Q.9.15 - Intermediate Algebra

Question

Solve $5{(a-5)}^2+4=104$

Step-by-Step Solution

Answer

The solutions of the quadratic equation $a=2\sqrt{5}+5, a=-2\sqrt{5}+5$

Step 1. Given information

Consider the quadratic equation

$5{(a-5)}^2+4=104$

The objective is to solve the quadratic equation by using square root property.

If $x^2=k$, then $x=\pm \sqrt{k}$

Subtract 4 from both sides to isolate the binomial term.

$$\begin{gathered} 5{(a-5)}^2=104-4 \\ 5{(a-5)}^2=100 \end{gathered}$$

Divide both sides by 5.

${(a-5)}^2=20$

Use the Square Root Property.

$(a-5)=\pm \sqrt{20}$

Step 2. Simplify the radical expression.

Simplify the radical and solve for $a$.

$$\begin{gathered} a-5=\pm \sqrt{4·5} \\ a=\pm 2\sqrt{5}+5 \end{gathered}$$

Rewrite to show two solutions.

$a=2\sqrt{5}+5, a=-2\sqrt{5}+5$

Step 3. Check the solutions.

Substitute $a=2\sqrt{5}+5$ in the original equation.

$$\begin{gathered} 5{(a-5)}^2+4=104 \\ 5{(2\sqrt{5}+5-5)}^2+4\stackrel{?}{=}104 \\ 5(4·5)+4\stackrel{?}{=}104 \\ 100+4\stackrel{?}{=}104 \\ 104=104 \end{gathered}$$

Substitute $a=-2\sqrt{5}+5$ in the original equation.

$$\begin{gathered} 5{(a-5)}^2+4=104 \\ 5{(-2\sqrt{5}+5-5)}^2+4\stackrel{?}{=}104 \\ 5(4·5)+4\stackrel{?}{=}104 \\ 100+4\stackrel{?}{=}104 \\ 104=104 \end{gathered}$$