

2

15. Solve the equation : x 4 – 29x 2 + 100 = 0

Ans:- equation is

x 4 – 29x 2 + 100 = 0

i.e. ( x 2 ) 2 – 29x 2 + 100 = 0

put x 2 = m

Thus the equation becomes

m 2 – 29 m + 100 = 0

i.e. m 2 – 25m – 4m + 100 = 0

i.e. m ( m – 25 ) – 4 ( m – 25 ) = 0

i.e. ( m – 4 ) ( m – 25 ) = 0

i.e. m = 4 or m = 25.

i.e. x 2 = 4 or x 2 = 25

∴ x = ± 2 or x = ± 5

Thus the roots are 2, - 2, 5, - 5

16. Solve : 7y 4 – 25y 2 + 12 = 0

Ans:- equation is

7y 4 – 25y 2 + 12 = 0

i.e. 7 ( y 2 ) 2 – 25y 2 + 12 = 0

put y 2 = m

i.e. 7m 2 – 25m + 12 = 0

7m 2 - 21m – 4m + 12 = 0

7m ( m – 3 ) – 4 ( m – 3 ) = 0

( 7m – 4 ) ( m – 3 ) = 0

Hence, 7m – 4 = 0 or m – 3 = 0

Thus, m = 4 / 7 or m = 3

2 2 4i.e. y 3 or y

7

2i.e. y 3 or y

7

22The roots are 3, 3, ,

7 7

17. Solve : 2

2

26m 7

m

Ans:- equation is

2

2

4 2

4 2

26m 7

m

i.e. 6m 2 7 m

i.e. 6m 7 m 2 0

i.e. 6 (m 2 ) 2 – 7m 2 + 2 = 0

put m 2 = x

thus the equation becomes

6x 2 – 7x + 2 = 0

6x 2 – 3x – 4x + 2 = 0

i.e. 3x ( 2x – 1 ) – 2 ( 2x – 1 ) = 0

i.e. ( 3x – 2 ) ( 2x – 1 ) = 0

∴ 3x – 2 = 0 or 2x – 1 = 0

i.e. x = 2 / 3 or x = 1 / 2

2 22 1m or m

3 2

2 1m or m

3 2

2 12 1Hence , , , are the roots.

3 3 2 2

18. Solve the equation : ( p 2 + p ) ( p 2 + p – 3 ) = 28

Ans:- equation is

( p 2 + p ) ( p 2 + p – 3 ) = 28

Put p 2 + p = m

Hence, the equation becomes

m ( m – 3 ) = 28

i.e. m 2 – 3m – 28 = 0

i.e. m 2 – 7m + 4m – 28 = 0

i.e. m ( m – 7 ) + 4 ( m – 7 ) = 0

i.e. ( m + 4 ) ( m – 7 ) = 0

i.e. m = - 4 or m = 7

∴ p 2 + p = - 4 or p 2 + p = 7

i.e. p 2 + p + 4 = 0 or p 2 + p – 7 = 0

consider p 2 + p + 4 = 0

comparing with ap 2 + bp + c = 0

a = 1, b = 1, c = 4

2b b 4acp

2a

1 1 4(1) (4) 1 15

2(1) 2

The roots are not real numbers.

