How do you solve #\sqrt { 8y ^ { 2} + 13y } = \sqrt { 5y ^ { 2} + 10}#?

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Answer 1 · 2017-08-07T02:49:28

Given: #\sqrt { 8y ^ { 2} + 13y } = \sqrt { 5y ^ { 2} + 10}#

You can say that we square both sides or assert that arguments must be equal:

#8y^2 + 13y = 5y^2+10#

Add #-5y^2-10# to both sides:

#3y^2 + 13y-10 = 0#

Factor:

#(y+5)(3y-2)=0#

#y = -5# and #y = 2/3 larr# answers

Check #y=-5# :

#sqrt(8(-5)^2 + 13(-5)) = sqrt(5(-5)^2+10)#

#sqrt(135) = sqrt(135)#

#y = -5# checks

Check #y = 2/3#

#sqrt(8(2/3)^2 + 13(2/3)) = sqrt(5(2/3)^2+10)#

#sqrt(12.bar2) = sqrt(12.bar2)#

#y = 2/3# checks