Question

which equation is the inverse of $2(x - 2)^2 = 8(7 + y)$?
$\circ -2(x - 2)^2 = -8(7 + y)$
$\circ y = \frac{1}{4}x^2 - x - 6$
$\circ y = -2 \pm \sqrt{28 + 4x}$
$\circ y = 2 \pm \sqrt{28 + 4x}$

Explanation:

Step1: Start with the original equation

The original equation is \( 2(x - 2)^2 = 8(7 + y) \). To find the inverse, we first solve for \( y \) in terms of \( x \), then swap \( x \) and \( y \) and solve for the new \( y \).

First, divide both sides of the original equation by 2:
\( (x - 2)^2 = 4(7 + y) \)

Step2: Solve for \( y \) in the original equation

Expand the right - hand side: \( (x - 2)^2=28 + 4y \)
Then, isolate \( y \):
\( 4y=(x - 2)^2-28 \)
\( y=\frac{(x - 2)^2-28}{4}=\frac{x^{2}-4x + 4-28}{4}=\frac{x^{2}-4x-24}{4}=\frac{1}{4}x^{2}-x - 6 \)

Step3: Find the inverse by swapping \( x \) and \( y \)

Now, swap \( x \) and \( y \) in the equation \( y=\frac{1}{4}x^{2}-x - 6 \) to get \( x=\frac{1}{4}y^{2}-y - 6 \)

Multiply both sides by 4 to eliminate the fraction: \( 4x=y^{2}-4y - 24 \)

Rearrange the equation to complete the square for \( y \):
\( y^{2}-4y=4x + 24 \)
Complete the square for the left - hand side. The coefficient of \( y \) is - 4, half of it is - 2, and squaring it gives 4. Add 4 to both sides:
\( y^{2}-4y + 4=4x + 24+4 \)
\( (y - 2)^2=4x + 28 \)

Take the square root of both sides:
\( y - 2=\pm\sqrt{4x + 28} \)
Solve for \( y \):
\( y=2\pm\sqrt{4x + 28} \)

Answer:

\( y = 2\pm\sqrt{28 + 4x} \) (the fourth option)