# Question-1

**Check whether the following equation is quadratic or not: x**

^{2 }- 6x - 4 = 0.**Solution:**

The degree of the equation is 2

âˆ´ x

^{2 }-6x - 4 = 0 is quadratic equation.

# Question-2

**Check whether the following equation is quadratic or not: 3x**

^{2}â€“ 7x â€“ 2 = 0.**Solution:**

The degree of the equation is 2

**âˆ´**3x

^{2}â€“ 7x â€“ 2 = 0 is quadratic equation.

# Question-3

**Check whether the following equation is quadratic or not:x**

^{3 }- 6x^{2 }+ 2x -1 = 0.**Solution:**

The degree of the equation is 3

**âˆ´**x

^{3 }- 6x

^{2 }+ 2x -1 = 0 is not a quadratic equation.

# Question-4

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

^{2}.**Solution:**

The degree of the equation 2x

^{2 }â€“

_{ }7x = 0 is 2

**âˆ´**7x = 2x

^{2}is a quadratic equation.

# Question-5

**Check whether the following equation is quadratic or not: x**

^{2}+ = 2(x â‰ 0).**Solution:**

The given equation also written as

x

^{4 }â€“ 2x

^{2 }+ 1= 0

â‡’The degree of the equation is 4

âˆ´ x

^{2}+= 2(x â‰ 0) is not a quadratic equation.

# Question-6

**Check whether the following equation is quadratic or not: 3x**

^{2}â€“ 4 = 0.**Solution:**

The degree of the given equation is 2

âˆ´ 3x

^{2}â€“ 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

âˆ´ (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 â€“ 1)(x â€“ 2).**

**Solution:**

By solving the given equation (2x + 1)(3x + 2) = 6(x â€“ 1)(x â€“ 2) we get,

6x

^{2 }+ 7x + 2 = 6x

^{2 }â€“ 18x + 12

25x â€“ 10 = 0

The degree of the given equation is 1

âˆ´ (2x + 1)(3x + 2) = 6(x â€“ 1)(x â€“ 2) is not a quadratic equation.

# Question-9

**Check whether the following equation is quadratic or not: x += x**

^{2}(x â‰ 0).**Solution:**

x += x

^{2 }â‡’ x

^{2 }+ 1 = x

^{4}

The degree of the given equation is 4.

âˆ´ x += x

^{2 }(x â‰ 0) is not a quadratic equation.

# Question-10

**Check whether the following equation is quadratic or not: 16x**

^{2 }â€“ 3 = (2x + 5)(5x â€“ 3).**Solution:**

By solving the given equation we get,

6x

^{2 }â€“ 19x + 12 = 0

âˆ´ 16x

^{2 }â€“ 3 = (2x + 5)(5x â€“ 3) is a quadratic equation.

# Question-11

**Determine whether the given value of x is a solution of the given equation or not: 3x**

^{2 }â€“ 2x â€“ 1 = 0; x = 1.**Solution:**

3x

^{2 }â€“ 2x â€“ 1= 0

Let x = 1, 3(1)

^{2 }â€“ 2(1)â€“ 1 = 3 â€“ 2 â€“ 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: 2x**

^{2 }â€“ 6x + 3 = 0; x = .**Solution:**

2x

^{2 }â€“ 6x + 3 = 0

Let, x = â‡’ 2()

^{2 }â€“ 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 = â‡’(2 Ã— + 3)(3 Ã— - 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: x**

^{2 }+ x + 1 = 0; x = -1.**Solution:**

x

^{2 }+ x + 1 = 0

Let x = -1 â‡’ (-1)

^{2 }+ (-1) + 1 = 1 â€“ 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:****x**^{2 }+ 6x + 5 = 0; x = -1, x = -5**.****Solution:**

**x**

^{2 }+ 6x + 5 = 0Put x = -1

â‡’ (-1)

^{2 }+ 6(-1) + 5 = 1

^{ }â€“ 6 + 5 = 0

Put x = -5

â‡’(-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: 9x**

^{2 }- 3x - 2 = 0; x = -, x = .**Solution:**

9x

^{2 }- 3x - 2 = 0

Put x = -

9(-)

^{2 }- 3(-) â€“ 2 = 1 + 1 â€“ 2 = 0

Put x =

9x

^{2}â€“ 3x â€“ 2 = 9()

^{2 }â€“ 3() â€“ 2 = 4 â€“ 2 â€“ 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.

(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 Ã— + 8)(2 Ã— + 5) = (8 + 8)(+5) = 16 Ã— = â‰ 0

Put x = 2 =

(3x + 8)(2x + 5) = (3 Ã— + 8)(2 Ã— +5) = ( + 8)(5 + 5) =Ã— 10 = 155 â‰ 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: 9x**

^{2 }- 16 = 0.**Solution:**

9x

^{2 }- 16 = 0

(3x)

^{2 }- 4

^{2}= 0

(3x

^{ }â€“ 4)(3x + 4) = 0

x = 4/3, -4/3.

# Question-20

**Using factorization,****find the roots of the quadratic equation: 64x**^{2}â€“ 9 = 0.**Solution:**

64x

^{2}â€“ 9 = 0

(8x)

^{2}â€“ 3

^{2}= 0

(8x â€“ 3)(8x + 3) = 0

**âˆ´**x = 3/8, -3/8.

# Question-21

**Using factorization, find the roots of the quadratic equation: (x â€“ 2)**

^{2}â€“ 25 = 0.**Solution:**

(x â€“ 2)

^{2}â€“ 25 = 0

(x â€“ 2)

^{2}â€“ 5

^{2}= 0

(x â€“ 2 â€“ 5)(x â€“ 2 + 5) = 0

(x â€“ 7)(x + 3) = 0

x = 7, â€“ 3.

