SOLUTION: Can you help me solve this equation? sqrt 2y+7 + 4 = y The 2y+7 is in the radical sign Thanks

Question 39890: Can you help me solve this equation?
sqrt 2y+7 + 4 = y
The 2y+7 is in the radical sign
Thanks
Found 2 solutions by fractalier, junior403:
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
First we isolate the radical and then square both sides...so from
sqrt(2y+7) + 4 = y
sqrt(2y+7) = y - 4
2y + 7 = (y - 4)^2
2y + 7 = y^2 - 8y + 16 now collect like terms and solve
y^2 - 10y + 9 = 0
(y-9)(y-1) = 0
y = 9 or y = 1
Yo have to check these, however, and only y = 9 checks out. y = 1 doesn't check. Thus,
y = 9
Answer by junior403(76) (Show Source): You can put this solution on YOUR website!
let's see...

First we need to isolate the radical...
So we can subtract 4 from both sides of the equation...

Then, in order to eliminate the radical sign, we need to square both sides...

The square root of a squared number, is just that number so...

Now we can work on the other side by distributing the power of 2...

Now we can write the quadratic equation in standard form on one side of the equation and equal to zero by subtracting the 2y and the 7 from both sides...

Now we can solve this equation using the quadratic formula...

Now we simply insert the coefficients of our equation such that a=1 b=-2 c=9...

and solve...

and...

use imaginary numbers to sove for a negative square root...

or...

the square root of 16 is 4 so that can come out from under the radical...

Now we can factor out a 2 from the numerator in order to cancel out the 2 in the denominator...

Which can be simplified as...

This is your answer in its simplist form.
So the solution set would be...
,
I hope this helps
Good Luck!

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