Chapter 1.6, Problem 43E

To determine

To calculate: The solution of the equation, $\sqrt{4\sqrt{4x+9}}=\sqrt{8x+2}$, and check the solution.

The solution of the given equation, $\sqrt{4\sqrt{4x+9}}=\sqrt{8x+2}$, is $x=\frac{7}{4}$.

Calculation:

Consider the provided equation,

$\sqrt{4\sqrt{4x+9}}=\sqrt{8x+2}$.

Square each side.

$4\sqrt{4x+9}=8x+2$

Square each side again and solve.

$\begin{array}{c}16\left(4x+9\right)={\left(8x+2\right)}^2\\ 16\left(4x+9\right)=64x^2+4+32x\\ 64x+144=64x^2+4+32x\\ 16x^2-8x-35=0\end{array}$

Now, factorize the above equation.

$\left(4x+5\right)\left(4x-7\right)=0$

Equate first factor to zero.

$\begin{array}{c}4x+5=0\\ x=\frac{-5}{4}\end{array}$

Equate second factor to zero.

$\begin{array}{c}4x-7=0\\ x=\frac{7}{4}\end{array}$

Check:

Put $x=\frac{-5}{4},\frac{7}{4}$ in the equation, $\sqrt{4\sqrt{4x+9}}=\sqrt{8x+2}$.

First put $x=\frac{-5}{4}$.

$\begin{array}{l}\sqrt{4\sqrt{4\left(\frac{-5}{4}\right)+9}}\stackrel{?}{=}\sqrt{8\left(\frac{-5}{4}\right)+2}\\ \sqrt{4\sqrt{-5+9}}\stackrel{?}{=}\sqrt{-10+2}\\ \sqrt{4\sqrt{4}}\ne \sqrt{-8}\end{array}$

Thus, it is false and therefore, $x=\frac{-5}{4}$ is an extraneous solution.

Now, put $x=\frac{7}{4}$.

$\begin{array}{l}\sqrt{4\sqrt{4\left(\frac{7}{4}\right)+9}}\stackrel{?}{=}\sqrt{8\left(\frac{7}{4}\right)+2}\\ \sqrt{4\sqrt{7+9}}\stackrel{?}{=}\sqrt{14+2}\\ \sqrt{4\sqrt{16}}\stackrel{?}{=}\sqrt{16}\\ 4=4\end{array}$

Thus, it is true.

Hence, the solution of given equation is $x=\frac{7}{4}$.