# Question-22

**Using factorization, find the roots of the quadratic equation: (x + 5)**

^{2}â€“ 36 = 0.**Solution:**

(x + 5)

^{2}â€“ 36 = 0

(x + 5)

^{2}â€“ 6

^{2}= 0

(x + 5 â€“ 6)(x + 5 + 6) = 0

(x â€“ 1)(x + 11) = 0

x = 1, â€“11

# Question-23

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

^{2}= 81.**Solution:**

(2x + 3)

^{2}= 81

(2x + 3)

^{2}â€“ 9

^{2}= 0

(2x + 3 â€“ 9) (2x + 3 + 9) = 0

(2x â€“ 6) (2x + 12) = 0

x = 3, â€“6.

# Question-24

**Using factorization, find the roots of the quadratic equation: y**

[Hint: 3 =()

^{2}-3 = 0[Hint: 3 =()

^{2}].**Solution:**

y

^{2 }- 3 = 0

y

^{2 }â€“ ()

^{ 2}= 0

(y - )(y + ) = 0

y = , -.

# Question-25

**Using factorization, find the roots of the quadratic equation: a**

^{2}z^{2}-b^{2}= 0.**Solution:**

a

^{2}z

^{2 }- b

^{2}= 0

(az)

^{2 }- b

^{2}= 0

(az - b)(az + b) = 0

z = b/a, - b/a.

# Question-26

**Using factorization, find the roots of the quadratic equation: 3z-z**

^{2}= 0.**Solution:**

3z - z

^{2}= 0

z(3 â€“ z) = 0

z = 0 or (3 - z) = 0

z = 0, z = 3.

# Question-27

**Using factorization, find the roots of the quadratic equation: 5z**

^{2}-30 = 0.**Solution:**

5z

^{2 }- 30 = 0

5(z

^{2 }â€“ 6) = 0

z

^{2 }â€“ ()

^{2}= 0

(z

^{ }â€“)(z

^{ }+) = 0

z = , -.

# Question-28

**Using factorization, find the roots of the quadratic equation: ax**

^{2}-2abx = 0.**Solution:**

ax

^{2 }- 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: 4y**

^{2}+4y+1 = 0.**Solution:**

4y

^{2 }+ 4y + 1 = 0

4y

^{2 }+ 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: y**

^{2}-8y+16 = 0.**Solution:**

y

^{2 }- 8y + 16 = 0

y

^{2 }- 4y â€“ 4y + 16 = 0

y(y

^{ }â€“ 4) â€“ 4(y â€“ 4) = 0

(y - 4)(y - 4) = 0

y = 4 or y = 4.

# Question-31

**Using factorization, find the roots of the quadratic equation: z**

^{2}-z+=0.**Solution:**

z

^{2 }â€“ z + = 0

z

^{2 }â€“ z - z + = 0

z(z

^{ }â€“ ) - (z - ) = 0

(z

^{ }â€“ )(z - ) = 0

z = or z = .

# Question-32

**Using**

**factorization, find the roots of the quadratic equation: x**

^{2}-x+1 = 0**.**

**Solution:**

x

^{2 }- x + 1 = 0

x

^{2 }- x â€“ x + 1 = 0

x(x â€“ 1) â€“ (x â€“ 1) = 0

(x â€“ 1)(x â€“ 1) = 0

x = 3 or x = 3.

# Question-33

**Using factorization, find the roots of the quadratic equation: y**

^{2}+2y + 3 = 0.**Solution:**

y

^{2 }+ 2y + 3 = 0

y

^{2 }+ 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: x**

^{2}-4qx+4q^{2}=0.**Solution:**

x

^{2 }â€“ 4qx + 4q

^{2 }= 0

x

^{2 }â€“ 2qx â€“ 2qx + 4q

^{2 }= 0

x(x

^{ }â€“ 2q) â€“ 2q(x â€“ 2q) = 0

(x

^{ }â€“ 2q)(x â€“ 2q) = 0

x = 2q or x = 2q.

# Question-35

**Using factorization, find the roots of the quadratic equation: z**

^{2}-2z-8 = 0.**Solution:**

z

^{2 }- 2z - 8 = 0

z

^{2 }â€“ 4z + 2z - 8 = 0

z(z

^{ }â€“ 4) + 2(z â€“ 4) = 0

(z

^{ }â€“ 4)(z + 2) = 0

z = 4 or z = -2.

# Question-36

**Using factorization, find the roots of the quadratic equation: 6z**

^{2}-5z-21 = 0.**Solution:**

6z

^{2 }- 5z - 21 = 0

6z

^{2 }- 14z + 9z - 21 = 0

2z(3z

^{ }â€“ 7) + 3(3z â€“ 7) = 0

(2z + 3)(3z

^{ }â€“ 7) = 0

(2z + 3) = 0 or (3z

^{ }â€“ 7) = 0

z = -3/2 or z = 7/3.

# Question-37

**Using factorization, find the roots of the quadratic equation: y**

^{2}+3y-18 = 0.**Solution:**

y

^{2 }+ 3y - 18 = 0

y

^{2 }+ 6y â€“ 3y - 18 = 0

y(y

^{ }+ 6)-3(y

^{ }+ 6) = 0

(y â€“ 3)(y

^{ }+ 6) = 0

y = 3 or y = -6.

# Question-38

**Using factorization, find the roots of the quadratic equation: y**

^{2}-3y-10 = 0.**Solution:**

y

^{2 }- 3y - 10 = 0

y

^{2 }- 5y + 2y - 10 = 0

y(y

^{ }â€“ 5) + 2(y â€“ 5) = 0

(y + 2)(y

^{ }â€“ 5) = 0

y = -2 or y = 5.

