Question

solve $x^2 + 8x - 3 = 0$ using the completing-the-square method.
$\bigcirc$ $x = 4 \pm \sqrt{3}$
$\bigcirc$ $x = 4 \pm \sqrt{19}$
$\bigcirc$ $x = -4 \pm \sqrt{19}$
$\bigcirc$ $x = -4 \pm \sqrt{3}$

Explanation:

Step1: Isolate the quadratic and linear terms

Start with the equation \( x^2 + 8x - 3 = 0 \). Add 3 to both sides to get \( x^2 + 8x = 3 \).

Step2: Complete the square

Take half of the coefficient of \( x \) (which is 8), so \( \frac{8}{2} = 4 \), then square it: \( 4^2 = 16 \). Add 16 to both sides: \( x^2 + 8x + 16 = 3 + 16 \).

Step3: Rewrite as a perfect square

The left side becomes \( (x + 4)^2 \), and the right side is \( 19 \), so \( (x + 4)^2 = 19 \).

Step4: Solve for \( x \)

Take the square root of both sides: \( x + 4 = \pm\sqrt{19} \). Subtract 4 from both sides: \( x = -4 \pm\sqrt{19} \).

Answer:

C. \( x = -4 \pm\sqrt{19} \) (assuming the third option is labeled C; if the original options had labels like A, B, C, D, adjust accordingly. Based on the options given, the correct one is \( x = -4 \pm\sqrt{19} \) which is the third option in the list)