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A rectangle has an area of $$\sqrt{2x^2 - x - 1}$$ Determine an expression for the width of the rectangle if the length is $$2x+1$$ The answer is x-1

I started with $$\sqrt{2x^2 - x - 1} = y(2x+1)$$ Then squared both sides $$2x^2 - x - 1 = y^2 (2x+1)^2$$ Then divided both sides by (2x+1)^2 $$(2x^2 - x - 1) / (2x+1)^2 = y^2$$ Then factored the top $$((2x-1)(2x+1))/((2x+1)(2x+1))$$ Then cancelled out the two (2x+1) $$(2x-1)/(2x+1) = y^2$$ Then square routed both sides $$\sqrt{(2x-1)/(2x+1)} = y$$

I don't know where to go from here, I think I might have done something wrong but I can't spot anything. Any help is greatly appreciated.

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  • $\begingroup$ Do you mean polygon? $\endgroup$ – David Diaz Jun 12 '18 at 23:07
  • $\begingroup$ "Then divided both sides by (2x+1)^2" Did you consider the case where $(2x + 1)^2 = 0$? $\endgroup$ – fleablood Jun 12 '18 at 23:07
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    $\begingroup$ "Then factored the top" . $(2x^2 - x -1)\ne (2x - 1)(2x + 1) = 4x^2 - 1$ $\endgroup$ – fleablood Jun 12 '18 at 23:09
  • $\begingroup$ "The answer is x-1" No it isn't $(x-1)(2x+1) = 2x^2 - x - 1 \ne \sqrt {2x^2 - x - 1}$. $\endgroup$ – fleablood Jun 12 '18 at 23:18
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$2x^2 - x - 1 \ne (2x + 1)(2x - 1)$

Instead:

$2x^2 - x - 1 = 2x^2 + x - 2x - 1 = (2x + 1)x - (2x -1) = (2x+1)(x-1)$

$y^2 =\frac {2x^2 - x-1}{(2x+ 1)^2} = \frac {x-1}{2x + 1}$

So $y = \sqrt \frac {x-1}{2x+1}$.

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I must assume you wrote the question incorrectly and that the question was that the area so $2x^2 - x -1$. In which case $y(2x+1)=(2x^2 -x -1)$

$y = \frac {2x^2 - x -1}{2x+1} = x-1$.

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$$ A= \sqrt{2x^2 - x - 1}$$

$$ L=2x+1$$

$$ W= A/L = \frac {\sqrt{2x^2 - x - 1}}{2x+1}$$

$$=\frac {\sqrt{(2x+1)(x - 1)}}{2x+1}$$

$$= \sqrt\frac {{x - 1}}{2x+1}$$

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