# Question-39

**Using factorization, find the roots of the quadratic equation: 6y**

^{2}-y-2 = 0.**Solution:**

6y

^{2 }- y - 2 = 0

6y

^{2 }â€“ 4y + 3y - 2 = 0

2y(3y

^{ }â€“ 2) + (3y â€“ 2) = 0

(2y + 1)(3y

^{ }â€“ 2) = 0

2y + 1 = 0 or 3y

^{ }â€“ 2 = 0

y = -1/2 or y = 2/3.

# Question-40

**Using factorization, find the roots of the quadratic equation: 9y**

^{2}-3y-2 = 0.**Solution:**

9y

^{2}-3y-2 = 0

9y

^{2 }- 6y + 3y - 2 = 0

3y(3y

^{ }â€“ 2) + (3y â€“ 2) = 0

(3y + 1)(3y

^{ }â€“ 2) = 0

y = -1/3 or y = 2/3.

# Question-41

**Using factorization, find the roots of the quadratic equation: 5z**

^{2}-3z-2 = 0.**Solution:**

5z

^{2 }- 3z - 2 = 0

5z

^{2 }- 5z + 2z - 2 = 0

5z(z

^{ }â€“ 1) + 2(z â€“ 1) = 0

(5z + 2)(z

^{ }â€“ 1) = 0

z = -2/5 or z = 1.

# Question-42

**Using factorization, find the roots of the quadratic equation: 2z**

^{2}+az-a^{2}= 0.**Solution:**

2z

^{2}+ az - a

^{2}= 0

2z

^{2}+ 2az - az - a

^{2}= 0

2z(z + a) â€“ a(z + a) = 0

(z + a)(2z â€“ a) = 0

z = -a or z = a/2.

# Question-43

**Using factorization, find the roots of the quadratic equation: 8x**

^{2}-22x-21 = 0.**Solution:**

8x

^{2 }- 22x - 21 = 0

8x

^{2 }+ 6x - 28x - 21 = 0

2x(4x

^{ }+ 3) â€“ 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: 5x**

^{2}-+ = 0.**Solution:**

5x

^{2 }- + = 0

5x

^{2 }- x - x + = 0

x(5x

^{ }â€“ 1) - (5x â€“ 1) = 0

(5x

^{ }â€“ 1) (x - ) = 0

x = 1/5 or x = .

# Question-45

**Using factorization, find the roots of the quadratic equation: x**

^{2}-+=0.**Solution:**

x

^{2 }- += 0

x

^{2 }- x - x + = 0

x(x

^{ }- ) - (x - ) = 0

(x

^{ }- )(x - ) = 0

x

^{ }= or x = .

# Question-46

**Using factorization, find the roots of the quadratic equation: x**

^{2}- x-=0.**Solution:**

x

^{2}- x - = 0

x

^{2}+ x - x - = 0

x(x + ) - (x + ) = 0

(x + )(x - ) = 0

x = - or x = .

# Question-47

**Using factorization, find the roots of the quadratic equation: 3x**

^{2}+10 = 11x.**Solution:**

3x

^{2 }+ 10 = 11x

3x

^{2 }- 11x + 10 = 0

3x

^{2 }- 6x â€“ 5x + 10 = 0

3x(x

^{ }- 2) â€“ 5(x â€“ 2) = 0

(3x â€“ 5)(x

^{ }- 2) = 0

x = 5/3 or x = 2.

# Question-48

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

**Solution:**

x + = -4, (x â‰ 0)

x

^{2}+ 4 = - 4x

x

^{2}+ 4x + 4 = 0

x

^{2}+ 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: abx**

^{2}+(b^{2}-ac)x - bc = 0**Solution:**

abx

^{2 }+ (b

^{2 }- ac)x - bc = 0

abx

^{2 }+ b

^{2}x

^{ }- acx - bc = 0

bx(ax

^{ }+ b) â€“ c(ax + b) = 0

(ax

^{ }+ b)(bx â€“ c) = 0

x = -b/a or x = c/b.

# Question-50

**Using factorization, find the roots of the quadratic equation: += 3(x â‰ 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 â€“ 4)

3(x

^{2}- 5x + 4 + x

^{2}- 5x + 6) = 10(x

^{2}- 6x + 8)

3(2x

^{2}- 10x + 10) = 10(x

^{2}- 6x + 8)

6x

^{2}- 30x + 30 = 10x

^{2}- 60x + 80

4x

^{2}- 30x + 50 = 0

2x

^{2}- 15x + 25 = 0

2x

^{2}â€“ 10x - 5x + 25 = 0

2x(x â€“ 5) â€“ 5(x â€“ 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 â‰ 3, â€“ 5**

**Solution:**

Multiplying throughout by (x â€“ 3)(x + 5), we get

6[(x + 5) â€“ (x - 3)] = (x â€“ 3)(x + 5)

6(x + 5 â€“ x + 3) = (x

^{2}+ 2x â€“ 15)

6(8) = (x

^{2}+ 2x â€“ 15)

48 = (x

^{2}+ 2x â€“ 15)

x

^{2}+ 2x â€“ 63 = 0

x

^{2}+ 9x â€“ 7x â€“ 63 = 0

x(x + 9) â€“ 7(x + 9) = 0

(x + 9)(x â€“ 7) = 0

x = â€“ 9 or x = 7.

