Exercise 1
1.
Which of the following is the quadratic equation? (i) 11 = - 4x2 – x3 7
(iii) z (v) Ans:- (i)
4 4
q2 q
2
4z
3
(ii)
4
y
y
(iv) 3 y 2
5
7
3 y
3
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 3 y
4
y
on multiplying both sides by y 3
4y
i.e. y2
y2 4y
3
0
This is a quadratic equation. { It is of the type ay 2 + by + c = 0, a ≠ 0 } (iii)
7
Given equation is z
4z
z
5
Multiply both sides by z to get z2 – 7 = 4z2 + 5z i.e. 3z2 + 5z + 7 = 0 This is of the type az 2 + bz + c = 0 ( a ≠ 0 )
∴ it is a quadratic equation. (iv)
Given equation is 3y 2
7
i.e. 3y 2
3 y 3 y
7
0
This is of the type ay2 + by + c = 0 ( a ≠ 0 )
∴ it is a quadratic equation. (v)
Given equation is
q2 q2
4
3
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 following quadratic equations in standard form ax2 + bx + c = 0 (i) p (p-6) = 0 5
(iii) 2z Ans:- (i)
z
z
7
(ii) n m2
(iv)
6
4
n 5
m2
3
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)
7
Given equation is n
n
4
i.e. n2 – 7 = 4n i.e. n2 – 4n – 7 = 0 This a quadratic quadratic equation equation in n in standard standard form. (iii)
2z
5 z
z
6
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)
m2
Given equation is
m
5 2
3
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.
1, Explain whether p 1,
1 2
,
3 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 t he 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) p
1 2
, LHS
2 2
1
2
2
1
5
1
5
4
2
2 3
3 1
5
2
2
3
0
RHS 1
As equation is satisfied , p iii)
2
is a root of equation
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 x 2 - 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 y 2 – (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.
20
Solve the quadratic equation x
Ans: Quadratic equation is x
20 x
12
x 12
0
0 by factorization method
2.
Solve the quadratic equation 10x2 + 3x – 4 = 0 by factorization method.
Ans:- Quadratic Equation is 10x2 + 3x – 4 = 0 Split middle middle term 3x as 8x – 5x
3.
Solve the quadratic equation x 2
method Ans:- Quadratic equation is x 2
3 3x
3 3x 6
6 0
0 using factorization
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
1
Thi Third term
2
coefficient of x 2
1 2 adding x
2
1
1
1
4
both sides
4
1
x
1
1
4
i.e.
4
2
1
x
2
5 4
taking taking square square roots x
1
5
2
4
2
1
2 1
∴ roots are
2.
2
5
1
x
5 5 2 5
2
and
1
5 2
Solve the equation 6m2 + m = 2 by completing square
Ans:- Quadratic equation is 6m 2 + m = 2 Divide both sides by 6 1
m2
6
1
m
3 1
Thi Third term
add
1 144
m2 i.e. m
2
2
coefficient of m
1
1
2
6
2
1 144
to both sides 1 6
m
1 12
2
1
1
1
144
3
144
48
1
49
144
144
Taking roots on both sides m m
1
7
12 1
12 7
12 12 1 7
i.e. m
12 6
i.e. m
12 1
i.e. m
2
12
or m or m
2
12 1
12 7
12
12
12 2
or m 1
7
8
or m
Thus Thus roots oots are are
1
3 and and
2 3
Exercise 6
1.
Find discriminant of the quadratic equation
Ans:- Quadratic equation is
3 x2
2
2 x
2 3
3 x2
2
2 x
2 3
0
0 ……… (1)
Compare it with ax 2 + bx + c = 0 to get a
3, b
2 2 and c
2 3
Now discriminant = ∆ = b2 – 4ac 2 2 4
2.
2
2
4 8
3
3
2
8
24
3 32
Determine the nature of the roots of the equation 3y 2 +9y + 4 = 0 , using discriminant.
Ans:- Quadratic equation is 3y 2 + 9y + 4 = 0 Compare it with ay 2 + 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 2 x 2 5 3 x 16 0 using discriminant.
Ans:- Quadratic Equation is 2 x 2
5 3x
16
0
Compare it with ax 2 + bx + c = 0 to get, a = 2, b = 5 3 , c = 16
The discriminant is ∆ = b2 – 4ac 5 3 25
2
3
4 2
16 16
128
75
53
As ∆ < 0, roots are not real.
128
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 ax 2 + 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 x 2 ≠ 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 real and equal roots.
Ans:Quadratic equation is k 2x
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 3k(k+1)–(k+1)=0 i.e. ( k + 1 ) ( 3k – 1 ) = 0
2
– 2 ( k – 1 ) x + 4 = 0 has
i.e. k + 1 = 0 or 3k – 1 = 0 1
i.e. k = -1 or k
3
Exercise 5
1.
x
Solve the equation x 2
1
0 , using formula method.
