SOLUTION: can someone please help me complete this? i have to solve and check. sqrt{{{2y+7}}}+4=y =sqrt{{{2y+7}}}=-4 =(sqrt{{{2y+7}}}^2=(-4)^2 2y+7=4 y=4

Algebra ->  Radicals -> SOLUTION: can someone please help me complete this? i have to solve and check. sqrt{{{2y+7}}}+4=y =sqrt{{{2y+7}}}=-4 =(sqrt{{{2y+7}}}^2=(-4)^2 2y+7=4 y=4      Log On


   

Question 78460: can someone please help me complete this? i have to solve and check.
sqrt2y%2B7+4=y
=sqrt2y%2B7=-4
=(sqrt2y%2B7^2=(-4)^2
2y+7=4
y=4
Answer by tutor_paul(519) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282y%2B7%29%2B4=y
First, get the radical expression alone on one side of the equation:
sqrt%282y%2B7%29=y-4 (looks like you tried to do this, but you forgot the y on the right).
Now, square both sides of the equation to get rid of the radical:
2y%2B7=%28y-4%29%5E2
Simplify the right hand side:
2y%2B7=y%5E2-8y%2B16
Combine like terms and equate to zero:
y%5E2-10y%2B9=0
Factor the expression:
%28y-9%29%28y-1%29=0
Equate each factor to zero, and solve:
y=9 and y=1
Now, this part is important...you need to plug these answers back in to
the original equation to be sure they are not "extraneous." An extraneous
root may be mathematically correct, but it is not the true answer. If you
plug y=1 back into the original equation, you will see that the equation
DOES NOT hold true. Hence, this is an extraneous root. If you plug y=9 back
into the original equation, you will see that the equation DOES hold true,
so that one is your answer.
Good Luck,
tutor_paul@yahoo.com