# Question-52

**Using factorization, find the roots of the quadratic equation: = 3, (x â‰ 1, â€“ 2).**

**Solution:**

Multiplying throughout by (x â€“ 1)(x + 2), we get

(x + 1)(x + 2) + (x â€“ 1)(x â€“ 2) = 3(x â€“ 1)(x + 2)

(x

^{2}+ 3x + 2) + (x

^{2}â€“ 3x + 2) = 3(x

^{2}+ x â€“ 2)

2x

^{2}+ 4 = 3x

^{2}+ 3x â€“ 6

x

^{2}+ 3x â€“ 10 = 0

x

^{2}+ 5x â€“ 2x â€“ 10 = 0

x(x + 5) â€“ 2(x + 5) = 0

(x + 5)(x â€“ 2) = 0

x = â€“ 5 or x = 2.

# Question-53

**Determine whether the given quadratic equation has root(s). If so, find the root(s). y**

^{2 }â€“ 4y â€“1 = 0.**Solution:**

a = 1, b = â€“4 and c = â€“1

Î” = b

^{2}â€“ 4ac = (â€“4)

^{2}â€“ 4(1)(â€“1) = 16 + 4 = 20

Therefore the quadratic equation has real roots.

Therefore the roots are 2 â€“

**âˆš5**and 2 +

**âˆš5**.

# Question-54

**Determine whether the given quadratic equation has root(s). If so, find the root(s). y**

^{2 }â€“ 6y+2=0.**Solution:**

a = 1, b = â€“6 and c = 2

Î” = b

^{2}â€“ 4ac = (â€“6)

^{2}â€“ 4(1)(2) = 36 â€“ 8 = 28

Therefore the quadratic equation has real roots.

Therefore the roots are 3 + âˆš7 and 3 â€“ âˆš7.

# Question-55

**Determine whether the given quadratic equation has root(s). If so, find the root(s). y**

^{2 }+ 8y + 4 = 0.**Solution:**

a = 1, b = 8 and c = 4

Î” = b

^{2}â€“ 4ac = (8)

^{2}â€“ 4(1)(4) = 64 â€“ 16 = 48

Therefore the quadratic equation has real roots.

Therefore the roots are â€“4 â€“ 2âˆš3 and â€“4 +2âˆš3.

# Question-56

**Determine whether the given quadratic equation has root(s). If so, find the root(s). y**

^{2}+ y â€“ 1 = 0.**Solution:**

a = 1, b = and c = -1

Î” = b

^{2}â€“ 4ac = ()

^{2}â€“ 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). 2y**

^{2 }+ 5y -3 = 0.**Solution:**

a = 2, b = 5 and c = â€“3

Î” = b

^{2}â€“ 4ac = 5

^{2}â€“ 4(2)(â€“3) = 25 + 24 = 49

Therefore the quadratic equation has real roots.

Therefore the roots are â€“3 and 1/2.

# Question-58

**Determine whether the given quadratic equation has root(s). If so, find the root(s) â€“2y**

^{2 }+ y + 1 = 0.**Solution:**

a = â€“2, b = 1 and c = 1

Î” = b

^{2}â€“ 4ac = 1

^{2}â€“ 4(â€“2)(1) = 1 + 8 = 9

Therefore the quadratic equation has real roots.

Therefore the roots are 1 and â€“1/2.

# Question-59

**Determine whether the given quadratic equation has root (s). If so, find the root (s): 3y**

^{2 }+ 9y + 4 = 0.**Solution:**

a = 3, b = 9 and c = 4

Î” = b

^{2 }â€“ 4ac = 9

^{2 }â€“ 4(3)(4) = 81 â€“ 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): 5y**

^{2 }â€“ 2y â€“ 2 = 0.**Solution:**

a = 5, b = â€“2 and c = â€“2

Î”= b

^{2}â€“ 4ac = (â€“2)

^{2}â€“ 4(5)(â€“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): 7z**

^{2 }+ 8z + 2 = 0.**Solution:**

a = 7, b = 8 and c = 2

Î” = b

^{2}â€“ 4ac = (8)

^{2}â€“ 4(7)(2) = 64 â€“ 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): z**

^{2 }+ z â€“ 3 = 0.**Solution:**

a = 1, b = and c = â€“3

Î” = b

^{2}â€“ 4ac = ()

^{2}â€“ 4(1)(-3) = + 12 = =

Therefore the quadratic equation has real roots.

Therefore the roots are â€“2 and 3/2.

# Question-63

**Determine whether the given quadratic equation has root(s). If so, find the root (s): z**

^{2 }+ 2z â€“ 8 = 0.**Solution:**

a = 1, b = 2 and c = â€“8

Î” = b

^{2}â€“ 4ac = (2)

^{2}â€“ 4(1)(â€“8) = 4 + 32 = 36

Therefore the quadratic equation has real roots.

Therefore the roots are â€“4 and 2.

# Question-64

**Determine whether the given quadratic equation has root(s). If so, find the root(s): z**

^{2 }â€“ 6z + 4 = 0.**Solution:**

a = 1, b = â€“6 and c = 4

Î” = b

^{2}â€“ 4ac = (â€“6)

^{2}â€“ 4(1)(4) = 36 â€“ 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): 6x**

^{2 }+ x â€“ 2 = 0.**Solution:**

a = 6, b = 1 and c = â€“2

Î” = b

^{2}â€“ 4ac = (1)

^{2}â€“ 4(6)(â€“2) = 1 + 48 = 49

Therefore the quadratic equation has real roots.

Therefore the roots are â€“2/3 and .

