Chapter 11.10, Problem 44WE

To determine

To find: The solution set of the given quadratic equation.

Answer to Problem 44WE

The solution set is $\left\{-4,1\right\}$ .

Explanation of Solution

Given:

The given quadratic equation is:

  $\sqrt{a^2+3a}=2\dots \left(1\right)$

To find the value of $a$ , first, solve the equation $\left(1\right)$ .

The solution of equation $\left(1\right)$ is given as:

  $\begin{array}{c}{a}^2+3a=4\left[\text{squaring}\text{both}\text{side}\right]\\ a^2+3a-4=0\\ a^2+4a-a-4=0\\ a\left(a+4\right)-1\left(a+4\right)=0\end{array}$

Further simplified as:

  $\begin{array}{c}\left(a-1\right)\left(a+4\right)=0\\ \left(a-1\right)=0\\ \text{or}\left(a+4\right)=0\end{array}$

Further simplified as:

  $\begin{array}{c}a=1\\ \text{or}a=-4\end{array}$

So $a=1$ or $a=-4$ .

Check:

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

  $\begin{array}{c}\sqrt{{\left(1\right)}^2+3\left(1\right)}=2\\ \sqrt{4}=2\\ 2=2\left[\text{True}\right]\end{array}$

If $a=-4$ , then the equation $\left(1\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(-4\right)}^2+3\left(-4\right)}=2\\ \sqrt{4}=2\\ 2=2\left[\text{True}\right]\end{array}$

Therefore the solution set is $\left\{-4,1\right\}$ .