Question

The quadratic equation x^2 - 6x = 12 is rewritten in the form (x - p)^2 = q, where q is a constant. What is the value of p? Hint: After you complete the square, move constant to right side to rewrite in the form (x - p)2 = q and constant term sign will change.

Answer 1

To solve the quadratic equation \( x^2 - 6x = 12 \) and rewrite it in the form \( (x - p)^2 = q \), we will first bring all terms involving \( x \) to one side and constants to the other side.

1. Start with the equation:
\[
x^2 - 6x = 12
\]

2. Move the constant term \( 12 \) to the left side:
\[
x^2 - 6x - 12 = 0
\]

3. Now, we want to complete the square for the expression \( x^2 - 6x \). To complete the square, we need to find the value that makes this a perfect square trinomial. We take half the coefficient of \( x \) (which is \( -6 \)), square it, and add it inside the equation.

Half of \( -6 \) is \( -3 \), and squaring it gives \( 9 \):
\[
x^2 - 6x + 9 - 9 - 12 = 0
\]

4. Rewrite the equation by factoring and rearranging:
\[
(x - 3)^2 - 21 = 0
\]

5. Move \( -21 \) to the right side:
\[
(x - 3)^2 = 21
\]

Now, we can see that this is in the form \( (x - p)^2 = q \), where \( p = 3 \) and \( q = 21 \).

Thus, the value of \( p \) is:
\[
\boxed{3}
\]