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=(9)/(2) \end{gathered}`
`\begin{gathered} 2x+9=0 \\ x=-(9)/(2) \end{gathered}`

The two solutions for the equation are:

`(9)/(2)\text{ and}-(9)/(2)`