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August 7th, 2009, 08:56 AM  #1 
Newbie Joined: Aug 2009 Posts: 1 Thanks: 0  help with a no solution
Studying up for a placement exam this fall to get back into calc one from which I had to withdraw two years ago. So far almost everything is coming back, but there are a few sticking points. This one is is 'no solution' in the answer key, and though I can come up with some crazy answers, they are still solutions. Could someone tell me what I'm missing? f(x)=(sqrt(52x))(x/2) find all input(s), x, suth that f(x)=2 
August 7th, 2009, 12:07 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
There is no real solution. The first term on the righthand side is nonnegative by definition, and so the minimum value of the righthandside is 5/4, occurring when x = 5/2. If you check what you did, it's almost certainly the case that you introduced extraneous solutions.

August 7th, 2009, 12:15 PM  #3  
Senior Member Joined: Mar 2009 Posts: 318 Thanks: 0  Quote:
[color=white]. . . . .[/color]2 = sqrt[5  2x]  x/2 [color=white]. . . . .[/color]4 = 2sqrt[5  2x]  x [color=white]. . . . .[/color]x  4 = 2sqrt[5  2x] [color=white]. . . . .[/color]x^2  8x + 16 = 4(5  2x) [color=white]. . . . .[/color]x^2  8x + 16 = 20  8x [color=white]. . . . .[/color]x^2  4 = 0 Solve the quadratic in the last line above for its two solution values. Then remember to check them in the original equation, to see which, if either, works there.  

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