3 x
Ans:- Quadratic equation is x 2
1
0
3
i.e. 3x2 + x – 1 = 0 Comparing it with ax 2 + bx + c = 0 To get, a = 3, b = 1, c = -1 Now b2 – 4ac = 12 – 4 ( 3 ) ( -1) = 1 + 12 = 13 b2 4 ac
b
The roots are x
2a 13
1 6
∴ The roots of the given equation are 2.
1
13 6
and
1
13 6
Solve the quadratic equation 3q2 = 2q + 8 using formula method.
Ans:- Quadratic equation is 3q 2 = 2q + 8 i.e. 3q2 – 2q – 8 = 0 Compare it with aq 2 + 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 b2
b
q
4ac
2a 2
100
2
2
3
10 6
∴ roots of quadratic equations are 2
10 6
and
i.e i.e. 2 and
2
10 6
4 3
i.e.
12 6
and
8 6
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, 5
coefficient of x coefficient of x 2 5 i.e. 5 k
i.e. 4
5 k
1 ....... 1 k
i.e.
Also product of roots 2 k
i.e. 4 . 1 k
Put 1 k k2
2
k
i.e.
2
constant constant k coeffici coefficien entt of x 2
. 2 4k
1 2k
from 1
1 1 i.e. 2 2k k
1 2k
2k
divide by k
2.
k
5 k
0
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
coeff coeffici icient ent of x coeffi coefficie cient nt of x 2
sum of roots 2m
5m
k 1
k
m
7m
k k 7
Also lso, prod roduct uct of roo roots 2m
40
m2
4 k 7
Put m k 7 k2 k 3.
constant coeffic coefficien ientt of x 2
40 1
5m
10m2
.
2
k2 4 i.e. 49
49
4
4
196
14
Find k, if one of the roots of quadratic quadrat ic equation kx 2 – 7x + 12 = 0 is 3
Ans:- The given quadratic equation is kx 2 – 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 x 2 + px + q = 0 Roots differ by 1
Let α = β + 1
coefficient of x coefficient of x 2
now,
p i.e. 2 1 1 1 p 2 1 p 2 Constant Also, Coefficient of x 2 q i.e. 1 . 1 1 p put 2 1 p 1 p 1 . q 2 2 i.e.
1
1 p 2
2
1 p
1 p 2 1 p
2 1 p2
i.e. p2 5
1
q q
2
4 1 p2 4q
p
q
4q
Find k, if the sum of the roots of the quadratic equation 4x2 + 8kx + k + 9 = 0 is equal to their t heir product.
Ans:- Quadratic Equation is 4x2 + 8kx + k + 9 = 0 We are given that Sum of roots = product of roots i.e.
Coeffi Coefficie cient nt of x Coef Coeffi fici cien entt of x
2
Cons Cons tan tan t Coef Coeffi ficcientof ientof x2
i.e. – (coefficient (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 t he roots of the equation x 2 – 5x + 6 = 0, find (i)
α2 + β2
(ii)
Ans:- Quadratic equation is x 2 – 5x + 6 = 0
As α and β are a re the roots of above equation, 5
5 and
1
1
6
α2 + β2 = ( α + β )2 - 2α β = 52 – 2 × 6 = 25 – 12 = 13
(i)
2
2
(ii)
7.
6
.
13 6
from i
If one root of the quadratic equation kx2 – 20x + 34 = 0 is 5
Ans:- Quadratic Equation is kx2 – 20 x + 34 = 0 a bove equation, As α = 5 2 2 is a root of above
β= 5
2
2 must be the other root
coefficient of x
Now, Now, sum of roots roots 5 2 i.e. 10
2
5
20
2 k
k
coeffic coefficient ient of x 2 20
2 20 10
k 2
Exercise - 8
1.
Form a quadratic equation with roots – 2 and
Ans:- The given roots are α = -2 and β = 2 2
11 2 11 2
4 11 2
11 2
7 and 2
11
Quadratic Equation is x2 – (sum of roots) x + product of roots = 0 i.e. x2 – (α + β) x + α . β = 0
11 . 2
2
2 find k
on putting the values 7 x2 x 11 0 2 Multiplying both sides by 2
∴ 2x2 – 7x – 22 = 0 This is required Q.E. 2.
5
Form a quadratic equation if its one of the roots is
Ans:- If one of the root of a quadratic equation is
5
is
5
3
3 , then the other root
3
Let
5
3, 5
.
5
3 3 .