consider p 2 + p - 7 = 0

comparing with ap 2 + bp + c = 0

a = 1, b = 1, c = - 7

2b b 4acp

2a

1 1 4(1) ( 7) 1 29

2(1) 2

Thus, the roots are 1 29

2 and

1 29

2

19. Solve the equation : 2

2

1 13 x 4 x 6 0

xx

Ans:- equation is

2

2

1 13 x 4 x 6 0

xx

2

2 2

2

1put x m

x

now

1 1x x 2 m 2

xx

Hence, the given equation becomes

3 ( m 2 + 2 ) – 4m – 6 = 0

i.e. 3m 2 + 6 – 4m – 6 = 0

i.e. 3m 2 – 4m = 0

i.e. m ( 3m – 4 ) = 0

i.e. m = 0 or 3m – 4 = 0

∴ m = 0 or m = 4 / 3

Hence,

2

2

2 2

1 1 4x 0 or x

x x 3

x 1 4i.e. x 1 0 or

x 3

i.e. x 1 or 3( x 1) 4x

∴ x = ± 1 or 3x 2 – 4x – 3 = 0

Consider 3x 2 – 4x – 3 = 0

comparing with ax 2 + bx + c = 0

a = 3, b = - 4, c = - 3

2

2

b b 4acx

2a

( 4) ( 4) 4(3) ( 3) 4 16 36

2(3) 6

4 52 4 2 13 2 (2 13 ) 2 13

6 6 6 3

Hence, the roots are 1, - 1, 2 13

3 ,

2 13

3


2

1 19 x 3 x 20 0

xx

Ans:- equation is

2

2

1 19 x 3 x 20 0

xx

2

2 2

2

1put x m

x

now

1 1x x 2 m 2

xx


9 ( m 2 + 2 ) – 3m – 20 = 0

i.e. 9m 2 + 18 – 3m – 20 = 0

i.e. 9m 2 – 3m – 2 = 0

i.e. 9m 2 – 6m + 3m – 2 = 0

i.e. 3m ( 3m – 2 ) +1 ( 3m – 2 ) = 0

i.e. ( 3m + 1 ) ( 3m – 2 ) = 0

i.e. ( 3m + 1 ) = 0 or ( 3m – 2 ) = 0

∴ m = - 1 / 3 or m = 2 / 3

2 2

2 2

2 2

11 1 2x or x

x 3 x 3

x 1 1 x 1 2i.e. or

x 3 x 3

i.e. 3( x 1) x or 3( x 1) 2x

i.e. 3x x 3 0 or 3x 2x 3 0

Consider 3x 2 + x – 3 = 0


a = 3, b = 1, c = - 3

2

2

b b 4acx

2a

1 1 4(3) ( 3) 1 1 36 1 37

2(3) 6 6

Consider 3x 2 - 2x – 3 = 0


a = 3, b = - 2, c = - 3

2

2

b b 4acx

2a

( 2) ( 2) 4(3) ( 3) 2 4 36 2 40

2(3) 6 6

2 2 10 2 ( 1 10 ) 1 10

6 6 3

Hence, the roots are 1 37

6 ,

1 10

3


2

1 130 x 77 x 12 0

xx

Ans:- equation is

2

2

1 130 x 77 x 12 0

xx

2

2 2

2

1put x m

x

now

1 1x x 2 m 2

xx


30 ( m 2 + 2 ) – 77m – 12 = 0

i.e. 30m 2 + 60 – 77m – 12 = 0

i.e. 30m 2 – 77m + 48 = 0

i.e. 30m 2 – 45m – 32m + 48 = 0

i.e. 15m ( 2m – 3 ) – 16 ( 2m – 3 ) = 0

i.e. ( 15m - 16 ) ( 2m – 3 ) = 0

i.e. ( 15m - 16 ) = 0 or ( 2m – 3 ) = 0

∴ m = 16 / 15 or m = 3 / 2

2 2

2 2

2 2

1 16 1 3x or x

x 15 x 2

x 1 x 116 3i.e. or

x 15 x 2

i.e. 15 ( x 1) 16x or 2( x 1) 3x

i.e. 15x 16x 15 0 or 2x 3x 2 0

Consider 15x 2 - 16x – 15 = 0

i.e. 15x 2 – 25x + 9x – 15 = 0

i.e. 5x ( 3x – 5 ) + 3 ( 3x – 5 ) = 0

i.e. ( 5x + 3 ) ( 3x - 5 ) = 0

i.e. 5x + 3 = 0 or 3x – 5 = 0

i.e. x = - 3 / 5 or x = 5 / 3

Consider 2x 2 - 3x – 2 = 0

i.e. 2x 2 - 4x + x – 2 = 0

i.e. 2x ( x – 2 ) + 1( x – 2 ) = 0

i.e. ( 2x + 1 ) ( x – 2 ) = 0

i.e. 2x + 1 = 0 or x – 2 = 0

i.e. x = - 1 / 2 or x = 2

Hence, the roots are – 3/ 5, 5 / 3, - 1/ 2, 2


2

1 112 x 56 x 89 0

xx

Ans:- equation is

2

