Maharashtra State Board Class 10 (SSC Board) Mathematics & Statistics Solutions Chapter 2 Quadratic Equations Practice Set 2.3 - Digest Answers | Maharashtra Board Solutions

Practice Set 2.3 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1 Maharashtra Board Solution

Solve the following quadratic equations by completing the square method. 1. x2 + x – 20 = 0 2. x2 + 2x – 5 = 0 3. m2 – 5m = -3 4. 9y2 – 12y + 2 = 0 5. 2y2 + 9y + 10 = 0 6. 5x2 = 4x + 7

Solution & Step-by-Step Answer:

1. x2 + x – 20 = 0 If x2 + x + k = (x + a)2, then x2 + x + k = x2 + 2ax + a2 Comparing the coefficients, we get 1 = 2a and k = a2 ∴ The roots of the given quadratic equation are 4 and -5.

2. x²+ 2x – 5 = 0
If x²+ 2x + k = (x + a)², then
x²+ 2x + k = x²+ 2ax + a²
Comparing the coefficients, we get
2 = 2a and k = a²
∴ a = 1 and k = (1)²= 1
Now, x²+ 2x – 5 = 0
∴ x²+ 2x + 1 – 1 – 5 = 0
∴ (x + 1)²– 6 = 0
∴ (x + 1)²= 6
Taking square root of both sides, we get
x + 1 = ± √6
∴ x + 1 √6 or x + 1 = √6
∴ x = √6 – 1 or x = -√6 – 1
∴ The roots of the given quadratic equation are √6 -1 and – √6 -1.

3. m²– 5m = -3
∴ m²– 5m + 3 = 0
If m²– 5m + k = (m + a)², then
m²– 5m + k = m²+ 2am + a²
Comparing the coefficients, we get
-5 = 2a and k = a²

4. 9y²– 12y + 2 = 0

5. 2y²+ 9y + 10 = 0

Taking square root of both sides, we get

∴ The roots of the given quadratic equation are -2 and .

6. 5x²= 4x + 7
∴ 5x²– 4x – 7 = 0

Comparing the coefficients, we get

Chapter 2 Quadratic Equations Practice Set 2.3 Solutions