# Question-66

**Determine whether the given quadratic equation has root(s). If so, find the root(s):4x**

^{2 }+ 12x + 9 = 0.**Solution:**

a = 4, b = 12 and c = 9

Î” = b

^{2 }â€“ 4ac = 12

^{2 }â€“ 4(4)(9) = 0

Therefore the quadratic equation has real and equal roots.

â‡’

âˆ´

**and**

Therefore the roots are â€“3/2 and â€“3/2.

# Question-67

**Determine whether the given quadratic equation has root(s). If so, find the root(s):2x**

^{2 }+ 5+ 16 = 0.**Solution:**

a = 2, b = 5and c = 16

Î” = b

^{2}â€“ 4ac = (5)

^{2}â€“ 4(2)(16) = 75 â€“ 128 = â€“ 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): 3x**

^{2 }+ 2â€“ 5 = 0.**Solution:**

a = 3, b = 2and c = -5

Î” = b

^{2}â€“ 4ac = (2)

^{2}â€“ 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): 3y**

^{2 }+ (6 + 4a)y + 8a = 0.**Solution:**

a = 3, b = (6 + 4a) and c = 8a

Î” = b

^{2}â€“ 4ac = (6 + 4a)

^{2}â€“ 4(3)(8a) = 36 + 16a

^{2 }+ 48a â€“ 96a

= 16a

^{2 }- 48a + 36 = (4a â€“ 6)

^{2}

Therefore the quadratic equation has real roots.

Therefore the roots are and â€“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:**

x

^{2 }+ 9x + 20 = 3(x

^{2}+ 3x + 2) + 2x

x

^{2 }+ 9x + 20 = 3x

^{2}+ 9x + 6 + 2x

2x

^{2}+ 2x â€“14 = 0

a = 2, b = 2 and c = â€“14

Î” = b

^{2}â€“ 4ac = 2

^{2}â€“ 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: px**

^{2 }+ 4x + 1 = 0.**Solution:**

Given the quadratic equation has real roots,

âˆ´ Î”

__>__0

Here, a = p, b = 4 and c = 1

b

^{2}- 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: 2x**

^{2 }+ 3x + p = 0.**Solution:**

a = 2, b = 3 and c = p

Given, Î”

__>__0

â‡’ b

^{2}- 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 5px**

^{2 }- 8x + 2 = 0 has real roots:**Solution:**

a = 5p, b = -8 and c = 2

Î” = b

^{2}â€“ 4ac = (-8)

^{2}â€“ 4(5p)(2) = 64 â€“ 40p

Î” â‰¥ 0

Therefore 64 â€“ 40p â‰¥ 0

64 â‰¥ 40p

p â‰¤ 64/40 â‰¤ 8/5 â‰¤ 1.

âˆ´ p â‰¤ 1.

# Question-74

**Determine the value(s) of p for which the given quadratic equation has real roots: 4x**

^{2 }+ 8x - p = 0.**Solution:**

a = 4, b = 8 and c = -p

Î” = b

^{2}â€“ 4ac = 8

^{2}â€“ 4(4)(-p) = 64 + 16p

Given, Î” â‰¥ 0

64 + 16p â‰¥ 0

4 + p â‰¥ 0

p Â³ -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 â€“ x

^{2}= 32

- x

^{2}+ 12x â€“ 32 = 0

- x

^{2}+ 8x + 4x â€“ 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 x

^{2}+ (12 - x)

^{2}= 74

x

^{2}+ (12 - x)

^{2}= 74

x

^{2}+ 144 - 24x + x

^{2}= 74

2x

^{2}- 14x â€“ 10x + 70 = 0

2x(x â€“ 7) â€“ 10(x â€“ 7) = 0

(2x â€“ 10)(x â€“ 7) = 0

x = 10/2 or x = 7

x = 5 or x = 7

Therefore the two products are 5 and 7.

Checking:

5

^{2}+ 7

^{2}= 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 + =

3x

^{2}+ 3 = 10x

3x

^{2}- 10x + 3 = 0

3x

^{2}- 9x - x + 3 = 0

3x(x - 3) - (x - 3) = 0

(x - 3)(3x - 1) = 0

x = 3 or x = 1/3

âˆ´ 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 cm**

^{2}, 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 Ã— breadth = (x + 2) Ã— x = x

^{2}+ 2x

x

^{2}+ 2x = 195

x

^{2}+ 2x â€“ 195 = 0

x

^{2}+ 15x â€“ 13x â€“ 195 = 0

x(x + 15) â€“ 13(x + 15) = 0

(x â€“ 13)(x + 15) = 0

x = 13 or x = â€“15

Therefore the sides of the rectangle is 13 cm and 15 cm.

Checking :

13 cm Ã— 15 cm = 195 cm

^{2.}

# 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(5x

^{2}+ 4x + 1) = 58(2x

^{2}+ x)

105x

^{2}+ 84x + 21 = 116x

^{2}+ 58x

11x

^{2}- 26x - 21 = 0

11x

^{2}- 33x + 7x - 21 = 0

11x(x â€“ 3) + 7(x â€“ 3) = 0

(11x + 7)(x â€“ 3) = 0

11x + 7 = 0 or x â€“ 3 = 0

x = -7/11 or x = 3

The required fraction is .

Checking:

.

# Question-80

**The product of Ramuâ€™s age (in years) five years ago and his age(in years) nine years later is 15. Determine Ramuâ€™s present age.**

**Solution:**

Ramuâ€™s present age = x years

Ramuâ€™s age five years ago = (x - 5) years

Ramuâ€™s age nine years later = (x + 9) years

Condition: Product of Ramuâ€™s age five years ago and his age nine

**years later is 15.**

(x - 5)(x + 9) = 15

x

^{2}+ 4x - 45 = 15

x

^{2}+ 4x - 45 - 15 = 0

x

^{2}+ 4x - 60 = 0

x

^{2}+ 10x - 6x - 60 = 0

x(x + 10) - 6(x + 10) = 0

(x + 10)(x - 6) = 0

x = -10 or x = 6

âˆ´ Ramuâ€™s present age is 6 years

Checking:

Ramuâ€™s age five years ago = (6 - 5) year = 1 year

Ramuâ€™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 tenâ€™s digit is .

