Chapter 8: Quadratic Equations Exercise – 8.5

Question: 1Find the discriminant of the following quadratic equations:

2x

^{2}- 5x + 3 = 0

Solution:2x

^{2}- 5x + 3 = 0The given equation is in the form of ax

^{2}+ bx + c = 0Here, a = 2, b = -5 and c = 3

The discriminant, D = b

^{2}- 4acD = (-5)

^{2}– 4 x 2 x 3D = 25 – 24 = 1

Therefore, the discriminant of the following quadratic equation is 1.

Question: 2Find the discriminant of the quadratic equations.

x

^{2}+ 2x + 4 = 0

Solution:x

^{2}+ 2x + 4 = 0The given equation is in the form of ax

^{2}+ bx + c = 0Here, a = 1, b = 2 and c = 4

The discriminant is:-

D = (2)

^{2}- 4 x 1 x 4D = 4 – 16 = - 12

The discriminant of the following quadratic equation is = - 12.

Question: 3Find the discriminant of the quadratic equations.

(x -1) (2x -1) = 0

Solution:(x -1) (2x -1) = 0

The provided equation is (x -1) (2x -1) = 0

By solving it, we get 2x

^{2}- 3x + 1 = 0Now this equation is in the form of ax

^{2}+ bx + c = 0Here, a = 2, b = -3, c = 1

The discriminant is:-

D = (-3)

^{2}- 4 x 2 x 1D = 9 - 8 = 1

The discriminant of the following quadratic equation is = 1.

Question: 4Find the discriminant of the quadratic equations.

x

^{2 }- 2x + k = 0

Solution:x

^{2 }- 2x + k = 0The given equation is in the form of ax

^{2}+ bx + c = 0Here, a = 1, b = -2, and c = k

D = b

^{2}- 4acD = (-2)

^{2}- 4(1)(k)= 4 - 4k

Therefore, the discriminant, D of the equation is (4 - 4k)

Question: 5Find the discriminant of the quadratic equations.

Solution:The given equation is in the form of ax

^{2}+ bx + c = 0The discriminant is, D = b

^{2}- 4acD = 8 + 24 = 32

The discriminant, D of the following equation is 32.

Question: 6Find the discriminant of the quadratic equations.

x

^{2}- x + 1 = 0

Solution:x

^{2}- x + 1 = 0The given equation is in the form of ax

^{2}+ bx + c = 0Here, a = 1, b = -1 and c = 1

The discriminant is D = b

^{2 }- 4ac(-1)

^{2}- 4 × 1 × 11 - 4 = - 3

Therefore, the discriminant D of the following equation is -3.

Question: 7Find the discriminant of the quadratic equations.

x

^{2}+ x + 2 = 0

Solution:x

^{2}+ x + 2 = 0The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2}- 4acHere, a = 1, b = 1 and c = 2.

Therefore, the discriminant is given as,

D = (1)

^{2}- 4(1)(2)= 1 - 8

= - 7

For a quadratic equation to have real roots, D ≥ 0.

Here we find that the equation does not satisfy this condition, hence it does not have real roots.

Question: 8Find the discriminant of the quadratic equations.

16x

^{2}= 24x + 1

Solution:16x

^{2}- 24x - 1 = 0The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acHere, a = 16, b = -24 and c = - 1

Therefore, the discriminant is given as,

D = (-24)

^{2}- 4(16)(-1)= 576 + 64

= 640

For a quadratic equation to have real roots, D ≥ 0.

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

Therefore, the roots of the following equation are as follows,

The values of x for both the cases will be:

Question: 9Find the discriminant of the quadratic equations.

x

^{2}+ x + 2 = 0

Solution:x

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

Therefore, the discriminant is given as,

D = (1)

^{2}- 4(1)(2)= 1 - 8

= - 7

For a quadratic equation to have real roots, D ≥ 0.

Here we find that the equation does not satisfy this condition, hence it does not have real roots.

Question: 10Find the discriminant of the quadratic equations.

Solution:The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acTherefore, the discriminant is given as,

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

Therefore, the roots of the equation are given as follows,

The values of x for both the cases will be:

Question: 11Find the discriminant of the quadratic equations.

3x

^{2}- 2x + 2 = 0

Solution:3x

^{2}- 2x + 2 = 0The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acHere, a = 3, b = -2 and c = 2.

Therefore, the discriminant is given as,

D = (-2)

^{2}- 4(3)(2)= 4 - 24 = – 20

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation does not satisfy this condition, hence it has no real roots.

