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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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    $\begingroup$ Your second and third expressions should have an "$=0$" in there, but yes. $\endgroup$ – Cameron Buie Jun 19 '12 at 17:35
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    $\begingroup$ Strictly, you want $(W-1)(W-7)=0$ [same for second line too] so that $W=1$ or $W=7$. The easiest way to check you haven't made a mistake is by substituting these values into the original equation and checking that they work. $\endgroup$ – Mark Bennet Jun 19 '12 at 17:37
  • $\begingroup$ Omission of $=0$ is a frequent fault. $\endgroup$ – Michael Hardy Jun 19 '12 at 17:39
  • $\begingroup$ @MichaelHardy: The OP just meant to check if the simplification is correct. I don't see any frequent faults. $\endgroup$ – Gigili Jun 19 '12 at 17:40
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    $\begingroup$ @Gigili : I didn't mean frequent in this posting; I meant frequent out there in the world. Zillions of students make this same mistake. I think they think what they're doing is pushing symbols around according to prescribed rules, as in long division, rather than at each step making a statement that should be true. $\endgroup$ – Michael Hardy Jun 19 '12 at 17:50
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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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    $\begingroup$ Note that the arrows between the various displayed equations should be bidirectional. $\endgroup$ – André Nicolas Jun 19 '12 at 18:01
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    $\begingroup$ @AndréNicolas: I don't think they should, since I'm showing one direction only. $\endgroup$ – Gigili Jun 19 '12 at 18:03
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    $\begingroup$ However, in principle you have only shown then that the original equation can have no roots other than (possibly) $1$ and/or $7$, but it doesn't show these are roots. Reversibility is key. (Or alternately check at the end that $1$ and $7$ actually satisfy the original.) The forward arrows only say that if $W$ is a root, then there are only two candidates. $\endgroup$ – André Nicolas Jun 19 '12 at 18:08
  • $\begingroup$ @AndréNicolas: It isn't a "forward arrow", (what's that?) It's used for conclusion, I've learned it here on SE with the same usage. $\endgroup$ – Gigili Jun 19 '12 at 18:12
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    $\begingroup$ It is the implication symbol, and it is widely misused by students. $\endgroup$ – André Nicolas Jun 19 '12 at 18:16
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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!

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