Chapter A.5, Problem 86E

To determine

Tofind:The solution of the equation ${\left(x^2-x-22\right)}^{3/2}=27$ .

Answer to Problem 86E

The solutions of the equation ${\left(x^2-x-22\right)}^{3/2}=27$ are $x=\frac{1-5\sqrt{5}}{2}$ and $x=\frac{1+5\sqrt{5}}{2}$ .

Explanation of Solution

Given information:

The given equation is ${\left(x^2-x-22\right)}^{3/2}=27$ .

Calculation:

Solve the given equation.

  $\begin{array}{c}{\left(x^2-x-22\right)}^{3/2}=27\\ \sqrt[]{\left(x^2-x-22\right)}=3\left\{\text{Rewrite}\text{equation}\text{in}\text{radical}\text{form}\right\}\\ \left(x^2-x-22\right)=9\left\{\text{Square}\text{each}\text{side}\right\}\\ x^2-x-31=0\end{array}$

Apply Quadratic formula to solve the equation.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Compare the equation with standard quadratic equation $a{x}^2+bx+c=0$ .

So, $a=1$ , $b=-1$ and $c=-31$ .

Substitute the values in the above equation.

  $\begin{array}{c}x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\times 1\times \left(-31\right)}}{2}\\ =\frac{1\pm \sqrt{125}}{2}\\ =\frac{1\pm 5\sqrt{5}}{2}\end{array}$

From the above, it is clear that the values of $x$ are $\frac{1\pm 5\sqrt{5}}{2}$ .

Check solution:

Substitute $\frac{1+5\sqrt{5}}{2}$ for $x$ in the given equation.

  $\begin{array}{c}{\left({\left(\frac{1+5\sqrt{5}}{2}\right)}^2-\left(\frac{1+5\sqrt{5}}{2}\right)-22\right)}^{3/2}=27\\ {\left(9\right)}^{3/2}=3^3\\ {\left(9\right)}^{1/2}=3\\ 9=9\end{array}$

Substitute $\frac{1-5\sqrt{5}}{2}$ for $x$ in the given equation.

  $\begin{array}{c}{\left({\left(\frac{1-5\sqrt{5}}{2}\right)}^2-\left(\frac{1-5\sqrt{5}}{2}\right)-22\right)}^{3/2}=27\\ {\left(9\right)}^{3/2}=3^3\\ {\left(9\right)}^{1/2}=3\\ 9=9\end{array}$

Hence, $x=\frac{1-5\sqrt{5}}{2}$ and $x=\frac{1+5\sqrt{5}}{2}$ are solutions of the equation.

Therefore, the solutions of the equation ${\left(x^2-x-22\right)}^{3/2}=27$ are $x=\frac{1-5\sqrt{5}}{2}$ and $x=\frac{1+5\sqrt{5}}{2}$ .