Question

completing the square
solve by completing the square.
$3x^2 + 36x + 24 = 0$
$x = \square\square \pm \square\sqrt{\square}$

Explanation:

Step1: Divide by coefficient of \(x^2\)

Divide the entire equation \(3x^2 + 36x + 24 = 0\) by 3:
\(\frac{3x^2}{3} + \frac{36x}{3} + \frac{24}{3} = \frac{0}{3}\)
Simplify: \(x^2 + 12x + 8 = 0\)

Step2: Isolate \(x^2\) and \(x\) terms

Subtract 8 from both sides:
\(x^2 + 12x = -8\)

Step3: Complete the square

Take half of 12 (\( \frac{12}{2} = 6 \)), square it (\(6^2 = 36\)), add to both sides:
\(x^2 + 12x + 36 = -8 + 36\)
Factor left side: \((x + 6)^2 = 28\)

Step4: Solve for \(x\)

Take square roots:
\(x + 6 = \pm\sqrt{28}\)
Simplify \(\sqrt{28} = 2\sqrt{7}\):
\(x = -6 \pm 2\sqrt{7}\)

Answer:

\(x = -6 \pm 2\sqrt{7}\) (So in the boxes: \(-6\), \(2\), \(7\))