Therefore the required number is Ã— 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 + x

^{2}+ 36x = 10x

^{2}+ 12

9x

^{2}- 36x â€“ 108 = 0

x

^{2}- 4x â€“ 12 = 0

x

^{2}- 6x + 2x â€“ 12 = 0

x(x â€“ 6) + 2(x â€“ 6) = 0

(x + 2)(x â€“ 6) = 0

(x + 2) = 0 or (x â€“ 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.

x

^{2}+ (x + 2)

^{2}= 202

x

^{2}+ x

^{2}+ 4x + 4 = 202

2x

^{2 }+ 4x + 198 = 0

x

^{2 }+ 2x + 99 = 0

x

^{2 }+ 11x â€“ 9x + 99 = 0

x(x

^{ }+ 11) â€“ 9(x + 11) = 0

(x â€“ 9)(x

^{ }+ 11) = 0

x â€“ 9 = 0 or x + 11 = 0

x = 9 or x = -11

Therefore the two consecutive odd natural numbers are 9 and 11.

Checking :

9^{2} + 11^{2} = 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.

x

^{2}+ (x + 2)

^{2}= 100

x

^{2}+ x

^{2}+ 4x + 4 = 100

2x

^{2}+ 4x â€“ 96 = 0

x

^{2}+ 2x â€“ 48 = 0

x

^{2}+ 8x â€“ 6x â€“ 48 = 0

x(x + 8) â€“ 6(x + 8) = 0

(x â€“ 6)(x + 8) = 0

(x â€“ 6) = 0 or (x + 8) = 0

x = 6 or x = â€“ 8

The two consecutive even positive integers is 6 and 8.

Checking:

6

^{2}+ 8

^{2}= 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 â€“ x.

+ =

=

=

15 = x(8 - x)

15 = 8x - x

^{2}

x

^{2}- 8x + 15 = 0

x

^{2}- 5x â€“ 3x + 15 = 0

x(x â€“ 5) â€“ 3(x â€“ 5) = 0

(x â€“ 3)(x â€“ 5) = 0

(x â€“ 3) = 0 or (x â€“ 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.

x

^{2}+ (x + 1)(x + 2) = 154

x

^{2}+ x

^{2}+ 3x + 2 = 154

2x

^{2}+ 3x - 152 = 0

2x

^{2}+ 19x â€“ 16x â€“ 152 = 0

x(2x + 19) â€“ 8(2x + 19) = 0

(x â€“ 8)(2x + 19) = 0

x = 8 or x = â€“ 19/2

âˆ´ The integers are 8, 9 and 10.

Checking:

8

^{2}+ 9 Ã— 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

x

^{2}+ (x + 7)

^{2}= (2x + 1)

^{2}

x

^{2}+ x

^{2}+ 14x + 49 = 4x

^{2}+ 4x + 1

2x

^{2}- 10x - 48 = 0

2x

^{2}- 16x + 6 x â€“ 48 = 0

2x(x â€“ 8) + 6(x â€“ 8) = 0

(2x + 6)(x â€“ 8) = 0

2(x + 3) = 0 or x â€“ 8 = 0

âˆ´ x = -3 or x = 8

The sides of the grasy land are 8, 15 and 17.

Checking:

8

^{2}= 64

15

^{2}= 225

17

^{2}= 289

8

^{2}+ 15

^{2 }= 64 + 225 = 289 = 17

^{2}.

# 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 (x

^{2}- 45).

Condition: The squares of the smaller number is 4 times the larger number.

(x

^{2}â€“ 45) = 4x

x

^{2}â€“ 4x â€“ 45 = 0

x

^{2}â€“ 9x + 5x â€“ 45 = 0

x(x â€“ 9) + 5(x â€“ 9) = 0

(x + 5)(x â€“ 9) = 0

(x + 5) = 0 or (x â€“ 9) = 0

x = â€“5 or x = 9

Therefore the numbers are 9 and

Checking:

6

^{2}= 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 â€“ x) years.

Five years ago fatherâ€™s age = x â€“ 5 years

Five years ago sonâ€™s age = (45 â€“ x) â€“ 5 years = (40 â€“ x) years

Condition: The product of their ages was 124.

(x - 5)(40 - x) = 124

x(40 - x) - 5(40 - x) = 124

40x â€“ x

^{2}- 200 + 5x = 124

- x

^{2}- 324 + 45x = 0

x

^{2}- 45x + 324 = 0

x

^{2}- 36x â€“ 9x + 324 = 0

x(x â€“ 36) â€“ 9(x â€“ 36) = 0

(x â€“ 9)(x â€“ 36) = 0

x â€“ 9 = 0 or x â€“ 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 Ã— 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.

++ 7âˆšx ++ 56 = x

++ 7âˆš x + 56 = x

2x + 9x + 18 Ã— 7âˆšx + 56 Ã— 18 = 18x

11x + 126âˆšx + 1008 = 18x

126âˆš x + 1008 = 7x

18âˆšx + 144 = x

(18âˆšx)

^{2}= (x â€“ 144)

^{2}

324x = x

^{2}â€“ 288x + 20736

x

^{2}â€“ 612x + 20736 = 0

x

^{2}- 576x â€“ 36x + 20736 = 0

x(x â€“ 576) â€“ 36(x â€“ 576) = 0

(x â€“ 36)(x â€“ 576) = 0

(x â€“ 36) = 0 or (x â€“ 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:

++ 7âˆš576 ++ 56 = 64 + 144 + 7âˆš576 + 144 + 56

= 64 + 144 + 7Ã— 24 + 144 + 56

= 64 + 144 + 168 + 144 + 56

= 576.