2

1 112 x 56 x 89 0

xx

2

2 2

2

1put x m

x

now

1 1x x 2 m 2

xx


12 ( m 2 - 2 ) – 56m + 89 = 0

i.e. 12m 2 - 24 – 56m + 89 = 0

i.e. 12m 2 – 56m + 65 = 0

i.e. 12m 2 – 30m – 26m + 65 = 0

i.e. 6m ( 2m – 5 ) – 13 ( 2m – 5 ) = 0

i.e. ( 6m - 13 ) ( 2m – 5 ) = 0

i.e. ( 6m - 13 ) = 0 or ( 2m – 5 ) = 0

∴ m = 13 / 6 or m = 5 / 2

2 2

2 2

2 2

1 13 1 5x or x

x 6 x 2

x 1 x 113 5i.e. or

x 6 x 2

i.e. 6 ( x 1) 13x or 2( x 1) 5x

i.e. 6x 13x 6 0 or 2x 5x 2 0

Consider 6x 2 - 13x + 6 = 0

i.e. 6x 2 – 9x - 4x + 6 = 0

i.e. 3x ( 2x – 3 ) - 2 ( 2x – 3 ) = 0

i.e. ( 3x - 2 ) ( 2x - 3 ) = 0

i.e. 3x - 2 = 0 or 2x – 3 = 0

i.e. x = 2 / 3 or x = 3 / 2

Consider 2x 2 - 5x + 2 = 0

i.e. 2x 2 - 4x - x + 2 = 0

i.e. 2x ( x – 2 ) – 1 ( x – 2 ) = 0

i.e. ( 2x - 1 ) ( x – 2 ) = 0

i.e. 2x - 1 = 0 or x – 2 = 0

i.e. x = 1 / 2 or x = 2

Thus the roots are 2 /3, 3 / 2, 1 / 2, 2

23. Solve the equation : ( y 2 + 5y ) ( y 2 + 5y – 2 ) – 24 = 0

Ans:- equation is

( y 2 + 5y ) ( y 2 + 5y – 2 ) – 24 = 0

Put y 2 + 5y = m

Hence, the equation becomes

m ( m – 2 ) – 24 = 0

i.e. m 2 – 2m – 24 = 0

i.e. m 2 – 6m + 4m – 24 = 0

i.e. m ( m – 6 ) + 4 ( m – 6 ) = 0

i.e. ( m + 4 ) ( m – 6 ) = 0

i.e. m = - 4 or m = 6

∴ y 2 + 5y = - 4 or y 2 + 5y = 6

i.e. y 2 + 5y + 4 = 0 or y 2 + 5y – 6 = 0

consider y 2 + 5y + 4 = 0

i.e. y 2 + 4y + y + 4 = 0

i.e. y ( y + 4 ) + 1 ( y + 4 ) = 0

i.e. ( y + 1 ) ( y + 4 ) = 0

i.e. y = - 1 or y = - 4

consider y 2 + 5y – 6 = 0

i.e. y 2 + 6y – y – 6 = 0

i.e. y ( y + 6 ) – 1 ( y + 6 ) = 0

i.e. ( y + 6 ) ( y – 1 ) = 0

i.e. y = - 6 or y = 1

hence, the roots are – 1, - 4, - 6, 1

24. Solve : 2( y 2 – 6y ) 2 – 8 ( y 2 – 6y + 3 ) – 40 = 0

Ans:- equation is

2( y 2 – 6y ) 2 – 8 ( y 2 – 6y + 3 ) – 40 = 0

Put y 2 – 6y = m

The equation becomes

2m 2 – 8 ( m + 3 ) – 40 = 0

i.e. 2m 2 – 8m - 24 – 40 = 0

i.e. 2m 2 – 8m - 64 = 0

i.e. m 2 – 4m - 32 = 0

i.e. m 2 – 8m + 4m - 32 = 0

i.e. m ( m – 8 ) + 4 ( m – 8 ) = 0

i.e. ( m + 4 ) ( m – 8 ) = 0

i.e. m + 4 = 0 or m – 8 = 0

i.e. m = - 4 or m = 8

Hence,

y 2 – 6y = - 4 or y 2 – 6y = 8

i.e. y 2 – 6y + 4 = 0 or y 2 – 6y – 8 = 0

consider y 2 – 6y + 4 = 0

comparing with ay 2 + by + c = 0 to get,

a = 1, b = - 6, c = 4

2

2

b b 4acy

2a

( 6) ( 6) 4(1) (4) 6 36 16

2(1) 2

6 20 6 2 5 2(3 5 )3 5

2 2 2

consider y 2 – 6y – 8 = 0

comparing with ay 2 + by + c = 0 to get,

a = 1, b = - 6, c = - 8

2

2

b b 4acy

2a

( 6) ( 6) 4(1) ( 8) 6 36 32

2(1) 2

6 68 6 2 17 2(3 17 )3 17

2 2 2

The roots are 3 5 and 3 17

25. Three consecutive odd natural numbers are such that the product of the

first and third is greater than four times the middle by 1. Find the

