Chapter 4.5, Problem 59HP

To determine

Find the expression of $x$ in terms of $b\&c$

Answer to Problem 59HP

  $x=\left\{-\frac{b}{2}\pm \sqrt{\frac{b^2}{4}-c}\right\}$

Explanation of Solution

Given information:

Solve $x^2+bx+c=0$ by completing the square method.

Calculation:

Solve through the formula

  ${\left(a+b\right)}^2=a^2+2ab+b^2$

Consider the equation,

  $x^2+bx+c=0$

Add $\frac{b^2}{4}$ on both the sides, we get

  $\begin{array}{l}{x}^2+bx+c=0\\ x^2+bx=-c\\ x^2+bx+{\left(\frac{b}{2}\right)}^2=-c+{\left(\frac{b}{2}\right)}^2\\ {\left(x+\frac{b}{2}\right)}^2=-c+\frac{b^2}{4}\end{array}$

  $\begin{array}{l}{\left(x+\frac{b}{2}\right)}^2=\frac{-4c+b^2}{4}\\ \left(x+\frac{b}{2}\right)=\pm \sqrt{\left(\frac{-4c+b^2}{4}\right)}\\ \left(x+\frac{b}{2}\right)=\pm \frac{\sqrt{\left(-4c^2+b^2\right)}}{2}\end{array}$

First take positive value of the right hand side,

  $\begin{array}{l}x+\frac{b}{2}=\frac{\sqrt{\left(-4c+b^2\right)}}{2}\\ x=\frac{\sqrt{\left(-4c+b^2\right)}}{2}-\frac{b}{2}\\ =\frac{\sqrt{\left(-4c+b^2\right)-b}}{2}\end{array}$

Nect take the negative value of right hand side,

  $x+\frac{b}{2}=-\frac{\sqrt{\left(-4c+b^2\right)}}{2}$

  $\begin{array}{l}x=-\frac{\sqrt{\left(-4c+b^2\right)}}{2}-\frac{b}{2}\\ =\frac{-\sqrt{\left(-4c+b^2\right)-b}}{2}\end{array}$

Hence, the solution set for $x$ is $x=\left\{-\frac{b}{2}\pm \sqrt{\frac{b^2}{4}-c}\right\}$ .