In this chapter, we provide RD Sharma Class 10 Ex 8.5 Solutions Chapter 8 Quadratic Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 8.5 Solutions Chapter 8 Quadratic Equations pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 8 |

Chapter Name | Quadratic Equations |

Exercise | 8.5 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **8**** Quadratic Equations** Ex 8.5 Download PDF

**Quadratic Equations**Ex 8.5 Download PDF

**Chapter 8: Quadratic Equations Exercise – 8.5**

**Question: 1**

Find the discriminant of the following quadratic equations:

2x^{2} – 5x + 3 = 0

**Solution:**

2x^{2} – 5x + 3 = 0

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

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

The discriminant, D = b^{2} – 4ac

D = (-5)^{2} – 4 x 2 x 3

D = 25 – 24 = 1

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

**Question: 2**

Find the discriminant of the quadratic equations.

x^{2} + 2x + 4 = 0

**Solution:**

x^{2} + 2x + 4 = 0

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

Here, a = 1, b = 2 and c = 4

The discriminant is:-

D = (2)^{2} – 4 x 1 x 4

D = 4 – 16 = – 12

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

**Question: 3**

Find 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 = 0

Now this equation is in the form of ax^{2} + bx + c = 0

Here, a = 2, b = -3, c = 1

The discriminant is:-

D = (-3)^{2} – 4 x 2 x 1

D = 9 – 8 = 1

The discriminant of the following quadratic equation is = 1.

**Question: 4**

Find the discriminant of the quadratic equations.

x^{2 }– 2x + k = 0

**Solution:**

x^{2 }– 2x + k = 0

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

Here, a = 1, b = -2, and c = k

D = b^{2} – 4ac

D = (-2)^{2} – 4(1)(k)

= 4 – 4k

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

**Question: 5**

Find the discriminant of the quadratic equations.

**Solution:**

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

The discriminant is, D = b^{2} – 4ac

D = 8 + 24 = 32

The discriminant, D of the following equation is 32.

**Question: 6**

Find the discriminant of the quadratic equations.

x^{2} – x + 1 = 0

**Solution:**

x^{2} – x + 1 = 0

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

Here, a = 1, b = -1 and c = 1

The discriminant is D = b^{2 }– 4ac

(-1)^{2} – 4 × 1 × 1

1 – 4 = – 3

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

**Question: 7**

Find the discriminant of the quadratic equations.

x^{2} + x + 2 = 0

**Solution:**

x^{2} + x + 2 = 0

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

The discriminant is given by the following equation, D = b^{2} – 4ac

Here, 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: 8**

Find the discriminant of the quadratic equations.

16x^{2} = 24x + 1

**Solution:**

16x^{2} – 24x – 1 = 0

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Here, 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: 9**

Find the discriminant of the quadratic equations.

x^{2} + x + 2 = 0

**Solution:**

x^{2} + x + 2 = 0

Here, 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: 10**

Find the discriminant of the quadratic equations.

**Solution:**

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Therefore, 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: 11**

Find the discriminant of the quadratic equations.

3x^{2} – 2x + 2 = 0

**Solution:**

3x^{2} – 2x + 2 = 0

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Here, 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: 12**

Find the discriminant of the quadratic equations.

**Solution:**

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Therefore, 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: 13**

Find the discriminant of the quadratic equations.

3a^{2}x^{2 }+ 8abx + 4b^{2} = 0

**Solution:**

3a^{2}x^{2 }+ 8abx + 4b^{2} = 0

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Here, 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: 14**

Find the discriminant of the quadratic equations.

**Solution:**

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Therefore, 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: 15**

Find the discriminant of the quadratic equations.

x^{2} – 2x + 1 = 0

**Solution:**

x^{2} – 2x + 1 = 0

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Here, 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: 16**

Find the discriminant of the quadratic equations.

**Solution:**

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Therefore, 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: 17**

Find the discriminant of the quadratic equations.

**Solution:**

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Therefore, 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: 18**

Find the discriminant of the quadratic equations.

**Solution:**

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Therefore, 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: 19**

Find the discriminant of the quadratic equations.

3x^{2} – 5x + 2 = 0

**Solution:**

3x^{2} – 5x + 2 = 0

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

The discriminant is given by the following equation, D = b^{2 }– 4ac

Here, 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: 20**

Solve for x:

**Solution:**

The above equation can be solved as follows:

6x^{2} – 30x + 30 = 10x^{2} – 60x +80

4x^{2} – 30x + 50 = 0

2x^{2} – 15x + 25 = 0

The above equation is in the form of ax^{2} + bx + c = 0

The discriminant is given by the equation, D = b^{2 }– 4ac

Here, 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: 21**

Solve for x:

**Solution: **

The above equation can be solved as follows:

x^{2} + 1 = 3x

x^{2} – 3x + 1 = 0

The above equation is in the form of ax^{2} + bx + c = 0

The discriminant is given by the equation, D = b^{2 }– 4ac

Here, 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: 22**

Solve for x:

**Solution:**

The above equation can be solved as follows:

(16 – x)(x + 1) = 15x

16x + 16 – x^{2} – x = 15x

15x + 16 – x^{2} – 15x = 0

16 – x^{2} = 0

X^{2} – 16 = 0

The above equation is in the form of ax^{2} + bx + c = 0

The discriminant is given by the equation, D = b^{2 }– 4ac

Here, 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,

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