**Question 1.Solve the following quadratic equations by completing the square method.1. x**

^{2}+ x – 20 = 02. x

^{2}+ 2x – 5 = 03. m

^{2}– 5m = -34. 9y

^{2}– 12y + 2 = 05. 2y

^{2}+ 9y + 10 = 06. 5x

^{2}= 4x + 7Solution:

1. x

^{2}+ x – 20 = 0

If x

^{2}+ x + k = (x + a)

^{2}, then

x

^{2}+ x + k = x

^{2}+ 2ax + a

^{2}

Comparing the coefficients, we get

1 = 2a and k = a

^{2}

∴ The roots of the given quadratic equation are 4 and -5.

2. x^{2} + 2x – 5 = 0

If x^{2} + 2x + k = (x + a)^{2}, then

x^{2} + 2x + k = x^{2} + 2ax + a^{2}

Comparing the coefficients, we get

2 = 2a and k = a^{2}

∴ a = 1 and k = (1)^{2} = 1

Now, x^{2} + 2x – 5 = 0

∴ x^{2} + 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. m^{2} – 5m = -3

∴ m^{2} – 5m + 3 = 0

If m^{2} – 5m + k = (m + a)^{2}, then

m^{2} – 5m + k = m^{2} + 2am + a^{2}

Comparing the coefficients, we get

-5 = 2a and k = a^{2}

4. 9y^{2} – 12y + 2 = 0

5. 2y^{2} + 9y + 10 = 0

Taking square root of both sides, we get

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

6. 5x^{2} = 4x + 7

∴ 5x^{2} – 4x – 7 = 0

Comparing the coefficients, we get