How do you solve #25v^2=1#?

3 Answers
Sep 4, 2016

Answer:

#v=+-1/5#

Explanation:

Divide both sides by 25 giving:

#v^2=1/25#

Thus:

#sqrt(v^2)=sqrt(1/25)#

#v=sqrt(1)/(sqrt(25)) = 1/sqrt(25)#

But as this is square root we should write #=+-1/sqrt(25)#

#v=+-1/5#

Sep 4, 2016

Answer:

#v=+-1/5#

Explanation:

Begin by isolating #v^2#. To do this we divide both sides of the equation by 25.

#rArr(cancel(25)^1 v^2)/cancel(25)^1=1/25rArrv^2=1/25#

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

#rArrsqrt(v^2)=+-sqrt(1/25)#

#rArrv=+-1/5#

Sep 4, 2016

Answer:

#v = +-1/5#

Explanation:

Apart from the method shown by other contributors, we can also follow the usual method for a quadratic equation.

#rarr "Make it equal to 0 " rarr " find the factors."#

#25v^2 = 1#

#25v^2 -1 = 0 larr" difference of squares"#

#(5v+1)(5v-1) = 0#

Either factor could be 0

#5v+1 = 0 " " rarr v = -1/5#

#5v-1 = 0 " " rarr v = 1/5#