Question

what is the solution to $2x^2 + 8x = x^2 - 16$?
\\(\bigcirc\\) $x = -4$
\\(\bigcirc\\) $x = -2$
\\(\bigcirc\\) $x = 2$
\\(\bigcirc\\) $x = 4$

Explanation:

Step1: Simplify the equation

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

$$
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Step2: Factor the quadratic equation

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

Step3: Solve for \(x\)

Set \(x + 4 = 0\), then \(x=-4\). We can also verify by plugging \(x = - 4\) back into the original equation:
Left - hand side: \(2\times(-4)^2+8\times(-4)=2\times16-32 = 32 - 32=0\)
Right - hand side: \((-4)^2-16 = 16 - 16 = 0\)
So \(x=-4\) satisfies the original equation.

Answer:

A. \(x = - 4\)