$\displaystyle 16x^2-8x+3=1+(ax+b)^2$
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Originally Posted by totalnewbie $\displaystyle 16x^2-8x+3=1+(ax+b)^2$ If 16x^2-8x+3=1+(ax+b)^2 then 16x^2-8x+3 = a^2 x^2+2 a b x+b^2+1 so a=+/-4, ab=-8, so b=-/+2, and so b^2+1=4+1=5 != 3, which is a contradiction so there is no such real transformation. RonL
Last edited by CaptainBlack; Dec 10th 2006 at 09:28 AM.
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