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

3 Answers

Answer:

#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#

Jul 17, 2018

Answer:

#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#

Jul 17, 2018

Answer:

#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!