Question

what is the only solution of $2x^2 + 8x = x^2 - 16$?

Explanation:

Step1: Simplify the equation

Subtract \(x^2\) and add 16 to both sides of the equation \(2x^2 + 8x = x^2 - 16\) to get a quadratic equation in standard form.
\(2x^2 - x^2 + 8x + 16 = 0\)
\(x^2 + 8x + 16 = 0\)

Step2: Factor the quadratic equation

Notice that \(x^2 + 8x + 16\) is a perfect square trinomial, which can be factored as \((x + 4)^2 = 0\)

Step3: Solve for \(x\)

Set the factor equal to zero:
\(x + 4 = 0\)
Subtract 4 from both sides:
\(x = -4\)

Answer:

\(-4\)