My Math Forum help with a no solution

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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(5-2x))-(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 right-hand side is non-negative by definition, and so the minimum value of the right-handside 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
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Quote:
 Originally Posted by leith00000 f(x)=(sqrt(5-2x))-(x/2) find all input(s), x, suth that f(x)=-2
Plug -2 in for f(x), and (attempt to) solve the resulting radical equation:

[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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