SOLUTION: I need to find the solution for: radical (2y + 7) + 4 = y I've started my solution by getting the radical by itself creating: radical (2y+7) = y - 4. Then I square both sides so I

Algebra ->  Radicals -> SOLUTION: I need to find the solution for: radical (2y + 7) + 4 = y I've started my solution by getting the radical by itself creating: radical (2y+7) = y - 4. Then I square both sides so I      Log On


   

Question 88216This question is from textbook Beginning Algebra
: I need to find the solution for: radical (2y + 7) + 4 = y
I've started my solution by getting the radical by itself creating: radical (2y+7) = y - 4. Then I square both sides so I can remove the radical and end up with: 2y + 7 = y^2 - 8y + 16 from here on out I'm lost. Thanks. This question is from textbook Beginning Algebra

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct so far. What you have 2y+%2B+7+=+y%5E2+-+8y+%2B+16+ is a quadratic equation. You must set it equal to zero, and solve for y. It's probably best to subtract 2y and 7 from each side:
2y+%2B+7+=+y%5E2+-+8y+%2B+16+
2y+%2B+7-2y-7+=+y%5E2+-+8y+%2B+16+-2y-7
0=y%5E2+-10y%2B+9+

IT ALWAYS FACTORS!!
0=%28y-9%29%28y-1%29+
y= 9 or y = 1

Now, check the answers to make sure you don't have any extraneous roots (that is, extra solutions that don't really work in the original equation!).

sqrt%282y+%2B+7%29+%2B+4+=+y
If y=9
sqrt%282%2A9+%2B+7%29+%2B+4+=+9
sqrt%2825%29+%2B+4+=+9 This one checks!

sqrt%282y+%2B+7%29+%2B+4+=+y
If y=1
sqrt%282%2A1+%2B+7%29+%2B+4+=+1
sqrt%289%29+%2B+4+=+1 This one does NOT check! Reject this answer.

The only solution is y=9.

I have a page for this on my website, if you want to click on my tutor name "Rapaljer" anywhere in algebra.com, go to my MATH IN LIVING COLOR page, and in Intermediate Algebra, look for Radical Equations in Chapter 3. In College Algebra, look for Radical Equations in Chapter 1. I hope this is helpful.

R^2