numbers.

Ans:- Let x be an odd natural number.

∴ x, x + 2, x + 4 are consecutive odd natural numbers.

We are given

x ( x + 4 ) = 4 ( x + 2 ) + 1

∴ x 2 + 4x = 4x + 8 + 1

∴ x 2 = 9

i.e. x = ± 3

i.e. x = 3 or x = - 3

but – 3 is not a natural number. Hence x = 3

thus, the numbers are 3, 5, 7

26. In garden there are some rows and columns. The number of trees in a

row is greater than that in each column by 10. Find the number of trees

in each row if the total number of trees is 200.

Ans:- Let the number of trees in each column = x

∴ the number of trees in each row = x + 10

We are given

x ( x + 10 ) = 200

i.e. x 2 + 10x – 200 = 0

i.e. x 2 + 20x – 10x – 200 = 0

i.e. x ( x + 20 ) – 10 ( x + 20 ) = 0

i.e. ( x – 10 ) ( x + 20 ) = 0

∴ x = 10 or x = - 20

Number of trees cannot be negative, hence we discard x = - 20

Hence, number of trees in each column = 10

And number of trees in each row = 20

27. One diagonal of a rhombus is greater than other by 4 cm. If the area of

the rhombus is 96 cm 2, find the side of the rhombus.

Ans:- Let one diagonal = x cm

∴ The other diagonal is ( x + 4 ) cm

1Area of the rhombus (product of diagoals )

2

196 x ( x 4)

2

∴ x (x + 4 ) = 192

∴ x 2 + 4x – 192 = 0

∴ x 2 + 16x – 12x – 192 = 0

∴ x ( x + 16 ) – 12 ( x + 16 ) = 0

∴ ( x – 12 ) ( x + 16 ) = 0

∴ x = 12 or x = - 16

As length cannot be negative, x ≠ -16

∴ x = 12

Let diagonal BD = 12

∴ diagonal AC = 12 + 4 = 16

Diagonals of a rhombus bisect each other at right angles.

Hence, AM = 8 and BM = 6

In right triangle AMB

AB 2 = AM 2 + BM 2

∴ AB 2 = 64 + 36 = 100

∴ AB = 10

Length of side of the rhombus is 10 cm

28. A man riding on a bicycle covers a distance of 60 km in a direction of

wind and comes back to his original position in 8 hours. If the speed of

the wind is 10 km/hr, find the speed of the bicycle.

Ans:- Let the speed of the bicycle be x km/hr

Speed of wind = 10 km/hr

∴speed in the direction of the wind = ( x + 10 ) km/hr

and speed in the direction opposite to wind = ( x - 10 ) km/hr

Now,

distance

timespeed

Thus, from the given conditions

60 608

x 10 x 10

60( x 10) 60( x 10)i.e. 8

( x 10) ( x 10)

i.e. 60 ( x 10 x 10) 8( x 10) ( x 10)

i.e. 15 ( 2x ) = 2 ( x 2 – 100 )

∴ 30x = 2x 2 – 200

∴ 2x 2 – 30x – 200 = 0

∴ x 2 – 15x – 100 = 0

∴ x 2 – 20x + 5x – 100 = 0

∴ x ( x – 20 ) + 5 ( x – 20 ) = 0

∴ ( x + 5 ) ( x – 20 ) = 0

∴ x + 5 = 0 and x – 20 = 0

∴ x = - 5 and x = 20

Now x cannot be negative, hence x = 20

∴ speed of bicycle = 20 km/hr

29. The sum of four times a number and three times its reciprocal is 7. Find

that number.