5
3
5
3
5
3
2 5 5
2
3
2
5
3
2
∴ Required quadratic equation is x2 – ( α + β ) x + α . β = 0 i.e. x2 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) + β 3 (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 α 3 – β3 = 215 ∴ 3 α β × 5 = 215 - 53 = 215 – 125 = 90 15 α β = 90 αβ=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 such quadratic quadratic equations equations . They are x2 – 7x+ 6 = 0 and x2 + 7x+ 6 = 0
Exercise 9
1.
Solve: (x2 + 2x) (x 2 + 2x – 11) + 24 = 0
Ans:- Equation is (x 2 + 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. x 2 + 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. x 2 + 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.
1 x2
Solve: 2 x2
1 x
9 x 1 x2
Ans:- Equation is 2 x2
14 1 x
9 x
0 14
0
We know the identity
1 x2
2
x
1 x
x
1 Note: x x x
2
2
x2 2x . 1 x2
2
2
x
1 x
1 x
1 x2
2
2
∴ given equation becomes 2
x
1 x
2
2
Take x 2 m2 2
9 x 1 m x 9 m 14 14
1 x
14 0
0
x2 2
1 x2
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 5 i.e. m = 2 or m 2 1 As m x x 1 2 m = 2 , gives x x x2 1 2 i.e. x2 + 1 = 2x i.e. x i.e. x2 – 2x + 1 = 0 i.e. ( x – 1)2 = 0 i.e. x – 1 = 0 i.e. x = 1
# 5 1 5 gives x 2 x 2 2 x 1 5 i.e. i.e. 2 x2 1 x 2
m
5x
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 1 i.e. x or x 2 2 1 , x 2 Ans: x 1, x 2 1 Or solution set = {1, , 2} 2
3.
12 y2
Solve : 35 y 2
44 12 y2
Ans:- Equation is 35 y 2
44
Take y2 = m ∴ equation becomes 12 35 m 44 m 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 2 6 or m i.e. m 5 7
# #
As m = y2 2 m gives y 2 5
2 5
y
2 5
6 gives y 2 7
6 7
y
6 7
m
Ans:
y
2 , 5
i.e. solution set =
4.
6 7
2 , 5
6 7
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 10 10 m gives x2 x 3 3 Ans: x
10 3
1,
15 12 y2 15 Ans:- Equation is 2 y 2 y2 5.
10 3
Solve : 2 y 2
12
Take y2 = m
∴ equation becomes 2m
15 m
12
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 am 2 + bm + c = 0 , to get a = 2, b = -12, c = 15 Now, b2 – 4ac = (-12)2 – 4 (2) (15) = 144 – 120 = 24
b
Roots are m
12
i.e. m
2 m m
6
2
2 6 4
6 2
6
As m #
b2 4ac 4ac 2a 24 12
6 2 y2
or m
6
6 2
6
m
6 2 6
y
6
gives y 2
6 2
6 2
# 6
m
6 2 6
y Ans:
6
gives y 2
6 2
6 2
6
y
6 2
,
6
6 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 i.e. n = 1
or
n–9=0
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 i.e. x = -8
or x – 3 = 0 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 S Find n if the sum (S) is 276.
n n 2
1
n ( n 1)
Ans:- We are given S
276
2
∴ 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 i.e. n = -24
or or
n -23 = 0 n = 23
As n is a natural number , n ≠ -24 ∴ n = 23 Note: We have
n n 1 2
276
i.e. n(n+1) = 2 × 276 = 552 LHS is a product of two consecutive natural numbers. 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 t he length of playground whose area is 420 sq. m. breadth
420 x
m
Now length is increased by 7m and breadth is decreased by 5m.
∴ New length = (x + 7) m New breadth
420 x
5 m
It is given that area remains the same. x 7 i.e.
x 7
420 x
5
420 5 x x
420 420
(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 Breadth
6.
420
20 m
21
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
840 x
New cost of banana after increase increase by 1 Re is Rs. (x + 1) per dozen doze n
∴ Number of dozens for Rs. 840
840 x 1
As per given condition 840
840
x
x 1
840 x 1 x x 1
2 840 x
2
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 i.e. x = -21
or
or x – 20 = 0 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 As weightage of that place is 10, the number formed formed is 10 ( x
2
2
+5
+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:Quadratic equation is (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)]
– 4 ( m + 1 ) (1) = 4 ( m 2 – 2m + 1 ) – 4m - 4 = 4m 2 – 8m + 4 – 4m – 4 ∆ = 4m 2 – 12m ---------- (1) 2
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 t he roots of the equation ax
2
+ bx + c = 0 find the value
of Ans:Quadratic equation is ax
2
+ bx + c = 0
As α and β are the roots of the equation b a
c
and
a 2
2
and We have ,
2
2 2
(
2
a
b2
c
b2
2c
a
a2
a
c
c
a
a
2 ac a2
2
2
2
2
b
b
2
(
b2
a c
2ac ac
2a c ac
4.
