Q 285 - Intermediate Algebra

Question

In the following exercises, solve:

$4m^2= 17m - 15$

Step-by-Step Solution

Answer

The solution is $m=3.$

Step 1. Given Information

The given quadratic equation is $4m^2= 17m - 15.$

We have to solve the quadratic equation by factoring.

Step 2. Factor the quadratic expression

First, write the given equation in standard form.

The standard form of the quadratic equation is

$a{x}^2+bx+c=0$

Therefore,

$$\begin{gathered} 4m^2= 17m - 15 \\ 4m^2-17m+15=0 \end{gathered}$$

Now, let's factor the quadratic expression 

$$\begin{gathered} 4m^2-17m+15=0 \\ 4m^2-12m-5m+15=0 \\ 4m(m-3)-5(m-3)=0 \\ (4m-5)(m-3)=0 \end{gathered}$$

Step 3. Use the Zero Product Property

The equations we get are 

$$\begin{gathered} 4m-5=0 \\ m-3=0 \end{gathered}$$

Thus,

$$\begin{gathered} 4m-5=0 \\ m=\frac{5}{4} \end{gathered}$$ or $$\begin{gathered} m-3=0 \\ m=3 \end{gathered}$$

Step 4. Verify the answers

To verify the answer we will substitute the solutions separately into the original equation.

Put $m=\frac{5}{4}$ in the given equation,

$$\begin{gathered} 4{\left(\frac{5}{4}\right)}^2=17\left(\frac{5}{4}\right)-15 \\ \frac{5}{4}=\frac{85}{4}-15 \\ \frac{5}{4}-\frac{85}{4}=-15 \\ -20=-15 \end{gathered}$$

It is false.

Now, put $m=3$ in the given equation

$$\begin{gathered} 4{\left(3\right)}^2=17\left(3\right)-15 \\ 36=51-15 \\ 36=36 \end{gathered}$$

It is true.

Hence the solution is $m=3.$