Chapter 7.4, Problem 21WE

To determine

Solve :

  $3{\left|x\right|}^2=7\left|x\right|+5$

Answer to Problem 21WE

  $x=\pm \left(\frac{7+\sqrt{109}}{6}\right)\approx \pm 2.91$

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 $3{\left|x\right|}^2=7\left|x\right|+5$

  $\begin{array}{l}\Rightarrow 3y^2=7y+5\\ \Rightarrow 3y^2-7y-5=\mathrm{0................}[\text{subtract 7y+5 from each side]}\end{array}$

Compare the equation with standard quadratic equation , a = 3 , b = -7 , c = -5.

So, the roots are :

  $\begin{array}{l}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow y=\frac{7\pm \sqrt{{(-7)}^2-4(3)(-5)}}{2(3)}\\ \Rightarrow y=\frac{7\pm \sqrt{49+60}}{6}\\ \Rightarrow y=\frac{7\pm \sqrt{109}}{6}\end{array}$

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

  $\begin{array}{l}y=\left|x\right|=\frac{7+\sqrt{109}}{6}\approx 2.91\\ \Rightarrow x=\pm \left(\frac{7+\sqrt{109}}{6}\right)\approx \pm 2.91\end{array}$