Chapter 9.4, Problem 12PPS

To determine

Find the value of c that makes the trinomial a perfect square.

Answer to Problem 12PPS

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

Explanation of Solution

Given:

The trinomial: $x^2-19x+c$

Concept Used:

The perfect square term: ${(x+\frac{b}{2})}^2=x^2+2·x·\frac{b}{2}+{(\frac{b}{2})}^2$

Calculation:

The perfect square term: ${(x+\frac{b}{2})}^2=x^2+2·x·\frac{b}{2}+{(\frac{b}{2})}^2$

In this trinomial $x^2-19x+c$ : $b=-19$

  $c={(\frac{b}{2})}^2={(\frac{-19}{2})}^2=\frac{361}{4}$

So, c must be $\frac{361}{4}$ to make the trinomial a perfect square.

Thus, c must be $\frac{361}{4}$ to make the trinomial a perfect square.