To solve the given equation:
[tex]\[
\frac{k}{k+6} = \frac{2}{k-2}
\][/tex]
we need to follow these steps:
1. Cross Multiply:
[tex]\[
k(k - 2) = 2(k + 6)
\][/tex]
2. Distribute:
[tex]\[
k^2 - 2k = 2k + 12
\][/tex]
3. Move all terms to one side of the equation:
[tex]\[
k^2 - 2k - 2k - 12 = 0
\][/tex]
which simplifies to:
[tex]\[
k^2 - 4k - 12 = 0
\][/tex]
4. Solve the quadratic equation:
Use the quadratic formula [tex]\( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -12 \)[/tex].
Calculate the discriminant:
[tex]\[
b^2 - 4ac = (-4)^2 - 4(1)(-12) = 16 + 48 = 64
\][/tex]
So, the solutions are:
[tex]\[
k = \frac{4 \pm \sqrt{64}}{2 \cdot 1} = \frac{4 \pm 8}{2}
\][/tex]
This gives us two solutions:
[tex]\[
k = \frac{4 + 8}{2} = \frac{12}{2} = 6
\][/tex]
and
[tex]\[
k = \frac{4 - 8}{2} = \frac{-4}{2} = -2
\][/tex]
5. Check the solutions:
Ensure the solutions do not make any denominator zero.
For [tex]\( k = 6 \)[/tex]:
[tex]\[
k + 6 = 6 + 6 = 12 \quad \text{(not zero)}
\][/tex]
[tex]\[
k - 2 = 6 - 2 = 4 \quad \text{(not zero)}
\][/tex]
For [tex]\( k = -2 \)[/tex]:
[tex]\[
k + 6 = -2 + 6 = 4 \quad \text{(not zero)}
\][/tex]
[tex]\[
k - 2 = -2 - 2 = -4 \quad \text{(not zero)}
\][/tex]
Both solutions [tex]\( k = 6 \)[/tex] and [tex]\( k = -2 \)[/tex] are valid.
Thus, the correct choice is:
A. The solution(s) is/are [tex]\( k =\)[/tex] 6, -2.