Ans:- Let the number be x.

Then we are given

2

14x 3 7

x

i.e. 4x 3 7x

i.e. 4x 2 – 7x + 3 = 0

i.e. 4x 2 – 4x – 3x + 3 = 0

i.e. 4x ( x – 1) – 3 ( x - 1 ) = 0

i.e. ( 4x – 3 ) ( x – 1 ) = 0

∴ 4x – 3 = 0 or x – 1 = 0

∴ x = 3 / 4 or x = 1

The numbers are 3

4 and 1

30. For doing some work Ganesh takes 10 days more than John. If both work

together they complete the work in 12 days. Find the number of days if

Ganesh worked alone.

Ans:- Let John take x days to complete the work, thus Ganesh will require

( x + 10 ) days to complete the work.

Now, work completed by John in one day = 1

x

and work completed by Ganesh in one day = 1

x 10

thus, work done by John and Ganesh in one day = 1 1

x x 10

now, they complete the work in 12 days

2

2

1 112 1

x x 10

x 10 x12 1

x ( x 10)

12 (2x 10) x ( x 10)

24x 120 x 10x

i.e. x 14x 120 0

i.e. x 2 – 20x + 6x – 120 = 0

i.e. x ( x – 20 ) + 6 ( x – 20 ) = 0

i.e. ( x + 6 ) ( x – 20 ) = 0

i.e. x + 6 = 0 or x - 20 = 0

∴ x = - 6 or x = 20

Now x cannot be negative, hence x = 20

Thus, number of days taken by Ganesh to complete the work alone is

20 + 10 = 30

31. A natural number is greater than three times its square root by 4. Find

the number.

Ans:- Let the natural number be x.

We are given

x= 3 x + 4

i.e. x – 4 = 3 x

on squaring both sides

( x – 4 ) 2 = ( 3 x ) 2

i.e. x 2 – 8x + 16 = 9x

i.e. x 2 – 17x + 16 = 0

i.e. x 2 – 16x – x + 16 = 0

i.e. x ( x - 16 ) – ( x – 16 ) = 0

i.e. ( x – 16 ) ( x – 1 ) = 0

∴ x – 16 = 0 or x – 1 = 0

∴ x = 16 or x = 1.

Now, x = 1 does not satisfy the given condition, hence the required

natural number is 16.

32. The sum of the areas of two squares is 400 sq. m. If the difference

between their perimeters is 16m, find the sides of two squares.

Ans:- Let the side of smaller square be x m. Then the perimeter of the square is

4x.

Now perimeter of the second square is 4x + 16

Hence a side of the greater square is

4x 16

4 = ( x + 4 ) m

Area of first square = x 2

Area of second square = ( x + 4 ) 2 = x 2 + 8x + 16

We are given

Sum of areas = 400

∴ x 2 + x 2 + 8x + 16 = 400

∴ 2x 2 + 8x – 384 = 0

∴ x 2 + 4x – 192 = 0

∴ x 2 + 16x - 12x – 192 = 0

∴ x ( x + 16 ) – 12 ( x + 16 ) = 0

∴ ( x – 12 ) ( x + 16 ) = 0

∴ x – 12 = 0 and x + 16 = 0

∴ x = 12 or x = - 16

Now side of a square cannot be negative

Hence, side of smaller square = 12 m and side of greater square = 16 m

33. Exterior angle of a regular polygon having n – sides is more that of the

polygon having n 2 sides by 50 ˚. Find the number of the sides of each

polygon.

Ans:- Sum of all exterior angles of any polygon is 360˚.

A n sided polygon has n exterior angles. As it is a regular polygon all

exterior angles are of equal measure.

Hence, the measure of each exterior angle of n sided polygon is 360

n

Similarly, the measure if each exterior angle of a regular polygon having

n 2 side is 2

360

n

We are given

2

2

2

360 36050

n n

360n 36050

n

360n 360 50n

∴ 50n 2 – 360n + 360 = 0

Divide by 10

5n 2 – 36n + 36 = 0

∴ 5n 2 – 30n – 6n + 36 = 0

5n ( n – 6 ) – 6 ( n 6 ) = 0

∴ ( 5n – 6 ) ( n – 6 ) = 0

∴ 5n – 6 = 0 or n – 6 = 0

∴ n = 6 / 5 or n = 6

Now, number of sides cannot be 6 / 5, hence n = 6

Number of sides of one polygon = 6

Number of sides of other polygon = 6 2 = 36

34. Tinu takes 9 days more than his father to do a certain piece of work.

Together they can do the work in 6 days. How many days will Tinu take to

do that work alone ?

