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I have the problem (4x-4)^2 = (2x+9)^2 + (2x)^2. I ended up with 4x^2-34-1 = 0. Is that right? and if so, how do I factor it correctly? thank you.
Guest
Jan 31, 2012
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(4x-4)^2 = (2x+9)^2 + (2x)^2
16x
^{2}
- 32x + 16 = 4x
^{2}
+ 36x +81 +4x
^{2}
8x
^{2}
- 68x - 65 = 0
Then you have to use the quadratic formula to get the 2 irrational roots.
Melody
Oct 9, 2013
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