If one root of the quadratic equation x
2
+ 6x + k = 0 is h
Ans:Quadratic equation is x 2 + 6x + k = 0 Given that h 2 6 is one root
∴ the second root is h ∴α= h
2 6
2 6 and β = h
∴α+β= h
2 6 + h
Comparing with ax
2
2 6
2 6 = 2h
----------- (1)
+ bx + c = 0, 0 , we get
a = 1, b = 6, c = k b a
6 1
∴ from (1) and (2)
6
(2)
2 6 , find k.
∴h=-3
2h = - 6 c
Again
a
h 2 6 h
2
24
( 3) 2 k
9
k
h 2 6
1
k 24
24
k 15
∴ h = – 3 and k = – 15 5.
Solve the quadratic equation
1 x
1 2
by factorization method.
x2
Ans:- Quadratic equation is 1 x
1 x2
2
On cross multiplying we get x
2
= x + 2 i.e. x
2
–x–2=0
– 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 i.e. x
2
Thus, roots of the quadratic equation are 2 and – 1 6.
Solve the quadratic equation ( 2y + 3 )
2
= 81 by factorization method
Ans:- Quadratic equation is ( 2y + 3 )
2
= 81
i.e. ( 2y + 3 )
2
i.e. ( 2y + 3 )
2
– 81 = 0 –92=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 =
and
3
5
1
2 5
2 15
2
2
5 2
be the given roots of a quadratic equation in x.
Now, quadratic equation is x
2
– ( sum of roots ) x + ( product of roots ) = 0 1
i.e. x 2
x
2
15 2
0
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 x 2 – ( sum of roots ) x + ( product of roots ) = 0 i.e. x 2 – 6x – 3 = 0 9.
If α, β are the roots of the equation 4x
2
– 5x + 2 = 0, 0 , find the equation
whose roots are
i) α + 3β and 3α + β ii)
and 2
iii)
2
and
1
iv)
1
and
Ans:As roots of 4x
i)
2
– 5x + 2 = 0 are α and β
( 5)
5 4
2
4 1
4
2
Required quadratic equation has roots α + 3β and β + 3α
sum of roots = α + 3β + β + 3α = 4α + 4β = 4 ( α + β ) =4× product of roots
5
=5
4
(
) 2
3[ (
25
3
4 2
16
2
( 5
3
75
2
4
4
32
2
9
2
2
(
2
3
1 2
107
16
16
Now, required quadratic equation is x
2
– ( sum of roots ) x + ( product of roots ) = 0
on putting values, we get x2
ii)
107
5x
0 i.e. 16x 2
16
80x
10 107
0
Required quadratic equation has roots 2
2
sum of roots roots 5
2
2
4
1
25
2
16
1
1
2
2
product of roots
)2
(
1
and
9 16
2
9
2
8
1
Required quadratic equation is x
2
– ( sum of roots ) x + ( product of roots ) = 0
i.e. x 2 i.e. 8x 2
9 8
x 9x
1 8
0 0 2
iii)
Required quadratic equation has roots
2
and
6
2
3
2
3
sum of roots roots 3
( 3
3
3
(
3
3
3
3
5
3
4
sum of roots
3
3
(
1
5
125
15
2
4
64
8
3
125
120
1
1
64 1
2
2
2
5
5
2
64
32
Also, 2
2
1
product product of roots
2
Required quadratic equation is x
2
– ( sum of roots ) x + ( product of roots ) = 0
i.e. x 2 i.e. 32 x 2
iv)
5 32
1
x
0
2
5x
16
0
1
Roots of required quadratic equation are 1
sum of roots
1
1
(
1
and
1
( 5
5/ 4
5
5
4
1/ 2
4
4
1
product product of roots roots
2
(5 / 4) 4 )2
2 9
2
8
2
10
15
4
4
4
2
1
25
1 2
29 8
Required quadratic equation is
)2
(
1
1/ 2
1
5
1
2
1
2
1
1
1/ 2
2
16 1/ 2
2
1
1 2
1
9
2
16
2
2
x
2
– ( sum of roots ) x + ( product of roots ) = 0
i.e. x 2
15 4
i.e. 8x 2
10.
x
30 x
29 8 29
0 0
If the difference difference between the roots 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
α3–β3=(α–β)3+3αβ(α–β) ∴ 189 = (3) 3 +3 α β ( 3) ∴ 189 = 27 + 9 α β ∴ 9 α β = 189 – 27 ∴αβ=
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 x2–(α+β)x+αβ=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 are α and β.
