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.
Best Answer
$2x^2 - x - 1 \ne (2x + 1)(2x - 1)$
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 $y(2x+1)=(2x^2 -x -1)$
$y = \frac {2x^2 - x -1}{2x+1} = x-1$.