My Math Forum Let f(x)=x^2. Find x such that f(x+1)=f(x+2).

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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!

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