Chapter 11.10, Problem 48WE

To determine

To find: The solution of the given quadratic equation.

Answer to Problem 48WE

The solution set is $\left\{49\right\}$ .

Explanation of Solution

Given:

The given quadratic equation is:

  $3\sqrt{a}+7=\sqrt{16a}\dots \left(1\right)$

To find the value of $a$ , first, Substitute $\sqrt{a}=y$ in equation $\left(1\right)$ .

The equation become after substitution is:

  $3y+7=\sqrt{16y^2}\dots \left(2\right)$

Solve the equation $\left(2\right)$ .

The solution of equation $\left(2\right)$ is given as:

  $\begin{array}{c}{\left(3y+7\right)}^2=16y^2\left[\text{squaring}\text{both}\text{side}\right]\\ 9y^2+42y+49-16y^2=0\\ 7y^2-42y-49=0\\ y^2-6y-7=0\end{array}$

Further simplified as:

  $\begin{array}{c}{y}^2-7y+y-7=0\\ y\left(y-7\right)+\left(y-7\right)=0\\ \left(y+1\right)\left(y-7\right)=0\end{array}$

Further simplified as:

  $\begin{array}{c}\left(y+1\right)=0\\ \text{or}\left(y-7\right)=0\\ y=-1\\ \text{or}y=7\end{array}$

So the value of $a$ is calculated as:

  $\begin{array}{c}\sqrt{a}=-1\\ a=1\left[\text{squaring}\text{both}\text{side}\right]\end{array}$

So $\begin{array}{c}\sqrt{a}=7\\ a=49\left[\text{squaring}\text{both}\text{side}\right]\end{array}$

Check:

If $a=49$ , then the equation $\left(1\right)$ is calculated as:

  $\begin{array}{c}3\sqrt{49}+7=\sqrt{16\left(49\right)}\\ 28=28\left[\text{True}\right]\end{array}$

If $a=1$ , then the equation $\left(1\right)$ is calculated as:

  $\begin{array}{c}3\sqrt{1}+7=\sqrt{16\left(1\right)}\\ 10=4\left[\text{False}\right]\end{array}$

Therefore the solution set is $\left\{49\right\}$ .