How do you solve #(s+1)^2/10-12/5=15/2#?

1 Answer
May 8, 2017

Answer:

#s = -1+-3sqrt(11)#

Explanation:

Given:

#((s+1)^2)/(10)-12/5=15/2#

We want to isolate the #(s+1)^2# term in order to solve for s

First add #12/5# to both sides:

#((s+1)^2)/(10)-12/5color(red)(+12/5)=15/2color(red)(+12/5)#

On the left hand side, the #-12/5# and #+12/5# add to #0#:

#((s+1)^2)/(10)cancel(-12/5)cancel(color(red)(+12/5))=15/2color(red)(+12/5)#

And on the right hand side, we need to find a common denominator to add the fractions:

#((s+1)^2)/(10)=15/2color(red)((5/5))+12/5color(red)((2/2))#

#((s+1)^2)/(10)=color(red)(75/10)+color(red)(24/10) = 99/10#

Next we multiple both sides by 10 to isolate the #(s+1)^2# term

#color(red)(10)((s+1)^2)/(10) = color(red)(10) 99/10#

Which gives

#(s+1)^2 = 99#

Square root both sides of the equation

#sqrt((s+1)^2) = sqrt(99)#

Which gives

#s+1 = +-sqrt(99)#

NOTE: When taking the square root of any term, the answer can be both negative (-) and positive (+). When you raise a negative number to the second (2) power, the negative multiplies out to give a positive answer.

Now, simplifying the answer, and subtracting 1 from both sides

#s +1 color(red)(-1) = +-sqrt((9)(11)) color(red)(-1)#

9 is a perfect square, with a square root value of 3, so the answer can be simplified to

#s = -1+-3sqrt(11)#