SOLUTION: Help! I am stuck on solving a radical equation. The original problem is this: sq.rt.2y+7 plus 4 = y. I know I have to isolate the radical to one side, then square both sides of

Algebra ->  Radicals -> SOLUTION: Help! I am stuck on solving a radical equation. The original problem is this: sq.rt.2y+7 plus 4 = y. I know I have to isolate the radical to one side, then square both sides of      Log On


   

Question 85864: Help! I am stuck on solving a radical equation.
The original problem is this: sq.rt.2y+7 plus 4 = y.
I know I have to isolate the radical to one side, then square both sides of the equation. Getting that far, I have: (sq.rt.2y+7)= (y-4).
After that I am completely stuck! Please help!
Thank you!
Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
%28sqrt%282y%2B7%29+%2B+4%29 = y
%28sqrt%282y%2B7%29+%2B+4-4%29 = -4 + y
%28sqrt%282y%2B7%29+%29 = -4 + y
%28%28sqrt%282y%2B7%29%29%5E2+%29 = %28%28-4+%2B+y%29%5E2%29 [square both sides]
2y+7= = (-4 + y ) (-4 + y)[use the FOIL method to multiply]
2y+7=%2816-8y%2By%5E2%29%29 [set the equation equal to zero]
2y-2y+7-7=%2816-7-8y-2y%2By%5E2%29%29
0=%289-10y%2By%5E2%29%29 [re-arrange the terms]
0=%28y%5E2-10y%2B9%29%29 [factor}
0=(y-9)(y-1) [set each factor equal to zero and solve for the x-term]
.
y-9=0
y=9
and
y-1=0
y=1
.
check by plugging either value of x back into the original equation and solve.