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Question-1

Check whether the following equation is quadratic or not: x2 - 6x - 4 = 0.

Solution:
The degree of the equation is 2
∴ x2 -6x - 4 = 0 is quadratic equation.

Question-2

Check whether the following equation is quadratic or not: 3x2 – 7x  2 = 0.

Solution:
The degree of the equation is 2 
 3x2 – 7x  2 = 0 is quadratic equation.

Question-3

Check whether the following equation is quadratic or not:x3 - 6x2 + 2x -1 = 0.

Solution:
The degree of the equation is 3
 x3 - 6x2 + 2x -1 = 0 is not a quadratic equation.

Question-4

Check whether the following equation is quadratic or not: 7x = 2x2.

Solution:
The degree of the equation 2x2  7x = 0 is 2
 7x = 2x2 is a quadratic equation.

Question-5

Check whether the following equation is quadratic or not: x2 +  = 2(x 0).

Solution:
The given equation also written as
x4 – 2x2 + 1= 0
⇒The degree of the equation is 4

∴ x2 += 2(x 0) is not a quadratic equation.

Question-6

Check whether the following equation is quadratic or not: 3x2 – 4 = 0.

Solution:
The degree of the given equation is 2
∴ 3x2 – 4 = 0 is a quadratic equation.

Question-7

Check whether the following equation is quadratic or not: (x + 1)(x + 3) = 0.

Solution:
The degree of the given equation is 2
∴ (x + 1)(x + 3) = 0 is a quadratic equation.

Question-8

Check whether the following equation is quadratic or not: (2x + 1)(3x + 2) = 6(x – 1)(x – 2).

Solution:
By solving the given equation (2x + 1)(3x + 2) = 6(x – 1)(x – 2) we get,
6x2 + 7x + 2 = 6x2 – 18x + 12
25x – 10 = 0
 
The degree of the given equation is 1
∴ (2x + 1)(3x + 2) = 6(x – 1)(x – 2) is not a quadratic equation.

Question-9

Check whether the following equation is quadratic or not: x += x2(x 0).

Solution:
x += x2  x2 + 1 = x4
The degree of the given equation is 4.
∴ x += x2 (x 0) is not a quadratic equation.

Question-10

Check whether the following equation is quadratic or not: 16x2 – 3 = (2x + 5)(5x – 3).

Solution:
By solving the given equation we get,
6x2 – 19x + 12 = 0
∴ 16x2 – 3 = (2x + 5)(5x – 3) is a quadratic equation.

Question-11

Determine whether the given value of x is a solution of the given equation or not: 3x2 – 2x – 1 = 0; x = 1.

Solution:
3x2 – 2x – 1= 0
Let x = 1, 3(1)2 – 2(1)– 1 = 3 – 2 – 1 = 0.
Therefore the given value of x is a solution of the given equation.

Question-12

Determine whether the given value of x is a solution of the given equation or not: 2x2 – 6x + 3 = 0; x = .

Solution:
2x2 – 6x + 3 = 0
Let, x =  ⇒ 2()2 – 6() + 3 =  - 3 + 3 = .
Therefore the given value of x is not a solution of the given equation.

Question-13

Determine whether the given value of x is a solution of the given equation or not: (2x + 3)(3x - 2) = 0; x = .

Solution:
(2x + 3)(3x - 2) = 0;
Let x =  ⇒(2 × + 3)(3 × - 2) = ( + 3)(2 -2)
        = (+ 3) 0 = 0.
Therefore the given value of x is a solution of the given equation.

Question-14

Determine whether the given value of x is a solution of the given equation or not: x2 + x + 1 = 0; x = -1.

Solution:
x2 + x + 1 = 0
Let x = -1  ⇒ (-1)2 + (-1) + 1 = 1 – 1 + 1 = 1.
Therefore the given value of x is not a solution of the given equation.

Question-15

Determine whether the given values of x are solutions of the given equation or not:
x2 + 6x + 5 = 0; x = -1, x = -5.

Solution:
x2 + 6x + 5 = 0
Put x = -1
⇒ (-1)2 + 6(-1) + 5 = 1 – 6 + 5 = 0

Put x = -5

(-5)2 + 6(-5) + 5 = 25 - 30 + 5 = 0

Therefore the given value of x is a solution of the given equation.

Question-16

Determine whether the given values of x are solutions of the given equation or not: 9x2 - 3x - 2 = 0; x = -, x = .

Solution:
9x2 - 3x - 2 = 0
Put x = -
9(-)2 - 3(-) – 2 = 1 + 1 – 2 = 0
Put x =
9x2
 3x  2 = 9()2  3() – 2 = 4  2 – 2 = 0
Therefore the given value of x is a solution of the given equation.

Question-17

Determine whether the given values of x are solutions of the given equation or not:
(x + 4)(x - 5) = 0; x = -4, x = 5.

Solution:
Put x = -4
(x + 4)(x - 5) = (-4 + 4)(-4 -5) = 0(-9 ) = 0
Put x = 5
(x + 4)(x - 5) = (5 + 4)(5 - 5) = 9(0) = 0
Therefore the given value of x is a solution of the given equation.

Question-18

Determine whether the given values of x are solutions of the given equation or not:
(3x + 8)(2x + 5) = 0; x = 2, x = 2.

Solution:
Put x = 2 =
(3x + 8)(2x + 5) = (3
× + 8)(2 × + 5) = (8 + 8)(+5) = 16 × = ≠ 0
Put x = 2 =
(3x + 8)(2x + 5) = (3
× + 8)(2 × +5) = ( + 8)(5 + 5) =× 10 = 155 ≠ 0
Therefore the given value of x is not the solution of the given equation.

Question-19

Using factorization, find the roots of the quadratic equation: 9x2 - 16 = 0.

Solution:
9x2 - 16 = 0
(3x)2 - 42 = 0
(3x – 4)(3x + 4) = 0
x = 4/3, -4/3.

Question-20

Using factorization, find the roots of the quadratic equation: 64x2 – 9 = 0.

Solution:
64x2 – 9 = 0
(8x)2 – 32 = 0
(8x – 3)(8x + 3) = 0

 x = 3/8, -3/8.

