Chapter 4.6, Problem 53SR

To determine

To find:

The value of $c$ by using the trinomial as a perfect square of given quadratic equation.

Answer to Problem 53SR

  ${\left(x+6.5\right)}^2$

Explanation of Solution

Given:

The quadratic equation

  $x^2+13x+c=0$

Concept used:

The number is the square root of the quadratic formula.

  $b^2-4ac$

The discriminant formula

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The quadratic equation

  $x^2+13x+c=\mathrm{0.......................}\left(1\right)$

Comparing the quadratic equation $a{x}^2+bx+c=0$

  $a=1,b=13,c=c$

To make a perfect square, add on ${\left(\frac{b}{2}\right)}^2$ as the c term

In this case

  $b=13$

From equation (1)

  $\begin{array}{l}{x}^2+13x+c=0\\ x^2+13x+{\left(\frac{13}{2}\right)}^2=0\\ x^2+13x+\frac{169}{4}=0\\ x^2+13x+42.25=0\\ {\left(x+6.5\right)}^2\end{array}$

Therefore,

The value of trinomial as a perfect square.