Question

solve equations by factoring:4x^2-81=0

Answer 1

We have the equation:

$4x^2-81=0$

We can see that there is a differnce of squares:

$4x^2-81=(2x)^2-9^2$

And then we can rewite the difference of squares as:

$(2x)^2-9^2=(2x-9)(2x+9)$

And this is equal to 0:

$(2x-9)(2x+9)=0$

Now, there are two possibilities:

$\begin{gathered} 2x-9=0 \\ Or_{\colon} \\ 2x+9=0 \end{gathered}$

Then, we need to solve this two equatiions:

$\begin{gathered} 2x-9=0 \\ x=\frac{9}{2} \end{gathered}$$\begin{gathered} 2x+9=0 \\ x=-\frac{9}{2} \end{gathered}$

The two solutions for the equation are:

$\frac{9}{2}\text{ and}-\frac{9}{2}$