How do you solve #5v ^ { 2} - 40= - 10v#?

1 Answer
Mar 26, 2017

#v= 2#
#v= -4#

Explanation:

First, make sure the equation is in the standard form
#ax^2 + bx +c = 0#

#5v^2 + 10v - 40 = 0#

If #a!= 1#, the #ac# method is applied.
In this case, #a= 5# so we have to use the ac method.

What is the ac method???

The ac method is simply the value of #a# times the value of #c#

#a= 5#
#b= 10#
#c= -40#

#a*c#= -200

Now we have to find the numbers that will give a product of -200 as well as a sum of 10.

#-10 * 20 = -200 (ac)#
#-10 + 20 = 10 (b)#

Bingo! Now that we have these numbers, we rewrite the equation substituting -10 and 20 for b.

5#v^2# - 10v + 20v -40 = 0 ...this is the same as the inital equation when simplified.

Now, we factor by grouping and pulling out the gcf.

#(5v^2 - 10v) + (20v - 40) = 0#

#5v(v - 2) + 20(v - 2) = 0#

#(5v + 20)(v - 2) = 0#

Solve for v
#5v + 20 = 0#
#5v = -20#

Dividing both sides by 5, we get
#(cancel(5)v)/cancel5 = (-20)/5#
#v = -4#

Solve the other also for v
#v - 2 = 0#
#v = 2#