Question

solving a quadratic equation when a ≠ 1 by completing the square
consider ( 8x^2 - 48x = -104 )
write the equation so that
a = 1 ( x^2 + square x = square )
complete the square
( x^2 - 6x + square = -13 + square )
factor the trinomial and simplify
( (x + square )^2 = square )
use the square root property of equality to solve ( (x - 3)^2 = -4 )
the solutions are
options:
-1 or 7
1 or 5
( 3 pm 2i )
( 3 pm 4i )

Explanation:

Step1: Divide by 8 to make a=1

$\frac{8x^2 - 48x}{8} = \frac{-104}{8}$
$x^2 - 6x = -13$

Step2: Complete the square

Take half of -6: $\frac{-6}{2}=-3$, square it: $(-3)^2=9$. Add 9 to both sides:
$x^2 - 6x + 9 = -13 + 9$

Step3: Rewrite as perfect square

$(x - 3)^2 = -4$

Step4: Apply square root property

$x - 3 = \pm\sqrt{-4} = \pm2i$

Step5: Solve for x

$x = 3 \pm 2i$

Answer:

$3 \pm 2i$ (corresponding to the option: $3 \pm 2i$)