Question

fill in the blank 1 point
complete the square. find the vertex and roots.
ex: 2(x+4)^2 - 12 = 0, (-4,12), -4 - 2sqrt(3), -4 + 2sqrt(3)
2x^2 + 12x + 10 = 0 type your answer...

Explanation:

Step1: Factor out the coefficient of \(x^2\)

First, factor out the 2 from the first two terms of the quadratic equation \(2x^2 + 12x + 10 = 0\). We get \(2(x^2 + 6x) + 10 = 0\).

Step2: Complete the square inside the parentheses

To complete the square for \(x^2 + 6x\), we take half of the coefficient of \(x\) (which is \(6\)), square it, and add and subtract it inside the parentheses. Half of \(6\) is \(3\), and \(3^2 = 9\). So we have:

$$
LATEXBLOCK0
$$

Step3: Simplify the equation

Now distribute the 2:

$$
LATEXBLOCK1
$$

Step4: Find the vertex

For a quadratic in vertex form \(y = a(x - h)^2 + k\), the vertex is \((h, k)\). From \(2(x + 3)^2 - 8 = 0\) (or \(y = 2(x + 3)^2 - 8\)), the vertex is \((-3, -8)\).

Step5: Find the roots

Set the equation \(2(x + 3)^2 - 8 = 0\) equal to zero and solve for \(x\):

$$
LATEXBLOCK2
$$

So \(x = -3 + 2 = -1\) or \(x = -3 - 2 = -5\).

Answer:

\(2(x + 3)^2 - 8 = 0\), \((-3, -8)\), \(-5\), \(-1\)