You notice that the following problem 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?

x2 −14x + c = 13


(1 point)
Responses

196
196

4
4

49
49

-7
-7

-14

1 answer

To make the left-hand side of the equation \(x^2 - 14x + c\) a perfect square trinomial, we need to determine the value of \(c\).

A perfect square trinomial can be expressed in the form of \((x - a)^2\), which expands to \(x^2 - 2ax + a^2\). Here, the coefficient for \(x\) is \(-14\), which corresponds to \(-2a\). From this, we can find \(a\):

\[ -2a = -14 \implies a = \frac{14}{2} = 7 \]

The constant term in the perfect square trinomial is \(a^2\):

\[ a^2 = 7^2 = 49 \]

So, we need \(c\) to equal \(49\) in order for \(x^2 - 14x + c\) to be a perfect square trinomial.

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

\[ \boxed{49} \]