Question

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

Answer 1

`x_1 = -1 + √(2)i\\\\x_1 = -1 - √(2)i`

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:

`-3x ^ 2 -6x -9 = 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 = -3`. Then:

`-3(x ^ 2 + 2x) = 9`

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

We have:

`(2)/(2) = 1\\\\((2)/(2))^2 = 1`

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

`-3(x ^ 2 + 2x + 1) = 9 -3(1)`

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

`-3(x + 1) ^ 2 = 6`

`(x + 1) ^ 2 = -(6)/(3)`

`(x + 1) ^ 2 >0`

Then, The equation has no solution in real numbers.

In the same way we can find the complex roots:

Now solve the equation:

`(x + 1) ^ 2 = -(6)/(3)\\\\x+1 = √(-2)\\\\x = -1 + √(-2)\\\\x_1 = -1 + √(2)i\\\\x_1 = -1 - √(2)i`

Answer 2

Hello!

The answer is: There is no solution in the real numbers for the given equation since we have as result complex roots.

Complex roots:

`x_(1)=-1+√(2)i \\x_(2)=-1-√(2)i`

Why?

We can re-write the given function in a simplest way:

`-3x^(2)-6x-9=0\\-3x^(2)-6x=9`

Then, dividing each side into -3, we have:

`x^(2)+2x=-3`

Finding each term:

`a=1\\b=2\\c=3`

Adding
`((b)/(2))^(2)` to each side, we have:

`x^(2)+2x+((2)/(2))^(2)=-3+((2)/(2))^(2)`

`x^(2)+2x+1=-3+1`

Then,

`(x+1)^(2)=-2\\\sqrt{(x+1)^(2)}=√(-2)\\x+1=√(-2)`

`x_(1)=-1+√(-2) \\x_(2)=-1-√(-2)`

Since there is no negative roots in the real numbers, there is no solution for the given equation.

Have a nice day!