# Thread: Radical Equations Dimensions Problem

1. $(\sqrt{x+1} + 2)^2 = (\sqrt{3x+1})^2$

$(x+1) + 4\sqrt{x+1} + 4 = 3x+1$

combine like terms on the left side ...

$4\sqrt{x+1} + x + 5 = 3x + 1$

$4\sqrt{x+1} = 2x - 4$

divide all terms by 2 ...

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

square again ...

$4(x+1) = x^2 - 4x + 4$

$4x + 4 = x^2 - 4x + 4$

$0 = x^2 - 8x$

$0 = x(x - 8)$

$x = 0$ ... makes no sense in the context of the problem

$x = 8$

2. Originally Posted by skeeter
$(\sqrt{x+1} + 2)^2 = (\sqrt{3x+1})^2$

$(x+1) + 4\sqrt{x+1} + 4 = 3x+1$

combine like terms on the left side ...

$4\sqrt{x+1} + x + 5 = 3x + 1$

$4\sqrt{x+1} = 2x - 4$

divide all terms by 2 ...

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

square again ...

$4(x+1) = x^2 - 4x + 4$

$4x + 4 = x^2 - 4x + 4$

$0 = x^2 - 8x$

$0 = x(x - 8)$

$x = 0$ ... makes no sense in the context of the problem

$x = 8$
Alright so in order to get the dimensions I have to input x into the original widths and heights correct?

Thanks alot for the help man.

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