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Did I solve the following quadratic equation correctly.

$W(W+2)-7=2W(W-3)$

I got.

$W^2-8W+7$ Then for my solution I got.

$(W-1)(W-7)$

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    @MichaelHardy: Umm, no idea about the real world, never have been there! Yes, you're quite right.2012-06-19

2 Answers 2

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$W(W+2)-7=2W(W-3)$

$\Downarrow$

$W^2+2W-7=2W^2-6W$

$\Downarrow$

$W^2-8W+7=0$

$\Downarrow$

$(W-1)(W-7)=0$

You're right, well done. The solutions are $W=7$ and $W=1$.

EDIT: It should be written as I showed above, you've omitted "$=0$" part of the equation, intentionally or unintentionally.

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    @David Andre is correct. To make the above correct and complete requires either making the arrows bidirectional or else *explictly* verifying that the derived *possible* solutions are *actual* solutions of the initial equation. But neither is done.2012-06-19
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$W (W+2) -7 = 2W (W-3)$ Expanding the brackets : $W^2+2W -7 = 2W^2-6W$ Then shift the right-hand side part of the equation to the left-hand side of the equation: $W^2+2W-7-(2W^2-6W) = (2W^2-6W)-(2W^2-6W)$ $W^2+2W-7-(2W^2-6W)= 0$ Expand the brackets: $W^2+2W-7-2W^2+6W= 0$ $W^2-2W^2+2W+6W-7=0$ $-W^2+8W-7 =0$ The above line is multiplied by (-1): $(-1)(-W^2+8W-7) =(-1)(0)$ Exapanding the brackets: $W^2-8W+7 =0$ This could be factorised as below $W^2-7W-1W+7 =0$ $W(W-7)-1(W-7)=0$ $(W-1)(W-7) =0$ Using Null-Factor law; Either $(W-1) =0$ or $(W-7) =0$

Therefore $W =1$ or $W =7$

So you have solved it correctly!