Q.9.19 - Intermediate Algebra

Question

Solve $9m^2-12m+4=25$

Step-by-Step Solution

Answer

The solutions of the quadratic equation: $m=\frac{7}{3}, m=-1$

Step 1. Given information

Consider the quadratic equation

$9m^2-12m+4=25$

Factor the perfect square trinomial.

${(3m-2)}^2=25$

Use the Square Root Property.

$$\begin{gathered} 3m-2=\pm \sqrt{25} \\ 3m-2=\pm 5 \end{gathered}$$

Add 2 to both sides.

$3m=\pm 5+2$

Solve for $m$.

$m=\frac{\pm 5+2}{3}$

Rewrite to show two solutions.

$$\begin{gathered} m=\frac{5+2}{3}, m=\frac{-5+2}{3} \\ m=\frac{7}{3}, m=-1 \end{gathered}$$

Step 2. Check for solutions.

Substitute $m=\frac{7}{3}$in the original equation.

$$\begin{gathered} 9m^2-12m+4=25 \\ 9{\left(\frac{7}{3}\right)}^2-12\left(\frac{7}{3}\right)+4\stackrel{?}{=}25 \\ 9\left(\frac{49}{9}\right)-28+4\stackrel{?}{=}25 \\ 49-28+4\stackrel{?}{=}25 \\ 25=25 \end{gathered}$$

Substitute $m=-1$ in the original equation.

$$\begin{gathered} 9m^2-12m+4=25 \\ 9{\left(-1\right)}^2-12\left(-1\right)+4\stackrel{?}{=}25 \\ 9+12+4\stackrel{?}{=}25 \\ 25=25 \end{gathered}$$