Question

You notice that the following quadratic equation cannot be factored, so you solve it by completing the square. What value of c would make the left-hand side of this equation a perfect square trinomial?: x^2 - 8x + c = 13

Answer 1

To make the left-hand side of the equation \( x^2 - 8x + c \) a perfect square trinomial, we need to complete the square.

The general form for completing the square for the expression \( x^2 + bx \) is:

\[
x^2 + bx = \left( x + \frac{b}{2} \right)^2 - \left( \frac{b}{2} \right)^2
\]

In our case, we have \( b = -8 \). We can find \( \frac{b}{2} \):

\[
\frac{-8}{2} = -4
\]

Next, we calculate \( \left( \frac{b}{2} \right)^2 \):

\[
(-4)^2 = 16
\]

To complete the square, we need to add and subtract this value. Thus, we rewrite the expression:

\[
x^2 - 8x + 16 - 16 + c
\]

To make the left-hand side a perfect square trinomial, we set \( c \) such that:

\[
c - 16 = 0 \implies c = 16
\]

Thus, the value of \( c \) that would make the left-hand side a perfect square trinomial is

\[
\boxed{16}
\]

Now, the left-hand side becomes:

\[
x^2 - 8x + 16 = (x - 4)^2
\]

So, the revised equation becomes:

\[
(x - 4)^2 = 13
\]