Question: 12Find the discriminant of the quadratic equations.

Solution:The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acTherefore, the discriminant is given as,= 24 - 24 = 0

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

Question: 13Find the discriminant of the quadratic equations.

3a

^{2}x^{2 }+ 8abx + 4b^{2}= 0

Solution:3a

^{2}x^{2 }+ 8abx + 4b^{2}= 0The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acHere, a = 3a

^{2}, b = 8ab and c = 4b^{2}Therefore, the discriminant is given as,

D = (8ab)

^{2}- 4(3a^{2})(4b^{2})= 64a

^{2}b^{2}- 48a^{2}b^{2 }= 16a^{2}b^{2}For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

Therefore, the roots of the equation are given as follows,

The values of x for both the cases will be:

Question: 14Find the discriminant of the quadratic equations.

Solution:The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acTherefore, the discriminant is given as,

= 20 + 60

= 80

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

The values of x for both the cases will be:

Question: 15Find the discriminant of the quadratic equations.

x

^{2}- 2x + 1 = 0

Solution:x

^{2}- 2x + 1 = 0The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acHere, a =1, b = - 2 and c = 1

Therefore, the discriminant is given as,

D = (-2)

^{2}- 4(1)(1)= 4 - 4

= 0

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

x = 2/2

x = 1

Therefore, the equation real roots and its value is 1

Question: 16Find the discriminant of the quadratic equations.

Solution:The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acTherefore, the discriminant is given as,

= 75 - 48

= 27

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,\

Therefore, the roots of the equation are given as follows

The values of x for both the cases will be

Question: 17Find the discriminant of the quadratic equations.

Solution:The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acTherefore, the discriminant is given as,

D = 49 - 40

D = 9

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

Therefore, the roots of the equation are given as follows,

Question: 18Find the discriminant of the quadratic equations.

Solution:The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acTherefore, the discriminant is given as,

= 8 - 8

= 0

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

Therefore, the roots of the equation are given as follows,

Question: 19Find the discriminant of the quadratic equations.

3x

^{2}- 5x + 2 = 0

Solution:3x

^{2}- 5x + 2 = 0The given equation can be written in the form of, ax

^{2 }+ bx + c = 0The discriminant is given by the following equation, D = b

^{2 }- 4acHere, a = 3, b = -5 and c = 2.

Therefore, the discriminant is given as,

D = (-5)

^{2}- 4(3)(2)= 25 - 24

= 1

For a quadratic equation to have real roots, D ≥ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

The roots of an equation can be found out by using,

Therefore, the roots of the equation are given as follows,

The values of x for both the cases will be:

x = 1

And,

x = 2/3

Question: 20Solve for x:

Solution:The above equation can be solved as follows:

6x

^{2}- 30x + 30 = 10x^{2}- 60x +804x

^{2}- 30x + 50 = 02x

^{2}- 15x + 25 = 0The above equation is in the form of ax

^{2}+ bx + c = 0The discriminant is given by the equation, D = b

^{2 }- 4acHere, a = 2, b = -15, c = 25

D = (-15)

^{2}- 4(2)(25)= 225 - 200

= 25

The roots of an equation can be found out by using,

Therefore, the roots of the equation are given as follows,

The values of x for both the cases will be:

x = 5

Also,

x = 5/2

Question: 21Solve for x:

Solution:The above equation can be solved as follows:

x

^{2}+ 1 = 3xx

^{2}- 3x + 1 = 0The above equation is in the form of ax

^{2}+ bx + c = 0The discriminant is given by the equation, D = b

^{2 }- 4acHere, a = 1, b = - 3, c = 1

D = (-3)

^{2 }- 4(1)(1)D = 9 - 4

D = 5

The roots of an equation can be found out by using,

The values of x for both the cases will be:

Question: 22Solve for x:

Solution:The above equation can be solved as follows:

(16 - x)(x + 1) = 15x

16x + 16 - x

^{2}- x = 15x15x + 16 - x

^{2}- 15x = 016 - x

^{2}= 0X

^{2}- 16 = 0The above equation is in the form of ax

^{2}+ bx + c = 0The discriminant is given by the equation, D = b

^{2 }- 4acHere, a = 1, b = 0, c = -16

D = (0)

^{2}- 4(1)(-16)D = 64

The roots of an equation can be found out by using,

Therefore, the roots of the equation are given as follows,