Question-21

Using factorization, find the roots of the quadratic equation: (x  2)2 – 25 = 0.

Solution:
(x  2)2 – 25 = 0
(x
  2)2 – 52 = 0
(x – 2
 5)(x – 2 + 5) = 0
(x – 7)(x + 3) = 0
x = 7, 
 3.

Question-22

Using factorization, find the roots of the quadratic equation: (x + 5)2 – 36 = 0.

Solution:
(x + 5)2 – 36 = 0
(x + 5)2 – 62 = 0
(x + 5 – 6)(x + 5 + 6) = 0
(x – 1)(x + 11) = 0
x = 1, 
11

Question-23

Using factorization, find the roots of the quadratic equation: (2x + 3)2 = 81.

Solution:
(2x + 3)2 = 81
(2x + 3)2 – 92 = 0
(2x + 3 – 9) (2x + 3 + 9) = 0
(2x – 6) (2x + 12) = 0
x = 3, 
6.

Question-24

Using factorization, find the roots of the quadratic equation: y2-3 = 0
        [Hint: 3 =()2]
.

Solution:
y2 - 3 = 0
y2 – () 2 = 0
(y - )(y + ) = 0
y = , -.

Question-25

Using factorization, find the roots of the quadratic equation: a2z2-b2 = 0.

Solution:
a2z2 - b2 = 0
(az)2 - b2 = 0
(az - b)(az + b) = 0
z = b/a, - b/a
.

Question-26

Using factorization, find the roots of the quadratic equation: 3z-z2 = 0.

Solution:
3z - z2 = 0
z(3 – z) = 0
z = 0 or (3 - z) = 0
z = 0, z = 3.

Question-27

Using factorization, find the roots of the quadratic equation: 5z2-30 = 0.

Solution:
5z2 - 30 = 0
5(z2 – 6) = 0
z2 – ()2 = 0
(z )(z +) = 0
z = , -.

Question-28

Using factorization, find the roots of the quadratic equation: ax2-2abx = 0.

Solution:
ax2 - 2abx = 0
ax(x - 2b) = 0
ax = 0 or (x - 2b) = 0
x = 0 or x = 2b.

Question-29

Using factorization, find the roots of the quadratic equation: 4y2+4y+1 = 0.

Solution:
4y2 + 4y + 1 = 0
4y2 + 2y + 2y + 1 = 0
2y(2y + 1) + (2y + 1) = 0
(2y + 1)(2y + 1) = 0
(2y + 1) = 0 or (2y + 1) = 0
y = -1/2 or y = -1/2
.

Question-30

Using factorization, find the roots of the quadratic equation: y2-8y+16 = 0.

Solution:
y2 - 8y + 16 = 0
y2 - 4y – 4y + 16 = 0
y(y – 4) – 4(y – 4) = 0
(y - 4)(y - 4) = 0
y = 4 or y = 4.

Question-31

Using factorization, find the roots of the quadratic equation: z2-z+=0.

Solution:
z2 – z + = 0
z2 z - z + = 0
z(z ) - (z - ) = 0
(z )(z - ) = 0
z = or z = .

Question-32

Using factorization, find the roots of the quadratic equation: x2-x+1 = 0.

Solution:
x2 - x + 1 = 0
x2 - x – x + 1 = 0
x(x – 1) – (x – 1) = 0
(x – 1)(x – 1) = 0
x = 3 or x = 3
.

Question-33

Using factorization, find the roots of the quadratic equation: y2+2y + 3 = 0.

Solution:
y2 + 2y + 3 = 0
y2 + y + y + 3 = 0
y(y + ) + (y + ) = 0
(y +)(y + ) = 0
y = - or y = -.

Question-34

Using factorization, find the roots of the quadratic equation: x2-4qx+4q2=0.

Solution:
x2 – 4qx + 4q2 = 0
x2 – 2qx – 2qx + 4q2 = 0
x(x – 2q) – 2q(x – 2q) = 0
(x – 2q)(x – 2q) = 0
x = 2q or x = 2q
.

Question-35

Using factorization, find the roots of the quadratic equation: z2-2z-8 = 0.

Solution:
z2 - 2z - 8 = 0
z2 – 4z + 2z - 8 = 0
z(z – 4) + 2(z – 4) = 0
(z – 4)(z + 2) = 0
z = 4 or z = -2.

Question-36

Using factorization, find the roots of the quadratic equation: 6z2-5z-21 = 0.

Solution:
6z2 - 5z - 21 = 0
6z2 - 14z + 9z - 21 = 0
2z(3z – 7) + 3(3z – 7) = 0
(2z + 3)(3z – 7) = 0
(2z + 3) = 0 or (3z – 7) = 0
z = -3/2 or z = 7/3.

Question-37

Using factorization, find the roots of the quadratic equation: y2+3y-18 = 0.

Solution:
y2 + 3y - 18 = 0
y2 + 6y – 3y - 18 = 0
y(
y + 6)-3(y + 6) = 0
(y – 3)(y + 6) = 0
y = 3 or y = -6.

Question-38

Using factorization, find the roots of the quadratic equation: y2-3y-10 = 0.

Solution:
y2 - 3y - 10 = 0
y2 - 5y + 2y - 10 = 0
y(y – 5) + 2(y – 5) = 0
(y + 2)(y – 5) = 0
y = -2 or y = 5
.

Question-39

Using factorization, find the roots of the quadratic equation: 6y2-y-2 = 0.

Solution:
6y2 - y - 2 = 0
6y2 – 4y + 3y - 2 = 0
2y(3y – 2) + (3y – 2) = 0
(2y + 1)(3y – 2) = 0
2y + 1 = 0 or 3y – 2 = 0
y = -1/2 or y = 2/3
.

Question-40

Using factorization, find the roots of the quadratic equation: 9y2-3y-2 = 0.

Solution:
9y2-3y-2 = 0
9y2 - 6y + 3y - 2 = 0
3y(3y – 2) + (3y – 2) = 0
(3y + 1)(3y – 2) = 0
y = -1/3 or y = 2/3.

