Question

kent multiplies both sides of the equation by an expression.
k + \frac{12}{k} = 8
then he moves all the terms to one side of the equal sign in the resulting equation. which equation must he solve now?
\boxed{k^2 - 8 + 12 = 0} \boxed{k^2 - 8k + 12 = 0} \boxed{k^2 + 8k + 12 = 0}
\boxed{k^2 - 8k - 12 = 0}

Explanation:

Step1: Multiply both sides by \( k \)

To eliminate the denominator \( k \) (assuming \( k
eq 0 \)), multiply each term in the equation \( k+\frac{12}{k}=8 \) by \( k \).
We get \( k \times k + k \times \frac{12}{k}=8 \times k \), which simplifies to \( k^{2}+12 = 8k \).

Step2: Move all terms to one side

Subtract \( 8k \) from both sides to move all terms to the left - hand side of the equation.
\( k^{2}+12 - 8k=0 \), and re - arranging the terms in standard quadratic form \( ax^{2}+bx + c = 0 \), we have \( k^{2}-8k + 12=0 \).

Answer:

\( k^{2}-8k + 12 = 0 \) (the second option among the given options)