Chapter 5.5, Problem 56PPS

To determine

To find: the value of c that makes perfect square and write the trinomial as a perfect square.

Answer to Problem 56PPS

The value of $\text{c=}0.81$ and perfect square is ${\left(\text{x+0}\text{.9}\right)}^2$ .

Explanation of Solution

Given:

  ${\text{x}}^{\text{2}}\text{-1}\text{.8x+c}$ .

Concept used:

An expression obtained from the square of binomial equation is a perfect square trinomial is in the form of ${\text{ax}}^{\text{2}}\text{+bx+c}$ is said to be a perfect square, if only if it satisfies the condition ${\text{b}}^{\text{2}}\text{=4ac}$ .

The perfect squares trinomial formula is given as:

  ${\left(\text{ax}\right)}^{\text{2}}{\text{+2abx+b}}^{\text{2}}\text{=}{\left(\text{ax+b}\right)}^{\text{2}}$ .

  ${\left(\text{ax}\right)}^{\text{2}}{\text{-2abx+b}}^{\text{2}}\text{=}{\left(\text{ax-b}\right)}^{\text{2}}$ .

Calculation:

  ${\text{x}}^{\text{2}}\text{-1}\text{.8x+c}$ .

To make the trinomial a perfect square, find the value of $\text{c}$ .

First Find the half of the coefficient of second term:

  $\frac{-1.8}{2}=-0.9$ .

Square the result of the half of the coefficient of the second term will be the $\text{c}$ :

  $\text{c=}{\left(-0.9\right)}^2=0.81$ .

The trinomial ${\text{x}}^{\text{2}}\text{-1}\text{.8x+0}\text{.81}$ can be written as:

  $\Rightarrow {\text{x}}^{\text{2}}\text{-2}\left(\text{x}\right){\left(0.9\right)}^2+{\left(0.9\right)}^2-{\left(0.9\right)}^2+0.81$ .

  $\Rightarrow {\left(\text{x+0}\text{.9}\right)}^2$ .

Hence the value of $\text{c=}0.81$ and perfect square is

  ${\left(\text{x+0}\text{.9}\right)}^2$ .