Question

which value of x is a solution to $x^2 + 8x + 16 = 0$?
$\bigcirc$ $x = -8$
$\bigcirc$ $x = -4$
$\bigcirc$ $x = 4$
$\bigcirc$ $x = 8$

Explanation:

Step1: Factor the quadratic equation

The quadratic equation \(x^2 + 8x + 16 = 0\) can be factored as a perfect square trinomial. Recall that \((a + b)^2 = a^2 + 2ab + b^2\). Here, \(a = x\), \(2ab = 8x\), so \(2b = 8\) which gives \(b = 4\), and \(b^2 = 16\). So, \(x^2 + 8x + 16=(x + 4)^2\). Thus, the equation becomes \((x + 4)^2 = 0\).

Step2: Solve for \(x\)

Taking the square root of both sides, we get \(x + 4 = 0\) (since the square root of 0 is 0). Solving for \(x\), we subtract 4 from both sides: \(x=-4\).

We can also verify by plugging in the options:

• For \(x = -8\): \((-8)^2+8\times(-8)+16 = 64 - 64 + 16=16

eq0\)

• For \(x = -4\): \((-4)^2+8\times(-4)+16 = 16 - 32 + 16 = 0\)
• For \(x = 4\): \(4^2+8\times4+16 = 16 + 32 + 16 = 64

eq0\)

• For \(x = 8\): \(8^2+8\times8+16 = 64 + 64 + 16 = 144

eq0\)

Answer:

\(x = -4\) (corresponding to the option with \(x=-4\))