SOLUTION: 2(2x+5/3x)^2 -5 = 9(2x+5/3x) This problem is supposed to be solved by substituting a variable like U for (2x+5/3x).Write in standard form. (2U^2 -9U -5=0) Factor. (2U+1)(U-5). S

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Question 531212: 2(2x+5/3x)^2 -5 = 9(2x+5/3x)
This problem is supposed to be solved by substituting a variable like U for (2x+5/3x).Write in standard form. (2U^2 -9U -5=0) Factor. (2U+1)(U-5). Set to zero. U=-1/2 and U=5. But, when I try to set these values to (2x+5/3x)to solve for x, the answers don't fit into the original equation as they should. What in the world is wrong? Thanks so much for your help.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
2(2x+5/3x)^2 -5 = 9(2x+5/3x)
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Let w = (2x+5)/(3x)
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2w^2 - 9w -5 = 0
2w^2-10w+w-5 = 0
2w(w-5)+(w-5) = 0
(w-5)(2w+1) = 0
w = 5 or w = -1/2
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solve for "x":
If w = 5 you get:
(2x+5)/(3x) = 5
2x+5 = 15x
13x = 5
x = 5/13
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If x = -1/2 you get:
(2x+5)/(3x) = -1/2
4x+10 = -3x
7x = -10
x = -10/7
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cheers,
Stan H.
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