Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Did I solve the following quadratic equation correctly.


I got.

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


share|cite|improve this question
Your second and third expressions should have an "$=0$" in there, but yes. – Cameron Buie Jun 19 '12 at 17:35
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. – Mark Bennet Jun 19 '12 at 17:37
Omission of $=0$ is a frequent fault. – Michael Hardy Jun 19 '12 at 17:39
@MichaelHardy: The OP just meant to check if the simplification is correct. I don't see any frequent faults. – Gigili Jun 19 '12 at 17:40
@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. – Michael Hardy Jun 19 '12 at 17:50








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.

share|cite|improve this answer
Note that the arrows between the various displayed equations should be bidirectional. – André Nicolas Jun 19 '12 at 18:01
@AndréNicolas: I don't think they should, since I'm showing one direction only. – Gigili Jun 19 '12 at 18:03
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. – André Nicolas Jun 19 '12 at 18:08
@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. – Gigili Jun 19 '12 at 18:12
It is the implication symbol, and it is widely misused by students. – André Nicolas Jun 19 '12 at 18:16

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.