Ans:- Let Tinu take x days to complete the work, thus her father will require

( x - 9 ) days to complete the work.

Now, work completed by Tinu in one day = 1

x

and work completed by her father in one day = 1

x 9

thus, work done by Tinu and her father in one day = 1 1

x x 9

now, they complete the work in 6 days

1 16 1

x x 9

x 9 x6 1

x ( x 9)

6 (2x 9) x ( x 9)

i.e. 12x – 54 = x 2 – 9x

i.e. x 2 – 21x + 54 = 0

i.e. x 2 – 18x - 3x + 54 = 0

i.e. x ( x – 18 ) - 3 ( x – 18 ) = 0

i.e. ( x - 3 ) ( x – 18 ) = 0

i.e. x – 3 = 0 or x - 18 = 0

∴ x = 3 or x = 18

Now x = 3 does not satisfy the given condition, hence x = 18

Thus, number of days taken by Tinu to complete the work alone = 18

35. The sum of the squares of five consecutive natural numbers is 1455. Find

the numbers.

Ans:- Let x, x + 1, x + 2, x + 3, x + 4 be the five consecutive numbers.

We are given

x 2 + ( x + 1 ) 2 + ( x + 2) 2 + ( x + 3 ) 2 + ( x + 4 ) 2 = 1455

∴ x 2 + x 2 + 2x + 1 + x 2 + 4x + 4 + x 2 + 6x + 9 + x 2 + 8x + 16 =

1455

∴ 5x 2 + 20x + 30 = 1455

∴ 5x 2 + 20x – 1425 = 0

∴ x 2 + 4x – 285 = 0

∴ x 2 + 19x – 15x – 285 = 0

∴ x ( x + 19 ) – 15 ( x + 19 ) = 0

∴ ( x – 15 ) ( x + 19 ) = 0

∴ x = 15 and x = - 19

As x is a natural number it cannot be negative, hence x ≠ -19

∴ x = 15

Thus the numbers are 15, 16, 17, 18 and 19.

36. The side of one regular hexagon is larger than that of the other regular

hexagon by 1 cm. If the product of their areas is 243, then find the sides

of both the regular hexagons.

Ans:- Let the side of the smaller regular hexagon be x cm.

Then the side of the larger hexagon is ( x + 1 ) cm

Area of smaller hexagon is

2 23 3 3 3(side ) x

2 2

Area of greater hexagon is

2 23 3 3 3(side ) ( x 1)

2 2

We are given

Product of areas = 243

2 2

2 2

2 2

2 2

3 3 3 3x ( x 1) 243

2 2

27x ( x 1) 243

4

4x ( x 1) 243

27

x ( x 1) 36

∴ x ( x + 1 ) = ± 6

∴ x ( x + 1 ) = 6 or x ( x + 1 ) = - 6

∴ x 2 + x = 6 or x 2 + x = - 6

∴ x 2 + x – 6 = 0 or x 2 + x + 6 = 0

Consider x 2 + x – 6 = 0

i.e. x 2 + 3x – 2x – 6 = 0

i.e. x ( x + 3 ) – 2 ( x + 3 ) = 0

i.e. ( x – 2 ) ( x + 3 ) = 0

∴ x – 2 = 0 or x + 3 = 0

∴ x = 2 or x = - 3

As side cannot be negative, x ≠ -3

∴ x = 2

Consider x 2 + x + 6 = 0

comparing with ax 2 + bx + c = 0 to get

a = 1, b = 1, c = 6

2 2here b 4ac 1 4(1) 6 1 24 23 0

roots are not real

∴ x = 2 is the only root

Answer :- Side of a smaller hexagon = 2 cm

Side of a greater hexagon = 2 + 1 = 3 cm

37. Two years ago my age was 41

2 times the age of my son. Six years ago,

my age was twice the square of the age of my son. What is the present

age of my son.

Ans:- Let the present age of the son be x years.

