Quadratic equations are widely used in Mathematics. In this exercise, the students will be able to identify quadratic equations and also verify the solution of a quadratic equation. The Selina Solutions for Class 10 Maths is the best resource students can rely on for learning the correct methods to solve these problems. The solutions of this exercise are available in PDF format in the Concise Selina Solutions for Class 10 Maths Chapter 5 Quadratic Equations Exercise 5(A) PDF in the links below.

## Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations Exercise 5(A) Download PDF

### Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations

### Access Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations Exercise 5(A)

**1. Find which of the following equations are quadratic: **

**(i) (3x – 1) ^{2} = 5(x + 8)**

**(ii) 5x ^{2} â€“ 8x = -3(7 â€“ 2x)**

**(iii) (x -4) (3x + 1) = (3x â€“ 1) (x + 2)**

**(iv) x ^{2} + 5x â€“ 5 = (x – 3)^{2}**

**(v) 7x ^{3} â€“ 2x^{2} + 10 = (2x – 5)^{2} **

**(vi) (x – 1) ^{2} + (x + 2)^{2} + 3(x + 1) = 0**

**Solution: **

(i) (3x – 1)^{2}Â = 5(x + 8)

â‡’Â (9x^{2}Â – 6x + 1) = 5x + 40

â‡’Â 9x^{2}Â – 11x – 39 = 0; which is of the general form ax^{2}Â + bx + c = 0.

Thus, the given equation is a quadratic equation.

(ii) 5x^{2}Â – 8x = -3(7 – 2x)

â‡’Â 5x^{2}Â – 8x = 6x – 21

â‡’Â 5x^{2}Â – 14x + 21 = 0; which is of the general form ax^{2}Â + bx + c = 0.

Thus, the given equation is a quadratic equation.

(iii) (x – 4) (3x + 1) = (3x – 1) (x +2)

â‡’Â 3x^{2}Â + x – 12x – 4 = 3x^{2}Â + 6x – x – 2

â‡’Â 16x + 2 = 0; which is not of the general form ax^{2}Â + bx + c = 0. And itâ€™s a linear equation.

Thus, the given equation is not a quadratic equation.Â

(iv) x^{2}Â + 5x – 5 = (x – 3)^{2}

â‡’Â x^{2}Â + 5x – 5 = x^{2}Â – 6x + 9

â‡’Â 11x – 14 = 0; which is not of the general form ax^{2}Â + bx + c = 0. And itâ€™s a linear equation.

Thus, the given equation is not a quadratic equation.Â

(v) 7x^{3}Â – 2x^{2}Â + 10 = (2x – 5)^{2}

â‡’Â 7x^{3}Â – 2x^{2}Â + 10 = 4x^{2}Â – 20x + 25

â‡’Â 7x^{3}Â – 6x^{2}Â + 20x – 15 = 0; which is not of the general form ax^{2}Â + bx + c = 0. And itâ€™s a cubic equation.

Thus, the given equation is not a quadratic equation.

(vi) (x – 1)^{2}Â + (x + 2)^{2}Â + 3(x +1) = 0

â‡’Â x^{2}Â – 2x + 1 + x^{2}Â + 4x + 4 + 3x + 3 = 0

â‡’Â 2x^{2}Â + 5x + 8 = 0; which is of the general form ax^{2}Â + bx + c = 0.

Thus, the given equation is a quadratic equation.

**2. (i) Is x = 5 a solution of the quadratic equation x ^{2}Â – 2x – 15 = 0?**

**Solution: **

Given quadratic equation, x^{2}Â – 2x – 15 = 0

We know that, for x = 5 to be a solution of the given quadratic equation it should satisfy the equation.

Now, on substituting x = 5 in the given equation, we have

L.H.S = (5)^{2}Â – 2(5) – 15

Â = 25 – 10 – 15

Â = 0

Â = R.H.S

Therefore, x = 5 is a solution of the given quadratic equation x^{2}Â – 2x – 15 = 0

**(ii) Is x = -3 a solution of the quadratic equation 2x ^{2}Â – 7x + 9 = 0?**

**Solution: **

Given quadratic equation, 2x^{2}Â – 7x + 9 = 0

We know that, for x = -3 to be solution of the given quadratic equation it should satisfy the equation.

Now, on substituting x = 5 in the given equation, we have

L.H.S = 2(-3)^{2}Â – 7(-3) + 9

= 18 + 21 + 9Â Â Â Â

=Â 48Â

â‰ Â R.H.S

Therefore, x = -3 is not a solution of the given quadratic equation 2x^{2}Â – 7x + 9 = 0.Â