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 October 12th, 2007, 08:06 PM #1 Newbie   Joined: Oct 2007 Posts: 2 Thanks: 0 Let f(x)=x^2. Find x such that f(x+1)=f(x+2). Is there a definite value for x that will make the two equations equal one another? I cannot seem to find a value that x will equal to make the two equations equal one another. I am tempted to say that there is no value for x that will do this. I've even tried graphing the equations but I see that the lines never intersect, therefore further egging me to state that there is no value for x to make the two equations equal. I have tried to solve for x and I've done it this way: (x+1)(x+1)=x^2+x+x+2=x^2+2x+2 (x+2)(x+2)=x^2+2x+2x+4=x^2+4x+4 x^2+2x+2=x^2+4x+4 -x^2 -x^2 2x+2=4x+4 -2 -2 2x=4x+2 -4x -4x -2x=2 --- --- -2 -2 x=-1 October 12th, 2007, 09:11 PM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 f(x+1)=(x+1)^2=f(x+2)=(x+2)^2 x^2+2x+1=x^2+4x+4 2x=-3 x=-3/2 October 12th, 2007, 09:17 PM #3 Senior Member   Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus f(x+1)=f(x+2) (x+1)^2=(x+2)^2 What you did wrong was the expansion of (x+1)^2. It's actually x^2+2x+1, not x^2+2x+2. Your method will solve for the solution, x=-3/2. No complex solutions. October 15th, 2007, 06:35 AM #4 Newbie   Joined: Oct 2007 Posts: 2 Thanks: 0 thank you! thank you both so much for your help! i cannot believe I made this kind of mistake! something so simple turned out to be a weekend's nightmare! but thank you both for waking me up!  Tags 1fx, find, fxx2 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post kingkos Algebra 5 November 24th, 2012 11:40 PM kevinman Algebra 8 March 8th, 2012 05:53 AM

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