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 - 6x + c = 15

1 answer

To determine the value of \( c \) that makes the left-hand side of the equation \( x^2 - 6x + c = 15 \) a perfect square trinomial, we focus on the expression \( x^2 - 6x + c \).

A quadratic expression \( x^2 + bx + c \) can be factored into a perfect square trinomial if \( c \) is equal to \(\left(\frac{b}{2}\right)^2\).

In this case, \( b = -6 \). Therefore, we calculate:

\[ \frac{b}{2} = \frac{-6}{2} = -3 \]

Next, we find \(\left(-3\right)^2\):

\[ \left(-3\right)^2 = 9 \]

Thus, for \( x^2 - 6x + c \) to be a perfect square trinomial, we need:

\[ c = 9 \]

The modified equation becomes:

\[ x^2 - 6x + 9 = 15 \]

So, the value of \( c \) that makes the left side a perfect square trinomial is:

\[ \boxed{9} \]