Question

How do you solve `7v^2+1=29`?

Answer 1

`v=\pm2`

Explanation:

Given that

`7v^2+1=29`

`7v^2=29-1`

`7v^2=28`

`v^2=28/7`

`v^2=4`

`\sqrt{v^2}=\sqrt4\quad(\text{taking square root on both sides})`

`|v|=2`

`v=\pm2`

Answer 2

`v=+-2`

Explanation:

`"subtract 1 from both sides"`

`7v^2=29-1=28`

`"divide both sides by 7"`

`v^2=28/7=4`

`color(blue)"take the square root of both sides"`

`v=+-sqrt4larrcolor(blue)"note plus or minus"`

`rArrv=+-2`

Answer 3

`v=+-2`

Explanation:

Remember, our end goal is to isolate `v`. We can start by subtracting `1` from both sides. We now have

`7v^2=28`

Dividing both sides by `7` gives us

`v^2=4`

Taking the `+-` square root of both sides, we get

`v=+-2`

Hope this helps!