Question-41

Using factorization, find the roots of the quadratic equation: 5z2-3z-2 = 0.

Solution:
5z2 - 3z - 2 = 0
5z2 - 5z + 2z - 2 = 0
5z(z – 1) + 2(z – 1) = 0
(5z + 2)(z – 1) = 0
z = -2/5 or z = 1.

Question-42

Using factorization, find the roots of the quadratic equation: 2z2 +az-a2 = 0.

Solution:
2z2 + az - a2 = 0
2z2 + 2az - az - a2 = 0
2z(z + a) – a(z + a) = 0
(z + a)(2z – a) = 0
z = -a or z = a/2
.

Question-43

Using factorization, find the roots of the quadratic equation: 8x2-22x-21 = 0.

Solution:
8x2 - 22x - 21 = 0
8x2 + 6x - 28x - 21 = 0
2x(4x + 3) – 7(4x + 3) = 0
(2x - 7)(4x + 3) = 0
x = 7/2 or x = -3/4.

Question-44

Using factorization, find the roots of the quadratic equation: 5x2-+ = 0.

Solution:
5x2 - + = 0
5x2 - x - x + = 0
x(5x – 1) - (5x – 1) = 0
(5x – 1) (x - ) = 0
x = 1/5 or x =
.

Question-45

Using factorization, find the roots of the quadratic equation: x2-+=0.

Solution:
x2 - += 0
x2 - x - x + = 0
x(x - ) - (x - ) = 0
(x - )(x - ) = 0
x = or x = .

Question-46

Using factorization, find the roots of the quadratic equation: x2 - x-=0.

Solution:
x2 - x - = 0
x2 + x - x - = 0
x(x + ) - (x + ) = 0
(x + )(x - ) = 0
x = - or x = .

Question-47

Using factorization, find the roots of the quadratic equation: 3x2+10 = 11x.

Solution:
3x2 + 10 = 11x
3x2 - 11x + 10 = 0
3x2 - 6x – 5x + 10 = 0
3x(x - 2) – 5(x – 2) = 0
(3x – 5)(x - 2) = 0
x = 5/3 or x = 2
.

Question-48

Using factorization, find the roots of the quadratic equation: x+= -4, (x 0).

Solution:
x + = -4, (x 0)
x2 + 4 = - 4x

x2 + 4x + 4 = 0
x2 + 2x + 2x + 4 = 0
x(x + 2) + 2(x + 2) = 0
(x + 2)(x + 2) = 0
x = - 2 or x = - 2.

Question-49

Using factorization, find the roots of the quadratic equation: abx2+(b2-ac)x - bc = 0  

Solution:
abx2 + (b2 - ac)x - bc = 0
abx2 + b2x - acx - bc = 0
bx(ax + b) – c(ax + b) = 0
(ax + b)(bx – c) = 0
x = -b/a or x = c/b.

Question-50

Using factorization, find the roots of the quadratic equation: += 3(x 2, 4)

Solution:
Multiplying throughout by (x - 2)(x - 4), we get
3[(x - 1)(x - 4) + (x - 2)(x - 3)] = 10(x - 2)(x – 4)
3(x2 - 5x + 4 + x2 - 5x + 6) = 10(x2 - 6x + 8)
3(2x2 - 10x + 10) = 10(x2 - 6x + 8)
6x2 - 30x + 30 = 10x2 - 60x + 80
4x2 - 30x + 50 = 0
2x2 - 15x + 25 = 0
2x2 – 10x - 5x + 25 = 0
2x(x – 5) – 5(x – 5) = 0
(2x - 5)(x - 5) = 0
x = 5/2 or x = 5.

Question-51

Using factorization, find the roots of the quadratic equation: = , x ≠ 3,  5

Solution:
Multiplying throughout by (x  3)(x + 5), we get
6[(x + 5) – (x - 3)] = (x
 3)(x + 5)
6(x + 5 – x + 3) = (x2 + 2x
– 15)
6(8) = (x2 + 2x
– 15)
48 = (x2 + 2x
– 15)
x2 + 2x
– 63 = 0
x2 + 9x
– 7x  63 = 0
x(x + 9) 
– 7(x + 9) = 0
(x + 9)(x
– 7) = 0
x = 
– 9 or x = 7.

Question-52

Using factorization, find the roots of the quadratic equation: = 3, (x 1,  2).

Solution:
Multiplying throughout by (x  1)(x + 2), we get
(x + 1)(x + 2) + (x
 1)(x  2) = 3(x  1)(x + 2)
(x2 + 3x + 2) + (x2
 3x + 2) = 3(x2 + x  2)
2x2 + 4 = 3x2 + 3x – 6
x2 + 3x – 10 = 0
x2 + 5x – 2x – 10 = 0
x(x + 5) – 2(x + 5) = 0
(x + 5)(x – 2) = 0
x =
 5 or x = 2.

Question-53

Determine whether the given quadratic equation has root(s). If so, find the root(s). y2 – 4y 1 = 0.

Solution:
a = 1, b = 4 and c = 1
Δ = b2 – 4ac = (4)2  4(1)(1) = 16 + 4 = 20
Therefore the quadratic equation has real roots.


Therefore the roots are 2
√5 and 2 +√5.

Question-54

Determine whether the given quadratic equation has root(s). If so, find the root(s). y2 – 6y+2=0.

Solution:
a = 1, b = 6 and c = 2
Δ = b2 – 4ac = (6)2  4(1)(2) = 36  8 = 28
Therefore the quadratic equation has real roots.


Therefore the roots are 3 + 
√7 and 3 – √7.

Question-55

Determine whether the given quadratic equation has root(s). If so, find the root(s). y2 + 8y + 4 = 0.

Solution:
a = 1, b = 8 and c = 4
Δ = b2 – 4ac = (8)2 – 4(1)(4) = 64 – 16 = 48
Therefore the quadratic equation has real roots.


Therefore the roots are
–4 – 2√3 and –4 +2√3.

Question-56

Determine whether the given quadratic equation has root(s). If so, find the root(s). y2+ y – 1  = 0.

Solution:
a = 1, b =  and c = -1
Δ = b2 – 4ac = ()2 – 4(1)(-1) = + 4 = =
Therefore the quadratic equation has real roots.


