Question

Find the value of c that makes each trinomial a perfect square
x^2-19x+c
Please Help! :)

Answer 1

The value of c is $\frac{361}{4}$

Explanation:

Given that the trinomial is $x^2-19x+c$

We need to determine the value of c such that the trinomial is a perfect square.

The value of c can be determined using the formula,

$c=(\frac{b}{2})^2$

From the trinomial, the value of b is given by

$b=-19$

Substituting the value of b in the above formula, we have,

$c=(\frac{-19}{2} )^2$

Squaring both the numerator and denominator, we have,

$c=\frac{361}{4}$

Thus, the value of c is $\frac{361}{4}$ which makes the trinomial a perfect square.