Chapter 2 Quadratic Equations Set 2.3

Question 1.
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:
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