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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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    Your second and third expressions should have an "$=0$" in there, but yes.2012-06-19
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    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.2012-06-19
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    Omission of $=0$ is a frequent fault.2012-06-19
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    @MichaelHardy: The OP just meant to check if the simplification is correct. I don't see any frequent faults.2012-06-19
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    @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.2012-06-19
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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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    Note that the arrows between the various displayed equations should be *bidirectional*.2012-06-19
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    @AndréNicolas: I don't think *they should*, since I'm showing one direction only.2012-06-19
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    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.2012-06-19
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    @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.2012-06-19
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    It is the implication symbol, and it is widely misused by students.2012-06-19
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    @AndréNicolas: That one is $\implies$, the one I used is $\Downarrow$; They must be different things. I have no idea about students and their mistakes.2012-06-19
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    "Downarrow" vs. "implies", that is!2012-06-19
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    @Gigili you and Andre are both right. "Conclusion" is one form of "implication" (A therefore B is another way of saying A implies B). Usually, the point of "posing" a question which is a quadratic equation, is to "find what the roots are". Andre is making the point that you have found what the roots have to be (if there really are any). "Checking one's solution" amounts to the "reverse arrows", that our deductions are appropriate to the original problem. This is especially important when "W" might represent some actual quantity of something, which may have to be positive.2012-06-19
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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!