# Question-90

**A farmer wishes to grow a 100 m**

[Hint: If the length of one side is x metres, the other side will be (30 - 2x) m. Therefore, x(30 - 2x) = 100]

^{2}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 - 2x

^{2}= 100

-2x

^{2}+ 30x â€“ 100 = 0

x

^{2}â€“ 15x + 50 = 0

x

^{2}â€“ 10x â€“ 5x + 50 = 0

x(x â€“ 10) â€“ 5(x â€“ 10) = 0

(x â€“ 5)(x â€“ 10) = 0

(x â€“ 5) = 0 or (x â€“ 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 Ã— 10 m = 100 m**

^{2}.

# Question-91

**The hypotenuse of a right triangle is 3âˆš10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be**

9âˆš 5 cm. How long are the legs of the triangle?

9âˆš 5 cm. How long are the legs of the triangle?

**Solution:**

Hypotenuse =cm

Let the smaller leg be x cm

Then the longer leg = Â±

= Â± = Â±

New Smaller leg = 3x cm

New Longer leg = 2cm

New hypotenuse = cm

()

^{2}= (3x)

^{2}+ (Â± 2)

^{2}

81Ã— 5 = 9x

^{2}+ 4(90 â€“ x

^{2})

405 = 9x

^{2}+ 360 â€“ 4x

^{2}

45 = 5x

^{2}

x

^{2}â€“ 9 = 0

(x â€“ 3)(x + 3) = 0

(x â€“ 3) = 0 or (x + 3) = 0

x = 3 or x = -3

The smaller leg is 3cm and the longer leg is == = 9 cm.

Checking:

3

^{2}+ 9

^{2}= 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 cm**

^{2}. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm^{2}. Determine the sides of the two squares.**Solution:**

Let the area of the smaller square be x

^{2 }cm

^{2}.

Condition 1: If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm

^{2}.

Then the area of the larger square = (14 + 2x

^{2}) cm

^{2}

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 cm

^{2}.

â‡’ 2(14 + 2x

^{2}) + 3 x

^{2}= 203

28 + 4x

^{2}+ 3 x

^{2}= 203

7x

^{2}= 175

x

^{2}= 25

x = Â± 5 cm

Therefore the sides of smaller square is 5 cm.

And the sides of the larger squares is 14 + 2Ã— 5

^{2}= 14 + 50 = 64 cm.

The sides of the two squares is 5 cm and 64 cm.

Checking:

a) 64 - 2(5

^{2}) = 64 â€“ 250 = 14

b) 2(64) + 3(5

^{2}) = 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 â€“ 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 â€“ 16) + 2x(x - 16)

192x = 192x â€“ 3072 + 2x

^{2}â€“ 32x

- 3072 + 2x

^{2}â€“ 32x = 0

2x

^{2}â€“ 32x â€“ 3072 = 0

x

^{2}â€“ 16x â€“ 1536 = 0

x

^{2}â€“ 48x + 32x â€“ 1536 = 0

x(x â€“ 48) + 32(x â€“ 48) = 0

(x + 32)(x â€“ 48) = 0

(x + 32) = 0 or (x â€“ 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 â€“ 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

â€“ 2 =

â€“ = 2

300 = 2

300 = 2

300 = 2

300[5/(x

^{2}+ 5x)] = 2

1500 = 2(x

^{2}+ 5x)

1500 = 2x

^{2}+ 10x

On rearranging

2x

^{2}+ 10x â€“ 1500 = 0

x

^{2}+ 5x â€“ 750 = 0

x

^{2}+ 30x â€“ 25x â€“ 750 = 0

x(x + 30) â€“ 25(x + 30) = 0

(x - 25)(x + 30) = 0

x - 25 = 0 or x + 30 = 0

x = 25 or x = â€“ 30

Since x is the usual speed it cannot be negative.

Therefore, we reject the solution x = â€“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)

3x

^{2}- 30x = 600x - 600x + 6000

On rearranging we have

-3x

^{2}+ 30x + 600x - 600x + 6000 = 0

-3x

^{2}+ 30x + 6000 = 0 [Dividing through out by 3]

-x

^{2}+ 10x + 2000 = 0

x

^{2}- 10x - 2000 = 0

x

^{2}- 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: x**

^{2 }- 6x - 4 = 0.**Solution:**

x

^{2 }- 6x - 4 = 0 is quadratic equation.

# Question-97

**Check whether the following equation is quadratic or not: 3x**

^{2}â€“ 7x â€“ 2 = 0.**Solution:**

3x

^{2}â€“ 7x â€“ 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 cm**

^{2}. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm^{2}. Determine the sides of the two squares.**Solution:**

Let the area of the smaller square be x

^{2 }cm

^{2}.

Condition 1: If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm

^{2}.

Then the area of the larger square = (14 + 2x

^{2}) cm

^{2}

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 cm

^{2}

2(14 + 2x

^{2}) + 3 x

^{2}= 203

28 + 4x

^{2}+ 3 x

^{2}= 203

7x

^{2}= 175

x

^{2}= 25

x = Â± 5 cm

âˆ´Therefore the sides of smaller square is 5 cm.