Therefore the roots are and .

Question-57

Determine whether the given quadratic equation has root (s). If so, find the root (s). 2y2 + 5y -3 = 0.

Solution:
a = 2, b = 5 and c = 3
Δ = b2 – 4ac = 52 – 4(2)(3) = 25 + 24 = 49
Therefore the quadratic equation has real roots.


Therefore the roots are
3 and 1/2.

Question-58

Determine whether the given quadratic equation has root(s). If so, find the root(s) –2y2 + y + 1 = 0.

Solution:
a = 2, b = 1 and c = 1
Δ = b2 – 4ac = 12 – 4(2)(1) = 1 + 8 = 9
Therefore the quadratic equation has real roots.


Therefore the roots are 1 and
1/2.

Question-59

Determine whether the given quadratic equation has root (s). If so, find the root (s): 3y2 + 9y + 4 = 0.

Solution:
a = 3, b = 9 and c = 4
Δ = b2  4ac = 92  4(3)(4) = 81  48 = 33
Therefore the quadratic equation has real roots.

Therefore the roots are and .

Question-60

Determine whether the given quadratic equation has root(s). If so, find the root(s): 5y2 – 2y – 2 = 0.

Solution:
a = 5, b = 2 and c = 2
Δ= b2 – 4ac = (2)2 – 4(5)(2) = 4 + 40 = 44
Therefore the quadratic equation has real roots.


Therefore the roots are and .

Question-61

Determine whether the given quadratic equation has root (s). If so, find the root (s): 7z2 + 8z + 2 = 0.

Solution:
a = 7, b = 8 and c = 2
Δ = b2 – 4ac = (8)2 – 4(7)(2) = 64 – 56 = 8
Therefore the quadratic equation has real roots.


Therefore the roots are and .

Question-62

Determine whether the given quadratic equation has root (s). If so, find the root (s): z2 z – 3 = 0.

Solution:
a = 1, b = and c = 3
Δ = b2 – 4ac = ()2 – 4(1)(-3) = + 12 = =
Therefore the quadratic equation has real roots.


Therefore the roots are –2 and 3/2.

Question-63

Determine whether the given quadratic equation has root(s). If so, find the root (s): z2 + 2z – 8 = 0.

Solution:
a = 1, b = 2 and c = 8
Δ = b2 – 4ac = (2)2 – 4(1)(8) = 4 + 32 = 36
Therefore the quadratic equation has real roots.


Therefore the roots are –4 and 2.

Question-64

Determine whether the given quadratic equation has root(s). If so, find the root(s): z2 – 6z + 4 = 0.

Solution:
a = 1, b = 6 and c = 4
Δ = b2 – 4ac = (6)2 – 4(1)(4) = 36  16 = 20
Therefore the quadratic equation has real roots.


Therefore the roots are and .

Question-65

Determine whether the given quadratic equation has root(s). If so, find the root(s): 6x2 + x – 2 = 0.

Solution:
a = 6, b = 1 and c = 2
Δ = b2 – 4ac = (1)2 – 4(6)(2) = 1 + 48 = 49
Therefore the quadratic equation has real roots.


Therefore the roots are –2/3 and
.

Question-66

Determine whether the given quadratic equation has root(s). If so, find the root(s):4x2 + 12x + 9 = 0.

Solution:
a = 4, b = 12 and c = 9
Δ = b2  4ac = 122 – 4(4)(9) = 0
Therefore the quadratic equation has real and equal roots.

 
∴    and

Therefore the roots are –3/2 and –3/2. 

Question-67

Determine whether the given quadratic equation has root(s). If so, find the root(s):2x2 + 5+ 16 = 0.

Solution:
a = 2, b = 5and c = 16
Δ = b2 – 4ac = (5)2 – 4(2)(16) = 75 – 128 =  53
Therefore the quadratic equation has no real roots.

Question-68

Determine whether the given quadratic equation has root(s). If so, find the root(s): 3x2 + 2 5 = 0.

Solution:
a = 3, b = 2and c = -5
Δ = b2 – 4ac = (2)2 – 4(3)(-5) = 20 + 60 = 80
Therefore the quadratic equation has real roots.


Therefore the roots are -and .

Question-69

Determine whether the given quadratic equation has root(s). If so, find the root(s): 3y2 + (6 + 4a)y + 8a = 0.

Solution:
a = 3, b = (6 + 4a) and c = 8a
Δ = b2 – 4ac = (6 + 4a)2 – 4(3)(8a) = 36 + 16a2 + 48a – 96a
   = 16a2 - 48a + 36 = (4a
 6)2
Therefore the quadratic equation has real roots.


Therefore the roots are and
2.

Question-70

Determine whether the given quadratic equation has root(s). If so, find the root(s). (x + 4)(x + 5) = 3(x + 1)(x + 2) + 2x.

Solution:
x2 + 9x + 20 = 3(x2 + 3x + 2) + 2x
x2 + 9x + 20 = 3x2 + 9x + 6 + 2x
2x2 + 2x
14 = 0
a = 2, b = 2 and c = 
14
Δ = b2 – 4ac = 22 – 4(2)(-14) = 4 + 112 = 116
Therefore the quadratic equation has real roots.


Therefore the roots are and .

Question-71

Determine the value(s) of p for which the given quadratic equation has real roots: px2 + 4x + 1 = 0.

Solution:
Given the quadratic equation has real roots, 
∴ Δ > 0
Here, a = p, b = 4 and c = 1
b2 - 4ac > 0
(4)2 - 4(p)(1) > 0
16 - 4p > 0
16 > 4p
4 > p
... p < 4
.

Question-72

Determine value(s) of p for which the given quadratic equation real roots: 2x2 + 3x + p = 0.

Solution:
a = 2, b = 3 and c = p
Given, Δ > 0
⇒ b2 - 4 ac > 0
(3)2 - 4(2)(p) > 0
9 - 8p > 0
9 > 8p
9/8 > p
... p < 9/8.

