Chapter 9.4, Problem 1CYU

To determine

Tofind the value of $c$ to make the given trinomial a perfect square.

Answer to Problem 1CYU

  $c=81$

Explanation of Solution

Given:

The given trinomial is:

  $x^2-18x+c$

Concept used:

A perfect square trinomial is of the form $a^2+2ab+b^2$ and the perfect square will be

  $a^2+2ab+b^2={\left(a+b\right)}^2$

Calculation:

Consider the given polynomial.

  $x^2-18x+c$

Now, compare $-18x$ with $2ab$ and $x^2$ with $a^2$ .

  $\begin{array}{c}a=x\\ 2ab=-18x\\ 2xb=-18x\\ b=-9\end{array}$

So, the value of $c$ should be

  $\begin{array}{c}c=b^2\\ ={\left(-9\right)}^2\\ =81\end{array}$

Hence, for $c=81$ , the trinomial $x^2-18x+c$ will be perfect square as ${\left(x-9\right)}^2$ .