How do you solve #v^ { 2} - 20v = 7#?

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Answer 1 · 2017-11-28T11:18:17

#v = 10 +- sqrt(107)#

We have

#v^2 - 20v= 7#

We will use "Completing The Square Method" to simplify.

Observe that our quadratic equation is in "Vertex Form".

#v^2 - 20v + square = 7 + square# #color(red)(Equation.1)#

What goes into the box above?

Divide the coefficient of our #v# term by 2 and square it. This is the value that goes into the box above.

Coefficient of our #v# term is #(-20)#. Divide this by 2 and square the value to get 100.

Now our #color(red)(Equation.1)# becomes

#v^2 - 20v + 100 = 7 + 100# #color(red)(Equation.2)#

Now, we can write #v^2 - 20v + 100# as #(v" " color(green)( - 10))^2#

We get #color(green)((-10))# by dividing the coefficient of our #v# term by 2.

This is the reason why we use "Completing The Square Method" to simplify.

Now, we have, using #color(red)(Equation.2),#

#(v - 10)^2 = 107#

Take square root on both sides

#sqrt((v-10)^2) = +-sqrt(107#

Hence, we get

#(v - 10) = +-sqrt(107)#

Therefore, our final answer is given by

#v = 10 +-sqrt(107)#

which gives us two values for #v#

#v = 20.3; v = -0.3#

Answer 2 · 2017-11-28T11:28:10

Solution: #v~~20.344 , v~~ -0.344 #

#v^2-20v=7 or v^2-20v+100 =100+7# or

# (v-10)^2 =107 or v-10 = +-sqrt107 # or

#v = 10+-sqrt107:. v=10 +10.344~~20.344 (3dp)# or

#v=10 -10.344~~-0.344 (3dp)#

Solution: #v~~20.344 , v~~ -0.344 # [Ans]