Question-73

Determine value(s) of p for which the given quadratic equation 5px2 - 8x + 2 = 0 has real roots:

Solution:
a = 5p, b = -8 and c = 2
Δ = b2 – 4ac = (-8)2 – 4(5p)(2) = 64 – 40p
Δ   0
Therefore 64 – 40p
0
64
40p
p
64/40 8/5 1.
∴ p  1.

Question-74

Determine the value(s) of p for which the given quadratic equation has real roots: 4x2 + 8x - p = 0.

Solution:
a = 4, b = 8 and c = -p
Δ = b2 – 4ac = 82 – 4(4)(-p) = 64 + 16p
Given, Δ  0
64 + 16p
0
4 + p
0
p
³  -4.

Question-75

Divide 12 into two parts such that their product is 32.

Solution:
Let one part be x.
Then the other part is 12 - x .

Condition: Product of two parts is 32.
Therefore x(12 - x) = 32
12x – x2 = 32
- x2 + 12x – 32 = 0
- x2 + 8x + 4x – 32 = 0
- x(x - 8) + 4(x - 8) = 0
(-x + 4)(x - 8) = 0
(-x + 4) = 0 or (x - 8) = 0
x = 4 or x = 8
Therefore the two products are 4 and 8.

Checking:
4(8) = 32.

Question-76

Divide 12 into two parts such that the sum of their squares is 74.

Solution:
Let one part be x.
Then the other part is 12 - x .

Condition:
Sum of the squares of two parts is 74.
Therefore x2 + (12 - x)2 = 74
x2 + (12 - x)2 = 74
x2 + 144 - 24x + x2 = 74
2x2 - 14x – 10x + 70 = 0
2x(x – 7) – 10(x – 7) = 0
(2x – 10)(x – 7) = 0
x = 10/2 or x = 7
x = 5 or x = 7
Therefore the two products are 5 and 7.

Checking:
52 + 72 = 25 + 49 = 74.

Question-77

The sum of a number and its reciprocal is , find the number(s).

Solution:
Let the number is x. Then its reciprocal is 1/x.
x + =
3x2 + 3 = 10x
3x2 - 10x + 3 = 0
3x2 - 9x - x + 3 = 0
3x(x - 3) - (x - 3) = 0
(x - 3)(3x - 1) = 0
x = 3 or x = 1/3

∴ The numbers are 3 and 1/3.
Checking :
.

Question-78

One side of a rectangle exceeds its other side by 2 cm. If its area is 195 cm2, determine the sides of the rectangle.

Solution:
Let the breadth of the rectangle be x cm.
Then the length of the rectangle is (x + 2) cm.
Area of the rectangle = length
× breadth = (x + 2) × x = x2 + 2x
x2 + 2x = 195
x2 + 2x – 195 = 0
x2 + 15x – 13x – 195 = 0
x(x + 15) – 13(x + 15) = 0
(x
– 13)(x + 15) = 0
x = 13 or x = 
15
Therefore the sides of the rectangle is 13 cm and 15 cm.

Checking :
13 cm
× 15 cm = 195 cm2.

Question-79

The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2, find the fraction.

Solution:
Let the numerator of the fraction be x.
Then the denominator of the fraction is 2x + 1.
The required fraction is .
+= 2
= 2
=
21(5x2 + 4x + 1) = 58(2x2 + x)
105x2 + 84x + 21 = 116x2 + 58x
11x2 - 26x - 21 = 0
11x2 - 33x + 7x - 21 = 0
11x(x – 3) + 7(x – 3) = 0
(11x + 7)(x – 3) = 0
11x + 7 = 0 or x – 3 = 0
x = -7/11 or x = 3
The required fraction is .
Checking:

.

Question-80

The product of Ramu’s age (in years) five years ago and his age(in years) nine years later is 15. Determine Ramu’s present age.

Solution:
Ramu’s present age = x years
Ramu’s age five years ago = (x - 5) years
Ramu’s age nine years later = (x + 9) years

Condition: Product of Ramu’s age five years ago and his age nine years later is 15.
(x - 5)(x + 9) = 15
x2 + 4x - 45 = 15
x2 + 4x - 45 - 15 = 0
x2 + 4x - 60 = 0
x2 + 10x - 6x - 60 = 0
x(x + 10) - 6(x + 10) = 0
(x + 10)(x - 6) = 0
x = -10 or x = 6

∴ Ramu’s present age is 6 years
 Checking:
Ramu’s age five years ago = (6 - 5) year = 1 year
Ramu’s age nine years later = (6 + 9) years = 15 years

Checking:
1(15) = 15.

Question-81

A two-digit number is such that the product of its digits is 12. When 36 is added to this number, the digits interchange their places. Find the number.

Solution:
Let the units digit be x.
Then the ten’s digit is
.
Therefore the required number is
× 10 + x = + x
The new number after interchanging their places is 10x + .

Condition: When 36 is added to this number, the digits interchange their places.
+ x + 36 = 10x + .
120 + x2 + 36x = 10x2 + 12
9x2 - 36x – 108 = 0
x2 - 4x – 12 = 0
x2 - 6x + 2x – 12 = 0
x(x – 6) + 2(x – 6) = 0
(x + 2)(x – 6) = 0
(x + 2) = 0 or (x – 6) = 0
x = -2 or x = 6
Therefore the required number is  + 6 = 20 + 6 = 26.
Checking:
Old number = 26.
New number = 62.
New number = 26 + 36 = 62.

Question-82

Find two consecutive odd natural numbers, the sum of whose squares is 202.

Solution:
Let the two consecutive odd natural numbers be x and x + 2.
x2 + (x + 2)2 = 202
x2 + x2 + 4x + 4 = 202
2x2 + 4x + 198 = 0
x2 + 2x + 99 = 0
x2 + 11x – 9x + 99 = 0
x(x + 11) – 9(x + 11) = 0
(x – 9)(x + 11) = 0
x – 9 = 0 or x + 11 = 0
x = 9 or x = -11
Therefore the two consecutive odd natural numbers are 9 and 11.

Checking :
92 + 112 = 81 + 121 = 202.


Question-83

Determine two consecutive even positive integers, the sum of whose squares is 100.