3
+ β 3 = 35, find the quadratic equation whose roots
Ans:- We are given
α + β = 5 and α 3 + β 3 = 35 α3+β3=(α+β)3-3αβ(α+β) ∴ 35 = 5 3 – 3 α β (5) ∴ 35 = 125 - 15 α β
∴ 15αβ = 90 ∴αβ=6 As α, β are roots of quadratic equation, the required required quadratic equation is
– (α + β ) x + α β = 0 i.e. x 2 – 5x + 6 = 0 x
12.
2
If the difference difference of the roots roots of the quadratic quadratic equation is 4 and difference difference of their cubes is 208, find the quadratic equation.
Ans:- Let α and β be the roots of the equation
∴α–β=4
α 3 – β 3 = 208
and
We have
α3–β3=(α–β)3+3αβ(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 x2–(α+β)x+αβ=0 ∴ x 2 – ( ± 8x ) + 12 = 0 ∴ x 2 – 8x + 12 = 0 and x 2 + 8x + 12= 0 are the required equations. equations. 13.
If one root of the quadratic equation ax the other, show that b3 + a
Ans:- Quadratic equation is ax
2
2
c +ac
2
2
+ bx + c = 0 is the square of
= 3 abc
+ bx + c = 0
Let first root be α. Then second root must be its square i.e. α
2
coeff of x
sum of of roots roots
coeff of x b
2
(1)
a
constant
product product of roots roots c
2
coeff of x 2 c
3
i.e.
a
2
( 2)
a
To get the required result, we have to eliminate α from (1) and (2) On cubing both sides of (1), we get 3
b
3
2
a 3
i.e.
2
3
using a 3
i.e.
b 3
3 6
now , 3
3
(
3
a3
( 3
)
2
(
b3
3
)
a3 b3
3ab ( a b ) 2
( 3
2
b
6
)
3
a3
)2 2
(
)
(
3
)
2
b3 a3
Put values from (1) and (2) to get c
3
a i.e.
b
c a
2
c
a
b3 a3
a
c
3bc
c2
a
a2
a2
b3 a3
multiply both sides by a a
2
c – 3abc + ac
2
3
to get
=-b3
on rearranging terms we get b3 + a 14.
2
c +ac
2
= 3 abc
If the sum of the roots of the quadratic equation Show that the product of the roots is
Ans:- Quadratic equation is
( p2
q2) 2
1
1
1
x p
x q
r
is zero.
1
1
1
x p
x q
r
i.e.
x q x p
1
x q
r
x p
i.e. ( 2x p q ) r
x q
i.e. 2x r
x2
i.e. x 2
pr
qr
x p px
( p q 2r 2r ) x
qx
pq
pr
pq qr
0
If α and β are the t he 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 r
p q
(1)
2
Now, pq pr qr
produc productt of roots roots pq
1 r (p q) p
pq pq
q 2
( p q)
( p q) 2
2pq
2
(p 2
......... from (1) (p 2
q2 2
q2) 2
15.
Solve the equation : x
4
– 29x 2 + 100 = 0
Ans:- equation is
– 29x 2 + 100 = 0 i.e. ( x 2 ) 2 – 29x 2 + 100 = 0 x
4
put x
2
=m
Thus the equation becomes
– 29 m + 100 = 0 i.e. m 2 – 25m – 4m + 100 = 0 i.e. m ( m – 25 ) – 4 ( m – 25 ) = 0 m
2
2pq )
i.e. ( m – 4 ) ( m – 25 ) = 0 i.e. m = 4 or m = 25. i.e. x
= 4 or x
2
= 25
2
∴ x = ± 2 or x = ± 5 Thus the roots are 2, - 2, 5, - 5 16.
Solve : 7y
4
– 25y 2 + 12 = 0
Ans:- equation is
– 25y 2 + 12 = 0 i.e. 7 ( y 2 ) 2 – 25y 2 + 12 = 0 7y
4
put y
=m
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 i.e. 7m
2
Thus, m = 4 / 7 i.e. y 2
3 or y 2
i.e. y
3 or y
or 4 7 2
The roots are 3,
17.
3,
2
Solve : 6 m 2
m=3
m2
7 2 7
,
2 7
7
Ans:- equation is 6m
2
2
m
7
2
i.e. 6 m 4
2
i.e. 6 m 4
7m 2
i.e. 6 (m put m
2
2
)
7m2
2
2
0
– 7m 2 + 2 = 0
=x
thus the equation becomes 6x
2
6x
2
– 7x + 2 = 0 – 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
m2
3
2
m 2
18.
3
2 1
or m
3
Hence
1
or m 2
2
,
,
3
2 1 2
1
,
2
Solve the equation : ( p
are the roots.
