Q 9.1. - Intermediate Algebra

Question

Solve $x^2-48=0$

Step-by-Step Solution

Answer

The solutions of the quadratic equation:$x=4\sqrt{3}, x=-4\sqrt{3}$.

Step 1. Given Information.

Consider the quadratic equation $x^2-48=0$

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

If $x^2=k$, then $x=\pm \sqrt{k}$.

Isolate the quadratic term and make its coefficient one. Add 48 both sides to get $x^2$by itself.

$$\begin{gathered} x^2-48=0 \\ x^2=48 \end{gathered}$$

Step 2. Use square root property.

Remember to write the $\pm$symbol.

$x=\pm \sqrt{48}$

Step 3. Simplify the radical.

Rewrite to show two solutions.

$$\begin{gathered} x=\pm \sqrt{16}·\sqrt{3} \\ x=\pm 4\sqrt{3} \\ x=4\sqrt{3}, -4\sqrt{3} \end{gathered}$$

Step 4. Check the solutions.

Substitute in $x=4\sqrt{3}$ and $x=-4\sqrt{3}$.

$$\begin{gathered} x^2-48=0 \\ {\left(4\sqrt{3}\right)}^2-48\stackrel{?}{=}0 \\ 48-48\stackrel{?}{=}0 \\ 0=0 \end{gathered}$$

And,

$$\begin{gathered} x^2-48=0 \\ {(-4\sqrt{3})}^2-48\stackrel{?}{=}0 \\ 48-48\stackrel{?}{=}0 \\ 0=0 \end{gathered}$$