Chapter 7.4, Problem 22WE

To determine

Solve :

  $\left|x\right|\cdot \left(\left|x\right|-3\right)=1$

Answer to Problem 22WE

  $x=\pm \left(\frac{3+\sqrt{13}}{2}\right)\approx \pm 3.30$

Explanation of Solution

Formula Used:

The roots of the quadratic equation $a{x}^2+bx+c=0$ are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Substitute $\left|x\right|=y$ in the equation $\left|x\right|\cdot \left(\left|x\right|-3\right)=1$

  $\begin{array}{l}\Rightarrow y\cdot \left(y-3\right)=1\\ \Rightarrow y^2-3y-1=\mathrm{0...............}[\text{subtract 1 from each side]}\end{array}$

Compare the equation with standard quadratic equation , a = 1 , b = -3 , c = -1.

So, the roots are :

  $\begin{array}{l}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow y=\frac{-(-3)\pm \sqrt{{(-3)}^2-4(1)(-1)}}{2(1)}\\ \Rightarrow y=\frac{3\pm \sqrt{9+4}}{2}\\ \Rightarrow y=\frac{3\pm \sqrt{13}}{2}\end{array}$

Since $y=\left|x\right|\ge 0$ , so,

  $\begin{array}{l}y=\left|x\right|=\frac{3+\sqrt{13}}{2}\approx 3.30\\ \Rightarrow x=\pm \left(\frac{3+\sqrt{13}}{2}\right)\approx \pm 3.30\end{array}$