+p)(p
2
2
+ p – 3 ) = 28
Ans:- equation is (p2+p)(p
2
+ p – 3 ) = 28
Put p 2 + p = m Hence, the equation becomes m ( m – 3 ) = 28
– 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
2
i.e. m = - 4 or m = 7
∴p2+p=-4 i.e. p
2
or
p
+p+4=0
consider p
2
2
+p=7
or
p
2
+p–7=0
+p+4=0
comparing with ap
2
+ bp + c = 0
a = 1, b = 1, c = 4 p
b2
b
4ac
2a 1
1 4 (1) (4)
1
2(1)
15 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 b2
b
p
4ac
2a 1
1 4 (1) ( 7)
1
2(1)
2 1
Thus, the roots are
19.
29
29 2
and
1
Solve the equation : 3 x 2
x2
1 2
4 x
Ans:- equation is 1
3 x2
x2 1
put x
1
4 x
6
0
m2
2
x
m
x
now x
1
2
x
1
x
2
2
2
x
Hence, the given equation becomes 3(m
2
i.e. 3m
+ 2 ) – 4m – 6 = 0 2
+ 6 – 4m – 6 = 0
– 4m = 0 i.e. m ( 3m – 4 ) = 0 i.e. 3m
2
i.e. m = 0
or
3m – 4 = 0
∴m=0
or
m=4/3
Hence, x
1 x
0
or
x x
i.e. x 2
1
i.e. x 2
1 or 3 ( x 2
0 or
2
1
4
x
3
1
x
4 3
1)
4x
∴ x = ± 1 or 3x 2 – 4x – 3 = 0 Consider 3x 2 – 4x – 3 = 0 comparing with ax
2
+ bx + c = 0
29
1 x
6
0
a = 3, b = - 4, c = - 3 b2
b
x
4ac
2a (
4)
4)2
(
4 (3) (
3)
4
16
2 (3) 4
52
4
6
2 13
6
2 (2
13 )
6 2
13 3
1
Solve the equation : 9 x 2
x
2
1 x 1
put x
x
3 x
2
1 x
20
0
m
now x
1
2
x
x
2
1 x
2
2
m2
2
Hence, the given equation becomes 9(m
2
i.e. 9m
+ 2 ) – 3m – 20 = 0 2
+ 18 – 3m – 20 = 0
– 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 i.e. 9m
2
,
3 x
Ans:- equation is 9 x2
2
6
Hence, the roots are 1, - 1,
20.
36
13 3
2
13 3
1 x
20
0
x i.e.
1
1 x x
or
3
2
1
1
x
i.e. 3 ( x
x or
3 2
i.e. 3x 2
1) x
x
1
2
x
3
2
1
2
x
3
x or 3 ( x 0 or 3x 2
3
Consider 3x
2
1)
2x
2x 3 0
+x–3=0
2
comparing with ax
+ bx + c = 0
2
a = 3, b = 1, c = - 3 b2
b
x
4ac
2a 12
1
4 (3) (
3)
1
1
2 (3)
Consider 3x
2
36
1
6
37 6
- 2x – 3 = 0
comparing with ax
2
+ bx + c = 0
a = 3, b = - 2, c = - 3 b2
b
x
4ac
2a ( 2) 2
( 2)
4 (3) (
3)
2
2 (3) 2
2 10
21.
10 )
1
6
Hence, the roots are
1 6
x
x2
77 x
1 x
12
10 3
1
77 x
2
0
1 x
40 6
10
1
,
Ans:- equation is 1
2
3 37
Solve the equation : 30 x 2
30 x 2
36 36
6
2 (1
6
4
12
0
1
put x
m
x
now x
1
2
x
x
2
2
1
2
x
m2
2
Hence, the given equation becomes 30 ( m
+ 2 ) – 77m – 12 = 0
2
i.e. 30m
2
+ 60 – 77m – 12 = 0
– 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 i.e. 30m
x i.e.
2
1
16
x
15
x2
1
x
or 16 15
i.e. 15 15 ( x
2
x or
1
3
x
2
x2 x
1
3 2
16 x or 2 ( x 2
1)
i.e. 15 15x 2 16 x
15
Consider 15x
2
0 or 2 x 2
1)
3x
3x 2 0
- 16x – 15 = 0
– 25x + 9x – 15 = 0 i.e. 5x ( 3x – 5 ) + 3 ( 3x – 5 ) = 0 i.e. 15x
2
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 i.e. 2x
2
2
- 3x – 2 = 0
- 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
22.
1
Solve the equation : 12 x 2
x
56 x
2
Ans:- equation is 1
12 x 2
x2 1
put x
1
56 x
89
x
0
m
x
now x
1
2
x
x
2
2
1
2
x
m2
2
Hence, the given equation becomes 12 ( m
- 2 ) – 56m + 89 = 0
2
i.e. 12m
2
- 24 – 56m + 89 = 0
– 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 2m – 5 ) = 0 ∴ m = 13 / 6 or m=5/2 i.e. 12m
x i.e.
