How do you solve #60v ^ { 2} + 6v - 18= 0#?

2 Answers
George C.
Apr 12, 2018

#v = 1/2" "# or #" "v = -3/5#

Explanation:

Given:

#60v^2+6v-18 = 0#

Note that all of the terms are divisible by #6#, so divide the whole equation by #6# to get:

#10v^2+v-3 = 0#

Use an AC method:

Find a pair of factors of #AC=10 * 3 = 30# which differ by #B=1#.

The pair #6, 5# works in that #6 * 5 = 30# and #6 - 5 = 1#

Use this pair to split the middle term and factor by grouping:

#0 = 10v^2+v-3#

#color(white)(0) = (10v^2+6v)-(5v+3)#

#color(white)(0) = 2v(5v+3)-1(5v+3)#

#color(white)(0) = (2v-1)(5v+3)#

Hence:

#v = 1/2" "# or #" "v = -3/5#

Daniel L.
Apr 12, 2018

See explanation.

Explanation:

First we can divide aboth sides of the equation by #6# to make the coefficients smaller (and still integer):

#10v^2+v-3=0# ##

Now to find the roots we calculate the discriminant:

#Delta=1^2-4*10*(-3)#

#Delta=121#

#sqrt(Delta)=11#

The discriminant is positive, so the equation has 2 real roots:

#x_1=(-1-11)/20=-3/5#

and

#x_2=(-1+11)/20=1/2#

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