Solution:
Let the two consecutive odd natural numbers be x and x + 2.
x2 + (x + 2)2 = 100
x2 + x2 + 4x + 4 = 100
2x2 + 4x
 96 = 0
x2 + 2x
 48 = 0
x2 + 8x – 6x
 48 = 0
x(x + 8) – 6(x + 8) = 0
(x – 6)(x + 8) = 0
(x – 6) = 0 or (x + 8) = 0
x = 6 or x = 
 8
The two consecutive even positive integers is 6 and 8.

Checking:
62 + 82 = 36 + 64 = 100.

Question-84

The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is .

Solution:
Let one number be x.
Then the other number is 8 – x.
 + =
=
=

15 = x(8 - x)
15 = 8x - x2
 x2 - 8x + 15 = 0
x2 - 5x – 3x + 15 = 0
x(x – 5) – 3(x – 5) = 0
(x – 3)(x – 5) = 0
(x – 3) = 0 or (x – 5) = 0
x = 3 or x = 5
Therefore the numbers are 3 and 5.

Checking:
+ ==.

Question-85

There are three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154. What are the integers?.

Solution:
Let the three consecutive positive integers be x, x + 1 and x + 2.
x2 + (x + 1)(x + 2) = 154
x2 + x2 + 3x + 2 = 154
2x2 + 3x - 152 = 0
2x2 + 19x – 16x
– 152 = 0
x(2x + 19) – 8(2x + 19) = 0
(x – 8)(2x + 19) = 0
x = 8 or x =
  19/2
∴ The integers are 8, 9 and 10.
Checking:
82 + 9
× 10 = 64 + 90 = 154.

Question-86

The hypotenuse of a grassy land in the shape of a right triangle is 1 metre more than twice the shortest side. If the third side is 7 metres more than the shortest side, find the sides of the grassy land.

Solution:
Let the shortest side be x.
Hypotenuse = (2x + 1) m
Third side = (x + 7) m
x2 + (x + 7)2 = (2x + 1)2
x2 + x2 + 14x + 49 = 4x2 + 4x + 1
2x2 - 10x - 48 = 0
2x2 - 16x + 6 x
 48 = 0
2x(x – 8) + 6(x – 8) = 0
(2x + 6)(x – 8) = 0
2(x + 3) = 0 or x – 8 = 0

∴ x = -3 or x = 8
The sides of the grasy land are 8, 15 and 17.

Checking:
82 = 64
152 = 225
172 = 289
82 + 152 = 64 + 225 = 289 = 172.

Question-87

The difference of the squares of two numbers is 45. The squares of the smaller number is 4 times the larger number. Determine the numbers.

Solution:
Let one number be x.
Then the other number is
(x2 - 45).
Condition: The squares of the smaller number is 4 times the larger number.
(x2
  45) = 4x
x2
  4x – 45 = 0
x2
 9x + 5x – 45 = 0
x(x – 9) + 5(x – 9) = 0
(x + 5)(x – 9) = 0
(x + 5) = 0 or (x – 9) = 0
x =
5 or x = 9
Therefore the numbers are 9 and

Checking:
62 = 36 = 4(9).

Question-88

The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.

Solution:
Let the present age of the father be x years.
Then the present age of the son is (45 – x) years.
Five years ago father’s age = x – 5 years
Five years ago son’s age = (45 – x) – 5 years = (40 – x) years

Condition: The product of their ages was 124.
(x - 5)(40 - x) = 124
x(40 - x) - 5(40 - x) = 124
40x – x2 - 200 + 5x = 124
- x2 - 324 + 45x = 0
x2 - 45x + 324 = 0
x2 - 36x – 9x + 324 = 0
x(x – 36) – 9(x – 36) = 0
(x – 9)(x – 36) = 0
x – 9 = 0 or x – 36 = 0
x = 9 or x = 36
Fathers age is 36 years and sons age is 9 years.

Checking:
a) 36 + 9 = 45.
b) 31
× 4 = 124.

Question-89

Out of a number of Saras birds, one fourth the number are moving about in lotus plants; th coupled(along) with th as well as 7 times the square root of the number move on a hill; 56 birds remain in vakula trees. What is the total number of birds?.

Solution:
Let the total number of birds be x.

Condition: One fourth the number are moving about in lotus plants; 
th coupled(along) with th as well as 7 times the square root of the number
move on a hill; 56 birds remain in vakula trees.
++ 7
x ++ 56 = x
++ 7
x + 56 = x
2x + 9x + 18
× 7x + 56 × 18 = 18x
11x + 126
x + 1008 = 18x
126
x + 1008 = 7x
18
x + 144 = x
(18
x)2 = (x – 144)2
324x = x2 – 288x + 20736
x2 – 612x + 20736 = 0
x2 - 576x – 36x + 20736 = 0
x(x – 576) – 36(x – 576) = 0
(x – 36)(x – 576) = 0
(x – 36) = 0 or (x – 576) = 0
x = 36 or x = 576
Since the remaining birds where 56 so we neglect the value x = 36.
Therefore the total number of birds are 576.

Checking:
++ 7
576 ++ 56 = 64 + 144 + 7576 + 144 + 56
                                        = 64 + 144 + 7
× 24 + 144 + 56
                                        = 64 + 144 + 168 + 144 + 56
                                        = 576
.

Question-90

A farmer wishes to grow a 100 m2 rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side-fence. Find the dimensions of his garden.
   [Hint: If the length of one side is x metres, the other side will be (30 - 2x) m. Therefore, x(30 - 2x) = 100]

Solution:
Let the length of one side is x m.
Then other side will be(30 - 2x) m.
Therefore, x(30 - 2x) = 100
30x - 2x2 = 100
-2x2 + 30x – 100 = 0
x2 
 15x + 50 = 0
x2 
 10x – 5x + 50 = 0
x(x – 10) – 5(x – 10) = 0
(x – 5)(x – 10) = 0
(x – 5) = 0 or (x – 10) = 0
x = 5 or x = 10

If one side is 5 m then the other side is 20 m, so the three side of the fence would be 25 m.
If one side is 10 m then the other side is 10 m, so the three side of the fence would be 30 m.
Since the fence wire is 30 m we consider the dimensions of his garden is 10 m and 10 m.