2
1
13
x
6
x2
1
or 13
x
i.e. 6 ( x
or
6 2
1
5
x
2
x2 x
1
5 2
13 x or 2 ( x 2
1)
i.e. 6 x 2 13 x
Consider 6x
x
6
2
0 or 2 x 2
1)
5x
5x 2 0
- 13x + 6 = 0
– 9x - 4x + 6 = 0 i.e. 3x ( 2x – 3 ) - 2 ( 2x – 3 ) = 0 i.e. 6x
2
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
1 x
89
0
Consider 2x i.e. 2x
2
2
- 5x + 2 = 0
- 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
Put y
2
2
+ 5y – 2 ) – 24 = 0
+ 5y = m
Hence, the equation becomes m ( m – 2 ) – 24 = 0
– 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
2
i.e. m = - 4 or m = 6
∴ y 2 + 5y = - 4 i.e. y
2
or
+ 5y + 4 = 0
consider y i.e. y
2
2
y
2
+ 5y = 6
or
y 2 + 5y – 6 = 0
+ 5y + 4 = 0
+ 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 or y = - 4 consider y i.e. y
2
2
+ 5y – 6 = 0
+ 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
– 6y ) 2 – 8 ( y 2 – 6y + 3 ) – 40 = 0 Put y 2 – 6y = m 2( y
2
The equation becomes
– 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 2m
2
i.e. m = - 4 or m = 8 Hence,
– 6y = - 4 or i.e. y 2 – 6y + 4 = 0 y
– 6y = 8 or y 2 – 6y – 8 = 0
y
2
consider y
2
2
– 6y + 4 = 0
comparing with ay
2
+ by + c = 0 to get,
a = 1, b = - 6, c = 4 b2
b
y
4a c
2a ( 6) 2
( 6)
4 (1) (4)
6
36 16
2(1) 6
20
6
2
consider y
2 2 5
2(3
2 2
5) 2
3
5
– 6y – 8 = 0
comparing with ay
2
+ by + c = 0 to get,
a = 1, b = - 6, c = - 8
b2
b
y
4ac
2a (
6) 6)
(
6) 6) 2
4 (1) ( 8)
6
36
2(1) 6
68
6
2
The roots are 3 25.
2 17
2 2(3
2
5 and 3
32
17 ) 2
3
17
17
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 ∴x2=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 are some some rows rows and columns. The number 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 t rees 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 rhombus is greater than other by 4 cm. 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 1
Area rea of the the rhom rhombu buss 96
1 2
2
( prod produc uctt of dia diagoa goals)
x ( x 4) 4)
∴ 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 distance of 60 km in a direction 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
time
speed
Thus, from the given conditions 60
60
x 10 10
x 10 10
i.e.
8
60 ( x 10 ) 60 ( x 10 ) ( x 10 ) ( x 10 )
i.e. 60 ( x 10 x 10 )
i.e. 15 ( 2x ) = 2 ( x
8
8 ( x 10 ) ( x 10 ) 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 sum of four times a number number and three three times its reciprocal is 7. Find that number.
Ans:- Let the number be x. Then we are given 4x
3
i.e. 4 x 2
1
7
x 3
7x
– 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 or x=1 ∴x=3/4 i.e. 4x
2
The numbers are
30.
3 4
and 1
For doing some work Ganesh takes 10 days more than John. John. If 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 Ga nesh 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 12
1
1
x
x 10
1
x 10 x
12
1
x ( x 10 )
12 ( 2 2x x 10 )
x ( x 10 ) 2
24 x
120
x
2
14 x
120
i.e. x
10 x 0
– 20x + 6x – 120 = 0 i.e. x ( x – 20 ) + 6 ( x – 20 ) = 0 i.e. ( x + 6 ) ( x – 20 ) = 0 i.e. x
2
i.e. x + 6 = 0
or
∴x=-6
x = 20
or
x - 20 = 0
Now x cannot be negative, hence x = 20 Thus, number of days taken by Ganesh Ga nesh 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
– 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. i.e. x
2
Now, x = 1 does not satisfy the given condition, hence the required natural number is 16. 32.
The sum sum of the areas of two squares is 400 sq. sq. m. If the differenc difference e between their perimeters is 16m, find the sides of two squares.