Age of son 2 years ago = ( x – 2 ) years

∴ Age of father 2 years back = 1 9

4 ( x 2) ( x 2)2 2

∴ present age of father =

9(x 2) 4 9x 18 4 9x 149( x 2) 2

2 2 2 2

Age of son 6 years ago = ( x – 6 ) years

Age of father 6 years ago = 9x 14 9x 14 12 9x 26

6 years2 2 2

We are given

9x 26

2 = 2 ( x – 6 ) 2

i.e. 9x – 26 = 4 ( x 2 – 12x + 36 )

i.e. 9x – 26 = 4x 2 – 48x + 144

i.e. 4x 2 – 57x + 170 = 0

i.e. 4x 2 – 40x – 17x + 170 = 0

i.e. 4x ( x – 10 ) – 17 ( x - 10 ) = 0

i.e. ( 4x – 17 ) ( x – 10 ) = 0

i.e. 4x – 17 = 0 or x – 10 = 0

i.e. x = 17 / 4 or x = 10

For x = 17/4, son’s age 6 years ago becomes negative !

∴ x ≠ 17/4

∴ x = 10

Ans: The present age of the son is 10 years.

38. A car covers a distance of 240 km with some speed. If the speed is

increased by 20 km / hr, it will cover the same distance in 2 hours less.

Find the usual speed of the car.

Ans:- Let the usual speed of the car be x km / hr.

Now,

distance 240

time hoursspeed x

Increased speed = ( x + 20 ) km / hr

Time taken with this speed to cover 240 km will be 240

x 20hours

We are given

240 2402

x x 20

240( x 20) 240x2

x ( x 20)

∴ 240 ( x + 20 – x ) = 2x ( x + 20 )

∴ 4800 = 2x ( x + 20 )

Divide both sides by 2

∴ 2400 = x ( x + 20 )

∴ x 2 + 20x = 2400

∴ x 2 + 20x – 2400 = 0

∴ x 2 + 60x – 40x – 2400 = 0

∴ x ( x + 60) – 40 ( x + 60 ) = 0

∴ ( x – 40 ) ( x + 60 ) = 0

∴ x – 40 = 0 or x + 60 = 0

∴ x = 40 or x = - 60

As speed cannot be negative, x ≠ -60

∴ x = 40

Thus usual speed of car = 40 km / hr

39. An express train takes 30 min less for a journey of 440 km, if its usual

speed is increased by 8 km / hr. Find its usual speed.

Ans:-

Let the usual speed of the train be x km / hr.

Now,

distance 440

time hoursspeed x

Increased speed = x + 8

Time taken with this speed to cover 440 km will be 440

x 8hour

We are given

440 440 1 130 min hr

x x 8 2 2

440 x 8 x440( x 8) 440x 1 1i.e.

x ( x 8) 2 x x 8 2

2 440 ( x 8 x ) x ( x 8)

∴ 880 ( 8 ) = x ( x + 8 )

∴ x 2 + 8x = 7040

∴ x 2 + 8x – 7040 = 0

∴ x 2 + 88x – 80x – 7040 = 0

∴ x ( x + 88) – 80 ( x + 88 ) = 0

∴ ( x – 80 ) ( x + 88 ) = 0

∴ x – 80 = 0 or x + 88 = 0

∴ x = 80 or x = - 88

As speed cannot be negative, x ≠ -88

∴ x = 80

Thus usual speed of train = 80 km / hr

40. The divisor and quotient of the number 6123 are same and the remainder

is half the divisor. Find the divisor.

Ans:- Let the divisor = quotient = x Also, dividend = 6123

Remainder = half the divisor = x

2

Now, dividend = ( quotient × divisor ) + remainder

∴ 6123 = x (x) + x

2

∴ 12246 = 2x 2 + x

∴ 2x 2 + x – 12246 = 0

∴ 2x 2 + 157x – 156x – 12246 = 0

∴ x ( 2x + 157 ) – 78 ( 2x + 157 ) = 0

∴ ( x – 78 ) ( 2x + 157 ) = 0

∴ x – 78 = 0 or 2x + 157 = 0

∴ x – 78 or x = - 157 / 2

We discard the negative fraction to get x = 78.

Hence the required divisor is 78.