And the sides of the larger squares is 14 + 2Ã— 5

^{2}= 14 + 50 = 64 cm.

The sides of the two squares is 5cm and 64 cm.

Checking:

a) 64 - 2(5

^{2}) = 64 â€“ 250 = 14

b) 2(64) + 3(5

^{2}) = 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 â€“ 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 â€“ 16) + 2x(x â€“ 16)

192x = 192x â€“ 3072 + 2x

^{2}â€“ 32x

â€“ 3072 + 2x

^{2}â€“ 32x = 0

2x

^{2}â€“ 32x â€“ 3072 = 0

x

^{2}â€“ 16x â€“ 1536 = 0

x

^{2}â€“ 48x + 32x â€“ 1536 = 0

x(x â€“ 48) + 32(x â€“ 48) = 0

(x + 32)(x â€“ 48) = 0

(x + 32) = 0 or (x â€“ 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 â€“ 16 = 32 km/hr.

Checking:

= + 2

= + 2

6 = 4 + 2

6 = 6.

# Question-100

**Solve the quadratic equation by factorization (3x + 1) (x â€“ 1) = 15.**

**Solution:**

(3x + 1) (x â€“ 1) = 15

=>3x

^{2}â€“ 2x â€“ 1 = 15

=>3x

^{2}â€“ 2x â€“ 1 â€“ 15 = 0

=>3x

^{2}â€“ 2x â€“ 16 = 0

=>(3x â€“ 8)(x + 2) = 0 âˆ´ x = â€“2, x = .

# Question-101

**Given that -3 and m are the roots of the quadratic equation kx**

^{2 }+ 5x â€“ 3 = 0. Find the value of m and of k.**Solution:**

Let us consider the given equation kx

^{2}+ 5x â€“ 3 = 0

**in the form ax**

^{2}+ 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

âˆ´ Sum of the roots = =

-3 + m =

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

âˆ´ Product of the roots = =

-3m =

=> -3km = -3

=> km = 1

Substitute km = 1 in (1)

-3k + 1 = -5

-3k = -5 â€“ 1 = -6 âˆ´ k = 2.

Thus by substituting k = 2 in km = 1 we get m = âˆ´ 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, 4x**^{2}+ kx + k â€“ 1 = 0.**Solution:**

Let us consider the given equation 4x

^{2}+ kx + k â€“ 1 = 0 in the form ax

^{2}+ bx + c = 0.

Here a = 4, b = k, c = k - 1 âˆ´ Sum of the roots = = âˆ´ Product of the roots = = .

# Question-103

**If the roots of the quadratic equation hx**

^{2}+ 21x + 10 = 0 where h â‰ 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 2Î± and 5Î± .

In the given equation hx

^{2}+ 21x + 10 = 0.

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

2Î± + 5Î± =

7Î± =

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

âˆ´ Product of the roots = =

(2Î± )(5Î±) =

10Î±

^{2}=

h = --------(2)

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

-3 = Î± =

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

# Question-104

**Without solving the quadratic equation, determine the nature of roots of the equation x**

^{2 }+ 3x +7 = 0.**Solution:**

In the given equation x

^{2 }+ 3x + 7 = 0. Here a = 1, b = 3, c = 7.

b

^{2}â€“ 4ac = 3

^{2}â€“ 4(1)(7)

= 9 â€“ 28

= -19

Since b

^{2}â€“ 4ac < 0, the given equation has no real roots.

# Question-105

**Find the value of "p", if the quadratic equation p**

^{2}x^{2}â€“ 12x + p + 7 = 0 has the root .**Solution:**

In the given equation p

^{2}x

^{2}â€“ 12x + p + 7 = 0, a = p

^{2}, b = -12, c = p + 7.

If one root is , let the other root be Î± . âˆ´ Sum of the roots = =

+ Î± = Î± = -

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

âˆ´ Product of the roots = =

Ã— Î± =

Î± = Ã—

= -----------(2)

Comparing (1) and (2)

=

3(24 â€“ 3p^{2}) = 4p + 28

72 â€“ 9p^{2} = 4p + 28

Thus we get 9p^{2 }+ 4p â€“ 44 = 0

Now by applying the formula: x =

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

âˆ´ p = =

= =

=

Thus p = or

= or

= 2 or âˆ´ The value of p is 2 or .

# Question-106

**Form the quadratic equation whose roots are**

**and**

**.**

**Solution:**

The quadratic equation is x

^{2 }â€“ (sum of the roots) x + (product of the roots) = 0

Sum of the roots = + = =

Product of the roots = Ã— =

âˆ´ x^{2 }â€“ (sum of the roots) x + (product of the roots) = 0

x^{2 }â€“ x + = 0

15x^{2} + 19x + 6 = 0.

# Question-107

**If H and K are the roots of the equation 2x**

(i) H + K

(ii) HK

(iii) discriminant.

^{2}+ 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) b

^{2}â€“ 4ac = 49 â€“ 4(2)(-4)

= 49 + 32

= 81.

# Question-108

**Frame the quadratic equations whose only root is -6.**

**Solution:**

**\**x

^{2 }â€“ (sum of the roots)x + (product of the roots) = 0

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

Product of the roots = -6 Ã— -6 = 36 âˆ´ x

^{2 }+ 12x + 36 = 0.

# Question-109

**Determine the nature of roots of the quadratic equation x**

^{2}â€“ 6x + 9 = 0.**Solution:**

In this equation a = 1, b = -6 and c = 9

b

^{2}â€“ 4ac = (-6)

^{2}â€“ 4(1)(9)

= 36 â€“ 36

= 0

Since b

^{2}â€“ 4ac = 0, the equation has two real and equal roots.