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 April 13th, 2011, 12:09 PM #1 Newbie   Joined: Apr 2011 Posts: 3 Thanks: 0 Help with radicals I found this problem in a math book: ?(y) + 2 = ?(y + 16) The answer in the back says that y = 9, and it works when you substitute it back in the equation, but I can't figure out how to solve it.
 April 13th, 2011, 12:32 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,571 Thanks: 931 Math Focus: Elementary mathematics and beyond Re: Help with radicals Square both sides and you'll be left with a single radical. Isolate the radical on one side of the equation and square both sides again, then solve for y.
 April 13th, 2011, 01:03 PM #3 Newbie   Joined: Apr 2011 Posts: 3 Thanks: 0 Re: Help with radicals When I square both sides, don't I get y + 4 = y +16 ?
 April 13th, 2011, 01:10 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs Re: Help with radicals No, let's start with the original: $\sqrt{y}+2=\sqrt{y+16}$ Square both sides: $$$\sqrt{y}+2$$^2=$$sqrt{y+16}$$^2$ $y+4\sqrt{y}+4=y+16$ Subtract y + 4 from both sides: $4\sqrt{y}=12$ Divide through by 4: $\sqrt{y}=3$ Square both sides: $$$\sqrt{y}$$^2=$$3$$^2$ $y=9$
 April 13th, 2011, 01:14 PM #5 Newbie   Joined: Apr 2011 Posts: 3 Thanks: 0 Re: Help with radicals Got it! No wonder I couldn't make it work. Thanks so much.
 April 13th, 2011, 01:19 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs Re: Help with radicals You're welcome, and welcome to the forum!
 April 13th, 2011, 03:40 PM #7 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 ?y + 2 = ?(y + 16) (?y + 2)² = ?(y + 16)² y + 4?y + 4 = y + 16 4?y = 16 - 4 = 12 ?y = 12/4 = 3 ?y² = 3² ? y = 9

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