Ans:- Let the side of smaller square be x m. Then T hen the perimeter of the square is 4x. Now perimeter of the second square is 4x + 16 Hence a side of the greater square is
4 x 16 4
=(x+4)m
Area of first square = x
2
Area of second square = ( x + 4 ) We are given Sum of areas = 400
∴ x 2 + x 2 + 8x + 16 = 400 ∴ 2x 2 + 8x – 384 = 0
2
=x
2
+ 8x + 16
∴ 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 or x = - 16 ∴ x = 12 Now side of a square cannot be negative Hence, side of smaller square = 12 m and a nd 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
360 n
2
side is
n
We are given 360 360
n2
n 360n
50
360
n2 360n
2
360
50 50 n 2
∴ 50n 2 – 360n + 360 = 0 Divide by 10
– 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 5n
2
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 34.
2
= 36
Tinu takes 9 days more than his father to do a certain certain piece 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 x
1 x
9
now, they complete the work in 6 days 6 6
1
1
x
x 9
1
x 9 x x(x
1
9)
6 (2 2x x 9)
x (x 9)
i.e. 12x – 54 = x
2
– 9x
– 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 i.e. x
2
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. numbers. We are given
x
2
+(x+1)
2
+ ( x + 2)
∴ x 2 + x 2 + 2x + 1 + x
2
+(x+3)
2
+ 4x + 4 + x
2 2
+(x+4)
2
= 1455
+ 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 side of one regular hexagon is larger than that of the other regular 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 3 3 2
3 3
( side ) 2
2
x2
Area of greater hexagon is 3 3 2
3 3
( side ) 2
2
( x 1) 2
We are given Product of areas = 243 3 3 2 27 4 x
2
3 3
x2
2
x2
( x 1) 2
( x 1) 2
( x 1)
x 2 ( x 1) 2
2
243
243 243 4 27
36
∴x(x+1)=±6 ∴ x ( x + 1 ) = 6 or x ( x + 1 ) = - 6
∴x2+x=6 or ∴x2+x–6=0 Consider x i.e. x
2
2
x
2
+x=-6
or
x
2
+x+6=0
+x–6=0
+ 3x – 2x – 6 = 0
i.e. x ( x + 3 ) – 2 ( x + 3 ) = 0 i.e. ( x – 2 ) ( x + 3 ) = 0
∴x–2=0 ∴x=2 or
or
x+3=0
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 here
b2
4ac
12
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 4
1 2
times the age of my son. Six years ago,
my age was twice the square of the th e 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 = 4
1 2
( x 2)
9 2
( x 2)
∴ present age of father = 9 ( x 2 ) 4 9x 18 18 4 9x 14 14 9 ( x 2) 2) 2 2 2 2 2 Age of son 6 years ago = ( x – 6 ) years 9 x 14 9x 14 12 9 x 26 6 Age of father 6 years ago = 2 2 2 We are given
years
9x 26 2
=2(x–6)
i.e. 9x – 26 = 4 ( x i.e. 9x – 26 = 4x
2
2
2
– 12x + 36 )
– 48x + 144
– 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. 4x
2
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 distance of 240 km km with with some some speed. If the speed 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, time
dist distan ance ce
240 240
speed
x
hours
Increased speed = ( x + 20 ) km / hr Time taken with this speed to cover 240 km will be We are given 240
240
x
x 20
240 ( x 20 )
2 240 x
x ( x 20 )
2
∴ 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
240 x 20
hours
∴ 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 express train takes 30 min less for a journey journey of 440 km, km, if if its its usual 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, time
dist distan ance ce
440 440
speed
x
hours
Increased speed = x + 8 Time taken with this speed to cover 440 km will be We are given 440
440
1
x
x 8
2
440 ( x 8 )
440 x
x ( x 8) 2
440 ( x 8 x )
30 min 1 2 x (x
i.e.
1 2
440 x
hr 8
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 = - 88 ∴ x = 80 or As speed cannot be negative, x ≠ -88 ∴ x = 80
8
x
1 2
440 x 8
hour
Thus usual speed of train = 80 km / hr 40.
The divisor and quotient of the number 6123 are are same and the remainder is half the divisor. Find the divisor.
Ans:- Let the divisor divisor = quotient = x Also, dividend 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 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 a nd 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 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 or x–4=0 ∴ 281x – 612 = 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 be filled filled up by two taps in 6 hours. hours. The smaller tap alone takes 5 hours more than the t he 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 = The work done by bigger tank in 1 hour = Thus, the work done by both in one hour =
1 x 1
x 5
1
1
x
x 5
Now, the tank fills in 6 hours by both the taps. That is both taps together fill
1 6
th part of the tank in one hour
1
1
1
x
x 5
6
i .e. 6 6
1
1
x
x 5
x 5 x
1
x ( x 5)
6 ( 2x 5 )
1
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 or x – 15 = 0 ∴x-2=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 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 We are given
2
+ 8x + 16 – x
2
= 8x + 16
area of footpath 8x
16
4 ( 8x
5
5 4
area of pool
x2
4 16 )
5x 2
∴ 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 or x–8=0 ∴ 5x + 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.
