# Solve x^{2} + 8x = 33 by completing the square. Which is the solution set of the equation?

{-11, 3}, {-3, 11}, {-4, 4}, {-7, 7}

**Solution:**

Given equation x^{2} + 8x = 33

We complete the square for making the LHS into a perfect square trinomial.

- Divide the coefficient of the x term by 2 then square the result.
- This number will be added to both sides of the equation.

For the quadratic equation x^{2} - 8x = 3, the coefficient of the x term is 8

So (8/2)^{2} = (4)^{2 }= 16

⇒ x^{2} + 8x +16 = 33 + 16

⇒ x^{2} + 2(4)(x) + 16 = 49

⇒ {x^{2} + 2(x)(4) + 4^{2}} = 49 [ since a^{2 }+ 2ab + b^{2} = (a + b)^{2}]

⇒ (x + 4)^{2 }= 49

Applying square root on both sides, we get

⇒ x + 4 = √49

⇒x = ±7 - 4

⇒ x = -11, 3

The solution set is {-11, 3}

## Solve x^{2} + 8x = 33 by completing the square. Which is the solution set of the equation?

**Summary:**

By solving the equation x^{2} + 8x = 33 by completing the square, we get solution set as {-11, 3}.

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