# Solution for a quadratic equation

The following is the solution from my book, but without the calculation path.

$$\frac{(18 \pm \sqrt{-180})}{6} = 3 \pm\sqrt{-5}$$

Why is it $-5$?

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Sorry for the format. I am on it to get it fixed – MartinVonMartinsgrün Aug 27 '11 at 15:17
It is unclear whether or not you mean: $\frac{1}{6}(18\pm\sqrt{-180})$ or $\frac{1}{6(18\pm\sqrt{-180})}$. – Asaf Karagila Aug 27 '11 at 15:18
first version, thx – MartinVonMartinsgrün Aug 27 '11 at 15:20

We have this:

$$\frac{18\pm\sqrt{-180}}6 = \frac{18}{6}\pm\frac{\sqrt{-180}}{6} = 3\pm\sqrt{\frac{-180}{36}}=3\pm\sqrt{-5}$$

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 Will the downvoter explain, or is he already gone into the sunset? – Asaf Karagila Aug 27 '11 at 19:29

\begin{align} \frac{1}{6}(18\pm\sqrt{-180}) &= \frac{1}{6} \times (18) \pm \frac{1}{6} \times \sqrt{-180}\\ &= \frac{1}{6} \times (18) \pm \frac{1}{6} \times \sqrt{-4 \times 9 \times 5}\\ &=\frac{1}{6} \times (18) \pm \frac{1}{6} \times (2 \times 3) \times\sqrt{-5}\\ &= 3 \pm\sqrt{-5} \end{align}

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 Your usage of implication arrows $\Rightarrow$ is a bit strange. They are commonly used for statements, e.g. equations. Equal signs $=$ would have been better here. – Calle Aug 27 '11 at 16:34 @Calle:You are right $=$ is better here. – Quixotic Aug 27 '11 at 16:43 I changed it to a more "correct" LaTeX code. I hope you don't mind. – Asaf Karagila Aug 27 '11 at 18:40 @Asaf :Not at all! – Quixotic Aug 28 '11 at 6:07

Note that no ingenuity is required. For clearing denominators reduces it to proving the equivalent equation $\:\sqrt{-180} = 6\ \sqrt{-5}\:,\:$ whose proof should now be obvious. In the same way, one can often reduce less trivial equations to a "simpler" form. This is especially helpful when one is attempting to follow proofs involving special functions whose laws may not be so familiar. Then, with luck, one may succeed in reducing an equation to a known basic law of the special function (e.g. an addition law). You may have already encountered such transformation-based proof strategies when attempting to prove trigonometric identities in high school.

In fact this was a method that I employed in high school to help prove trig identities. On the left, starting from the identity to be proved, I worked forwards, applying known laws. On the right, starting from known trig identities, I worked backwards, hoping to connect the two chains somewhere in the middle - hence concluding the sought proof. When there are many possible transformations to apply at each step (i.e. a large branching factor) this method of simultaneously forward and backward chaining - combined with intelligent pruning - can help to constrain combinatorial explosion due to the exponential growth of the search tree. My high school teacher liked this method so much that he incorporated it into his lessons. Later, when I arrived at MIT, I learned that such simultaneous forward and backward chaining is a standard technique for searching such trees. For example, it proves helpful in solving certain chess problems. It is always helpful to keep such "meta-level" techniques in mind when searching for proofs.

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 The following proof technique is called in my trigonometry text book "analytical method" $\frac{1+\sin a}{\cos a}=\frac{\cos a}{1-\sin a}\Leftrightarrow (1+\sin a)(1-\sin a)=\cos a\cdot \cos a$ $\Leftrightarrow 1-\sin ^{2}a=\cos ^{2}a\Leftrightarrow 1=\cos ^{2}a+\sin ^{2}a$. – Américo Tavares Aug 27 '11 at 19:19

$\sqrt{-180} = \sqrt{36\cdot(-5)} = \sqrt{36}\cdot\sqrt{-5} = 6\sqrt{-5}.$

(I think Asaf Karagila's answer is somewhat more complicated than it needs to be.)

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I don't really see how this is more or less complicated than my answer. – Asaf Karagila Aug 27 '11 at 16:04
Your answer makes it look as if you couldn't have done anything without the denominator to cancel something in the numerator. – Michael Hardy Aug 27 '11 at 18:48
You are aware that formally this is pretty much that case, right? :-) However, since many many mistakes come by manipulating minus under the root I think it is best to avoid it until one is comfortable with the basic laws of arithmetics (I'm referring, of course to $1 = (-1)^2=\sqrt{-1}\sqrt{-1}=i^2=-1$) – Asaf Karagila Aug 27 '11 at 18:53
I don't understand your initial question. I think there's something to be said for separating the largest square factor under the radical from the square-free factor and then simplifying the first factor, quite apart from whether it cancels anything in the denominator. – Michael Hardy Aug 27 '11 at 20:17
Oh, that was a rhetorical device, since formally you did not answer the question but rather shown an important step towards the solution. Yes, there is something to say. I do not see, regardless to that, how in this specific case it is more or less complicated than what I'd done. – Asaf Karagila Aug 27 '11 at 21:11