Checking:
a) Area of the rectangular vegetable garden = 10 m × 10 m = 100 m2.

Question-91

The hypotenuse of a right triangle is 310 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be
9 5 cm. How long are the legs of the triangle?

Solution:
Hypotenuse =cm
Let the smaller leg be x cm

Then the longer leg  = ±  
                              = ± = ±
New Smaller leg = 3x cm
New Longer leg = 2cm
New hypotenuse = cm
()2 = (3x)2 + (
± 2)2
81
× 5 = 9x2 + 4(90 – x2)
405 = 9x2 + 360 – 4x2
45 = 5x2
x2 – 9 = 0
(x – 3)(x + 3) = 0
(x – 3) = 0 or (x + 3) = 0
x = 3 or x = -3
The smaller leg is 3cm and the longer leg is == = 9 cm.

Checking:
32 + 92 = 9 + 81 = 90 = ()2
.

Question-92

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm2. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Determine the sides of the two squares.

Solution:
Let the area of the smaller square be x2 cm2.
Condition 1: If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm2.
Then the area of the larger square = (14 + 2x2) cm2

Condition 2: If twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2.
⇒ 2(14 + 2x2) + 3 x2 = 203
28 + 4x2 + 3 x2 = 203
7x2 = 175
x2 = 25
x =
± 5 cm
Therefore the sides of smaller square is 5 cm.
And the sides of the larger squares is 14 + 2
× 52 = 14 + 50 = 64 cm.
The sides of the two squares is 5 cm and 64 cm.

Checking:
a) 64 - 2(52) = 64 – 250 = 14
b) 2(64) + 3(52) = 128 + 75 = 203
.

Question-93

A journey of 192 km from Mumbai to Pune takes 2 hours less by a superfast train than that by an ordinary passenger train. If the average speed of the slower train is 16 km/h less than that of the faster train, determine their a average speeds.

Solution:
Let average speed of fast train = x
Therefore, average speed of slow train = x – 16
Time taken for journey by fast train = 192/x hr.
Time taken for journey by slow train = 192/(x - 16) hr
= + 2
192x = 192(x – 16) + 2x(x - 16)
192x = 192x – 3072 + 2x2 – 32x
- 3072 + 2x2 – 32x = 0
2x2 – 32x – 3072 = 0
x2 – 16x – 1536 = 0
x2 – 48x + 32x – 1536 = 0
x(x – 48) + 32(x – 48) = 0
(x + 32)(x – 48) = 0
(x + 32) = 0 or (x – 48) = 0
x = -32 or x = 48
Therefore the average speed of the fast train = 48 km/hr
and the speed of the slow train = x – 16 = 32 km/hr.
Checking:
= + 2
= + 2
6 = 4 + 2
6 = 6.

Question-94

A passenger train takes 2 hours less for a journey of 300 km, if its speed is increased by 5 km/h from its usual speed. Find its usual speed.

Solution:
Let the usual speed be x km/hr.
Distance travelled by a passenger train = 300 km.
We know that
Time = Distance / Speed
Time taken by a passenger train to travel with the speed of x km/hr =  hr
Time taken by a passenger train to travel with the speed of (x + 5) km/hr=  hr
By the given condition
 – 2 = 

 – = 2

300  = 2
300
   = 2
300
   = 2
300[5/(x2 + 5x)] = 2
1500 = 2(x2 + 5x) 
1500 = 2x2 + 10x
On rearranging
2x2 + 10x
– 1500 = 0
x2 + 5x
 – 750 = 0
x2 + 30x
 – 25x – 750 = 0
x(x + 30)
 – 25(x + 30) = 0
(x - 25)(x + 30) = 0
x - 25 = 0 or x + 30 = 0
x = 25 or x =
 – 30 
Since x is the usual speed it cannot be negative.
Therefore, we reject the solution x = 
30. 
Thus the usual speed is 25 km/hr.

Checking:
Time taken by a passenger train to travel with the speed of x km/hr = =  hr
Time taken by a passenger train to travel with the speed of (x + 5) km/hr = =  hr
Thus, a passenger train takes 2 hours less for a journey if its speed is increased by 5 km/hr from its usual speed.

Question-95

A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find the speeds of the two trains.

Solution:
Let the average speed of fast train be x km/hr.
Therefore, average speed of slow train is (x - 10) km/hr.
Time taken for journey by fast train = 
 hr

Time taken for journey by slow train =  hr
By the given condition
Now fast train takes 3 hours less than a slow train.

i.e., slow train time - fast train time = 3
Therefore, 
3 =  
 -Multiplying both sides by x(x - 10) we get
3x(x -10)= 600x - 600(x-10)
3x2 - 30x = 600x - 600x + 6000
On rearranging we have
-3x2 + 30x + 600x - 600x + 6000 = 0
-3x2 + 30x + 6000 = 0 [Dividing through out by 3]
-x2 + 10x + 2000 = 0
x2 - 10x - 2000 = 0
x2 - 50x + 40x - 2000 = 0
x(x - 50) + 40(x - 50) = 0
(x + 40)(x - 50) = 0
x + 40 = 0 or x - 50 = 0
x = - 40 or x = 50
Since the speed of the train cannot be negative. 
Therefore we neglect the solution x = - 40.
Thus, the average speed of the fast train is 50 km/hr
and the average speed of the slow train is 40 km/hr.

Checking:
Time taken for journey by fast train = 
  (600/50) = 12 hr

Time taken for journey by slow train =(600/40) = 15 hr
Thus, fast train takes 3 hours less for the journey and slow train speed is 10km/hr less than the fast train.

Question-96

Check whether the following equation is quadratic or not: x2 - 6x - 4 = 0.

Solution:
x2 - 6x - 4 = 0 is quadratic equation.

Question-97

Check whether the following equation is quadratic or not: 3x2  7x – 2 = 0.

Solution:
3x2  7x – 2 = 0 is quadratic equation.

Question-98

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm2. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Determine the sides of the two squares.

