SOLUTION: Solve the following equation. Be sure to check your answer by plugging your answer back into the equation. (sqrt2y+7)+4=y I have gotten to the quadratic equation of y^2-10

Algebra ->  Algebra  -> Functions -> SOLUTION: Solve the following equation. Be sure to check your answer by plugging your answer back into the equation. (sqrt2y+7)+4=y I have gotten to the quadratic equation of y^2-10      Log On

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Question 66981: Solve the following equation. Be sure to check your answer by plugging your answer back into the equation.
(sqrt2y+7)+4=y
I have gotten to the quadratic equation of y^2-10y+9=0, but I am stuck at that point. Can someone please lend me a hand here. Thanks.
Found 2 solutions by stanbon, rapaljer:
Answer by stanbon(57984) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt(2y+7)+4=y
sqrt(2y+7)=y-4
Square both sides to get:
2y+7=y^2-8y+16
y^2-10y+9=0
(y-9)(y-1)=0
y=9 or y=1
Cheers,
Stan H.







Answer by rapaljer(4667) About Me  (Show Source):
You can put this solution on YOUR website!
Stanbon forgot to CHECK the answers! Whenever you square both sides of an equation, you must check the answers to see if they really work according to the Principle Value Definition of Square Root. In this case, the answer of x=9 works, but x=1 results in sqrt%289%29%2B4=+1, (where sqrt%289%29+=+3 by Principle Value Definition) which is NOT true, so it must be rejected as an extraneous root! The solution is ONLY x=9.

R^2 at SCC