SOLUTION: Help..I can't figure out how to do this problem: the square of 2y+7, plus 4 = y. I keep trying to isolate y, but because it is on both sides of the equal sign, I can't figure out h

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Help..I can't figure out how to do this problem: the square of 2y+7, plus 4 = y. I keep trying to isolate y, but because it is on both sides of the equal sign, I can't figure out h      Log On


   

Question 27320: Help..I can't figure out how to do this problem: the square of 2y+7, plus 4 = y. I keep trying to isolate y, but because it is on both sides of the equal sign, I can't figure out how to do it and my book doesn't explain it well at all. I simplified it by doing the square of 2y plus the square of 7 = y-4, but then I'm not putting the y on the left. Can someone help straighten me out on this?
Answer by bmauger(101) About Me  (Show Source):
You can put this solution on YOUR website!
%282y%2B7%29%5E2%2B4=y First you should square the binomial. How ever you can't just square both terms when you evaluate:
%282y%2B7%29%5E2 it is NOT equal to %284y%5E2%2B49%29
Instead squaring a binomial involves you multiplying the binomial by itself and using FOIL like:
%282y%2B7%29%5E2=%282y%2B7%29%282y%2B7%29=4y%5E2%2B28y%2B49
Thus: 2y%2B7%5E2%2B4=y can be rewritten:
4y%5E2%2B28y%2B49%2B4=y
Now you can combine terms to get:
4y%5E2%2B27y%2B53=0
Now you can solve with the quadratic equation
y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29
y+=+%28-27+%2B-+sqrt%2827%5E2-4%2A4%2A53%29%29%2F%282%2A4%29
y+=+%28-27+%2B-+sqrt%28729-848%29%29%2F8
y+=+%28-27+%2B-+sqrt%28-119%29%29%2F8
y+=+%28-27+%2B-+i%2Asqrt%28119%29%29%2F8