Question

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

Answer 1

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:

$x_{1}=-1+\sqrt{2}i \\x_{2}=-1-\sqrt{2}i$

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 $(\frac{b}{2})^{2}$ to each side, we have:

$x^{2}+2x+(\frac{2}{2})^{2}=-3+(\frac{2}{2})^{2}$

$x^{2}+2x+1=-3+1$

Then,

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

$x_{1}=-1+\sqrt{-2} \\x_{2}=-1-\sqrt{-2}$

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

Have a nice day!

Answer 2

Answer:

$x_1 = -1 + \sqrt{2}i\\\\x_1 = -1 - \sqrt{2}i$

Step-by-step 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:

$\frac{2}{2} = 1\\\\(\frac{2}{2})^2 = 1$

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:

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

$(x + 1) ^ 2 = -\frac{6}{3}$

$(x + 1) ^ 2 >0$

Then, The equation has no solution in real numbers.

In the same way we can find the complex roots:

Now solve the equation:

$(x + 1) ^ 2 = -\frac{6}{3}\\\\x+1 = \sqrt{-2}\\\\x = -1 + \sqrt{-2}\\\\x_1 = -1 + \sqrt{2}i\\\\x_1 = -1 - \sqrt{2}i$