Solution:
Let the area of the smaller square be x2 cm2.
Condition 1: If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm2.
Then the area of the larger square = (14 + 2x2) cm2

Condition 2: If twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2
2(14 + 2x2) + 3 x2 = 203
28 + 4x2 + 3 x2 = 203
7x2 = 175
x2 = 25
x =
± 5 cm
Therefore the sides of smaller square is 5 cm.
And the sides of the larger squares is 14 + 2
× 52 = 14 + 50 = 64 cm.
The sides of the two squares is 5cm and 64 cm.

Checking:
a) 64 - 2(52) = 64 – 250 = 14
b) 2(64) + 3(52) = 128 + 75 = 203
.

Question-99

A journey of 192 km from Mumbai to Pune takes 2 hours less by a superfast train than that by an ordinary passenger train. If the average speed of the slower train is 16 km/h less that that of the faster train, determine their a average speeds.

Solution:
Let average speed of fast train = x
Therefore, average speed of slow train = x – 16
Time taken for journey by fast train = 192/x hr
Time taken for journey by slow train = 192/(x - 16) hr
= + 2
192x = 192(x – 16) + 2x(x
  16)
192x = 192x – 3072 + 2x2 – 32x

 3072 + 2x2 – 32x = 0
2x2 – 32x – 3072 = 0
x2 – 16x – 1536 = 0
x2 – 48x + 32x – 1536 = 0
x(x – 48) + 32(x – 48) = 0
(x + 32)(x – 48) = 0
(x + 32) = 0 or (x – 48) = 0
x = -32 or x = 48
Therefore the average speed of the fast train = 48 km/hr
and the speed of the slow train = x – 16 = 32 km/hr.

Checking:
= + 2
= + 2
6 = 4 + 2
6 = 6.

Question-100

Solve the quadratic equation by factorization (3x + 1) (x  1) = 15.

Solution:
(3x + 1) (x  1) = 15
=>3x2 – 2x – 1 = 15
=>3x2 – 2x – 1 – 15 = 0
=>3x2 – 2x – 16 = 0
=>(3x   8)(x + 2) = 0 x = 2, x = .

Question-101

Given that -3 and m are the roots of the quadratic equation kx2 + 5x – 3 = 0. Find the value of m and of k.

Solution:
Let us consider the given equation kx2 + 5x – 3 = 0 in the form ax2 + bx + c = 0.
Here a = k, b = 5, c = -3
The two roots are -3 and m. Thus sum of the roots are  

and the product of the roots are
Sum of the roots =  =
-3 + m =

-3k + km = -5 -------(1)

Product of the roots =  =

-3m =


=> -3km = -3

=> km = 1

Substitute km = 1 in (1)

-3k + 1 = -5

-3k = -5 – 1 = -6
k = 2.
Thus by substituting k = 2 in km = 1 we get m = 
The value of k = 2 and m = 

Question-102

Find the sum and product of the roots of the quadratic equation in terms of k, 4x2 + kx + k – 1 = 0.

Solution:
Let us consider the given equation 4x2 + kx + k – 1 = 0 in the form ax2 + bx + c = 0.
Here a = 4, b = k, c = k - 1 Sum of the roots = = Product of the roots = = .

Question-103

If the roots of the quadratic equation hx2 + 21x + 10 = 0 where h 0 are in the ratio 2:5, find the possible values of h.

Solution:
Since the roots of the equation are in the ratio 2:5, the roots are 2α and 5α .
In the given equation hx2 + 21x + 10 = 0.

a = h, b = 21, c = 10 Sum of the roots = =

2α + 5α =

7α =

h = = ------- (1)

Product of the roots =  

(2α )(5α) =  

10α 2 = 

h =     --------(2)
Comparing (1) and (2) we get, =

-3 = α =

Substituting α = in (1) we get, h = 9.

Question-104

Without solving the quadratic equation, determine the nature of roots of the equation x2 + 3x +7 = 0.

Solution:
In the given equation x2 + 3x + 7 = 0. Here a = 1, b = 3, c = 7.

b2 – 4ac = 32 – 4(1)(7)

            = 9 – 28

            = -19

Since b2 – 4ac < 0, the given equation has no real roots.

Question-105

Find the value of "p", if the quadratic equation p2x2 – 12x + p + 7 = 0 has the root .

Solution:
In the given equation p2x2 – 12x + p + 7 = 0, a = p2, b = -12, c = p + 7.

If one root is , let the other root be α . Sum of the roots = =

+ α = α = -

   = -------- (1)

Product of the roots = =

× α =

α = ×

   = -----------(2)

Comparing (1) and (2)

=

3(24 – 3p2) = 4p + 28

72 – 9p2 = 4p + 28

Thus we get 9p2 + 4p – 44 = 0

Now by applying the formula: x =

In this problem, a = 9, b = 4, c = -44

p = =

      = =

      =

Thus p = or

          = or

          = 2 or The value of p is 2 or .


Question-106

Form the quadratic equation whose roots are and .

Solution:
The quadratic equation is x2 – (sum of the roots) x + (product of the roots) = 0

Sum of the roots = + = =

Product of the roots = × =

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

x2 x + = 0

15x2 + 19x + 6 = 0.


Question-107

If H and K are the roots of the equation 2x2 + 7x = 4. Obtain the value of
(i) H + K
(ii) HK
(iii) discriminant.

Solution:
In the given problem a = 2, b = 7 and c = -4. Thus if H and K are the roots of the equation:
(i) Sum of the roots = H + K
                                  = =

(ii) Product of the roots = HK
                                 = = = -2

(iii) b2 – 4ac = 49 – 4(2)(-4)

                  = 49 + 32

                  = 81.

Question-108

Frame the quadratic equations whose only root is -6.

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

Sum of the roots = -6 + (-6) = -12

Product of the roots = -6 × -6 = 36 x2 + 12x + 36 = 0.

Question-109

Determine the nature of roots of the quadratic equation x2 – 6x + 9 = 0.

Solution:
In this equation a = 1, b = -6 and c = 9
b2 – 4ac = (-6)2 – 4(1)(9)

            = 36 – 36

            = 0
Since b2 – 4ac = 0, the equation has two real and equal roots.




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