Chapter 9.2, Problem 61SR

To determine

Solve the equation and check the solutions.

Answer to Problem 61SR

The solutions are: $-3+\sqrt{5}\text{and}-3-\sqrt{5}$

Explanation of Solution

Given:

The equation: ${(x+3)}^2=5$

Concept Used:

In algebra, a quadratic equation is any equation that can be rearranged in standard form $a{x}^2+bx+c=0$ as, where x represents an unknown, and a, b, and c represent known numbers, where a ? 0.

Calculation:

The equation: ${(x+3)}^2=5$

  $\begin{array}{l}$ {(x+3)}^2=5\text{Original}\text{Equation}$\sqrt{{(x+3)}^2}=\sqrt{5}\text{Take}\text{the}\text{square}\text{root}\text{of}\text{both}\\ (x+3)=\pm \sqrt{5}\text{Simplify}\end{array}$

Solve for x using the zero product property.

  $\begin{array}{l}x+3=\sqrt{5}x+3=-\sqrt{5}\\ x=-3+\sqrt{5}x=-3-\sqrt{5}\end{array}$

The solutions are: $-3+\sqrt{5}\text{and}-3-\sqrt{5}$

Check:

  $\begin{array}{}$ {(x+3)}^2=5\text{and}{(x+3)}^2=5$$ {(-3+\sqrt{5}+3)}^2=5\text{and}{(-3-\sqrt{5}+3)}^2=5$$ {(\sqrt{5})}^2=5\text{and}{(-\sqrt{5})}^2=5$$ 5=5\text{and}5=5$\end{array}$

Thus, the solutions are: $-3+\sqrt{5}\text{and}-3-\sqrt{5}$