41. From the same place at 7 am ‘A’ started walking in the north at the speed

of 5 km / hr. After 1 hour ‘B’ started cycling in the east at a speed of 16

km/hr. At what time they will be at distance of 52 km apart from each

other.

Ans:-

Let P be the common starting point. Let ‘A’ reach point A and B reach

point ‘B’ when they are at a distance of 52 km from each other.

Let the time taken by ‘A’ to reach the point A be x hours.

Then the time taken by ‘B’ to reach the point B is x – 1 hours.

Speed of ‘A’ is 5 km/hr and speed of ‘B’ is 16 km / hr, hence the

distances covered by ‘A’ and ‘B’ in km are

PA = 5x and PB = 16 ( x – 1 ) respectively

In right triangle APB,

AB 2 = PA 2 + PB 2

∴ ( 52 ) 2 = ( 5x ) 2 + [ 16 ( x – 1) ] 2

∴ 2704 = 25x 2 + 256 ( x 2 – 2x + 1 )

∴ 2704 = 25x 2 + 256 x 2 – 512x + 256

∴ 281x 2 – 512x – 2448 = 0

∴ 281x 2 – 1124x + 612x – 2448 = 0

∴ 281x ( x – 4) + 612 ( x – 4 ) = 0

∴ ( 281x + 612) ( x – 4 ) = 0

∴ 281x – 612 = 0 or x – 4 = 0

∴ x = - 612 / 281 or x = 4

As time taken cannot be negative, x ≠ -612/281

∴ x = 4

∴ they are 52 km apart after 4 hours from 7 a.m.

i.e. they are 52 km apart at 11 a.m.

Thus time taken by ‘A’ to reach at point A = 4 hours

‘A’ starts at 7 am, hence they meet at 7 + 4 = 11 am

42. One tank can be filled up by two taps in 6 hours. The smaller tap alone

takes 5 hours more than the bigger tap alone. Find the time required by

each tap to fill the tank separately.

Ans:- Let the smaller tank take x hours to fill the tank.

Hence, time taken by the bigger tank to fill the tap = x - 5

The work done by smaller tank in 1 hour = 1

x

The work done by bigger tank in 1 hour = 1

x 5

Thus, the work done by both in one hour = 1 1

x x 5

Now, the tank fills in 6 hours by both the taps.

That is both taps together fill 1

th6

part of the tank in one hour

1 1 1

x x 5 6

1 1i.e. 6 1

x x 5

x 5 x6 1

x ( x 5)

6( 2x 5) x ( x 5)

∴ 12x - 30 = x 2 - 5x

∴ x 2 – 17x + 30 = 0

∴ x 2 – 15x - 2x + 30 = 0

∴ x ( x – 15 ) - 2 ( x – 15 ) = 0

∴ ( x - 2 ) ( x – 15 ) = 0

∴ x - 2 = 0 or x – 15 = 0

∴ x = 2 or x = 15

Now x = 2 does not satisfy given condition. Hence x = 15

The smaller tape takes 15 hours and the bigger tap takes 15 – 5 = 10

hours to fill the tank alone.

43. Around a square pool there is a footpath of width 2m. If the area of the

footpath is 5

4 times that of the pool, find the area of the pool.

Ans:-

Let side of the pool be x m, then side of the footpath is ( x + 4 ) m

Area of pool = x 2

Area of outer square = ( x + 4 ) 2 = x 2 + 8x + 16

Area of footpath = area of outer square – area of pool

= x 2 + 8x + 16 – x 2 = 8x + 16

We are given

2

2

5area of footpath area of pool

4

58x 16 x

4

4 (8x 16 ) 5x

∴ 32x + 64 = 5x 2

∴ 5x 2 – 32x – 64 = 0

∴ 5x 2 – 40x + 8x – 64 = 0

∴ 5x ( x – 8 ) + 8 ( x – 8 ) = 0

∴ ( 5x + 8 ) ( x – 8 ) = 0

∴ 5x + 8 = 0 or x – 8 = 0

∴ x = - 8 / 5 or x = 8

As length cannot be negative, x ≠ - 8 / 5

∴ x = 8

i.e. side of the square pool = 8m

Its area = 8 2 = 64 sq. m.

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Quadrtic Equations Exercises