Chapter 1.6, Problem 44E

To determine

To calculate: The solution of the equation, $\sqrt{16+9\sqrt{x}}=4+\sqrt{x}$ and check the solution.

The solution of the given equation is $x=0,1$.

Calculation:

Consider the provided equation,

$\sqrt{16+9\sqrt{x}}=4+\sqrt{x}$

Square each side and solve.

$\begin{array}{c}{\left(\sqrt{16+9\sqrt{x}}\right)}^2={\left(4+\sqrt{x}\right)}^2\\ 16+9\sqrt{x}=16+x+8\sqrt{x}\\ x=\sqrt{x}\end{array}$

Again, square each side.

$\begin{array}{c}{\left(x\right)}^2={\left(\sqrt{x}\right)}^2\\ x^2-x=0\end{array}$

Now factorize the above equation.

$x\left(x-1\right)=0$

Put the first factor equal to zero.

$x=0$

Put the second factor equal to zero.

$\begin{array}{c}x-1=0\\ x=1\end{array}$

Check,

Put $x=0,1$ in the equation,$\sqrt{16+9\sqrt{x}}=4+\sqrt{x}$.

First put $x=0$.

$\begin{array}{l}\sqrt{16+9\sqrt{0}}\stackrel{?}{=}4+\sqrt{0}\\ \sqrt{16}\stackrel{?}{=}4\\ 4=4\end{array}$

Which is true.

Now put $x=1$.

$\begin{array}{l}\sqrt{16+9\sqrt{1}}\stackrel{?}{=}4+\sqrt{1}\\ \sqrt{25}\stackrel{?}{=}5\\ 5=5\end{array}$

Which is true.

Hence, the solution of the given equation is $x=0,1$.