Question

Solve each quadratic equation by completing the square. Give exact answers--no decimals.

Answer 1

x = -2+2i or x = -2-2i

Explanation:

We have given the equation:

7x² +28x+56 = 0

We have to solve it by completing the square.

7x² +28x+56 = 0

7( x² + 4x + 8) = 0

( x² + 4x + 8) = 0

x²+2(x)(2) +2² = -8+2²

(x+2)² = -8 +4

(x+2)² = -4

Taking square root on both sides we get,

x+2 = √-4

(x+2) = ±2i as (√i = -1)

x+2 = 2i or x+2 = -2i

x = -2+2i or x = -2-2i is the answer.

Answer 2

`x_1 = -2 + 2i\\\\x_2 = -2 -2i`

Explanation:

In this problem we have the equation of the following quadratic equation and we want to solve it using the method of square completion:

`7x ^ 2 + 28x +56 = 0`

The steps are shown below:

For any equation of the form:
`ax ^ 2 + bx + c = 0`

1. If the coefficient a is different from 1, then take a as a common factor.

In this case
`a = 7`. Then:

`7(x ^ 2 + 4x) = -56`

2. Take the coefficient b that accompanies the variable x. In this case the coefficient is 4. Then, divide by 2 and the result squared it.

We have:

`(4)/(2) = 2\\\\((4)/(2))^2 = 4`

3. Add the term obtained in the previous step on both sides of equality, remember to multiply by the common factor
`a = 7`:

`7(x ^ 2 + 4x + 4) = -56 + 7(4)`

4. Factor the resulting expression, and you will get:

`7(x + 2) ^ 2 = -28`

Now solve the equation:

Note that the term
`7(x + 2) ^ 2` is always
`> 0` therefore it can not be equal to -28.

The equation has no solution in real numbers.

In the same way we can find the complex roots:

`7(x+2)^2 = -28\\\\(x+2)^2 = -(28)/(7)\\\\x+2 = \sqrt{-(28)/(7)}\\\\x = -2 + √(-4)\\\\x = -2 + 2√(-1)\\\\x_1 = -2 + 2i\\\\x_2 = -2 -2i`