Question

How do you solve ${\left(2r-5\right)}^2=81$?

Answer 1

Step-1:Simplify the expression:

Square root property: If $x^2=C$ then $x=\pm \sqrt{C}$.

So,

$$\begin{gathered} {\left(2r-5\right)}^2=81 \\ \Rightarrow 2r-5=\pm \sqrt{81} \\ \Rightarrow 2r-5=\pm \sqrt{9^2} \\ \therefore 2r-5=\pm 9 \end{gathered}$$

Step-2: Solve for $r$.

If $2r-5=9$,

$\begin{array}{rcl}2r-5& =& 9\\ & \Rightarrow & 2r=9+5\left(\mathrm{Add}5\mathrm{on}\mathrm{each}\mathrm{side}\right)\\ & \Rightarrow & 2r=14\\ & \Rightarrow & r=\frac{14}{2}\left(\mathrm{Divide}\mathrm{both}\mathrm{sides}\mathrm{by}2\right)\\ \therefore r& =& 7\end{array}$

And if $2r-5=-9$,

$\begin{array}{rcl}2r-5& =& -9\\ & \Rightarrow & 2r=-9+5\left(\mathrm{Add}5\mathrm{on}\mathrm{each}\mathrm{side}\right)\\ & \Rightarrow & 2r=-4\\ & \Rightarrow & r=-\frac{4}{2}\left(\mathrm{Divide}\mathrm{both}\mathrm{sides}\mathrm{by}2\right)\\ \therefore r& =& -2\end{array}$

Hence, the solutions are $r=-2$ and $r=7$.