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I was solving a question from my book.

If $\alpha$, $\beta$ are the roots of $pt^2+qt+q=0$, $p \neq 0$ and $q \neq 0$ then show that, $\sqrt{\frac{q}{p}}+\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=0$.

For simplification one would multiply the second term and third term at the left with $\frac{\sqrt{\beta}}{\sqrt{\beta}}$ and $\frac{\sqrt{\alpha}}{\sqrt{\alpha}}$ respectively and then simply apply the law,

$$\sqrt{a}\sqrt{b}=\sqrt{ab}$$

without caring that this law not always holds. I want to know that what should we do before applying this property so that we can surely say $\sqrt{\alpha}\sqrt{\beta}=\sqrt{\alpha\beta}$. I mean what exclusions or suppositions should we make before applying this property.

My explanation to this problem.

I think when we used the property $\sqrt{a}\sqrt{b}=\sqrt{ab}$, we had in other words said that $\alpha$, $\beta$ are two numbers or roots such that they satisfy the applied property necessarily and this excludes the values from the domains of $\alpha$ and $\beta$ which do not satisfy the property. Therefore, we do not need to show any suppositions or exclusions directly that is; "applying that property" on two unknowns itself excludes those values from the domains of these unknowns which do not satisfy the property.

The thing which motivates me to make such explanation.

We often encounter this situation while solving the equations and apply this property without caring the condition.

For example, while solving the following equation.

$\sqrt{y^2+3y+2}+\sqrt{y^2+3y+8}-3=0$. We can put $y^2+3y=x$ and then taking the square of both side would lead us to $2x + 1=-2\sqrt{(x+2)(x+8)}$ and then again taking square and simplifying would lead us to solution set which would have all the values for $y$ such that they satisfy the given equation along with the property, we applied earlier, $\sqrt{y^2+3y+2}\sqrt{y^2+3y+8}=\sqrt{(y^2+3y+2)(y^2+3y+8)}$.

This case motivated me to think that when we applied the property $\sqrt{a}\sqrt{b}=\sqrt{ab}$, every value in domain of $y$ which could not make to satisfy this property would have been automatically removed from the domain of $y$. However I don't know I am right to say it or not.

Problem in my explanation about the proof my book asks for.

If I accept for a moment the statement given earlier that "applying that property on two unknowns itself excludes those values from the domains of these unknowns which do not satisfy the property", one may claim that maybe if we exclude such values from domains of roots of a particular type of equation ($pt^2+qt+q$, in this case) then maybe roots of such TYPE or SOME REAL EQUATIONS ANALOGOUS TO SUCH TYPE do not exist.

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    $\begingroup$ The rule $\sqrt{ab}=\sqrt a \sqrt b$ holds when $a,b\geqslant 0$. $\endgroup$ – Pedro Tamaroff Mar 25 '16 at 21:10
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Whatever book you're using, it better have in it somewhere a discussion of what it (the book) means by the square root symbol when applied to numbers that are not positive reals. Otherwise, what it's asking you to show doesn't make sense.

For example, consider the case $p=2$, $q=-1$, where $2t^2-t-1=(2t+1)(t-1)$ has roots $-1/2$ and $1$. You are asked to show that

$$\sqrt{q\over p}+\sqrt{\alpha\over\beta}+\sqrt{\beta\over\alpha}=\sqrt{-1\over2}+\sqrt{-1\over2}+\sqrt{-2}=0$$

Now this is nonsense if the square root symbol is given its conventional meaning when applied to negative numbers, namely $\sqrt{x}=\sqrt{|x|}i$ if $x\lt0$, because in that convention we have

$$\sqrt{-1\over2}+\sqrt{-1\over2}=2\sqrt{1\over2}i=\sqrt{2}i$$

and thus

$$\sqrt{-1\over2}+\sqrt{-1\over2}+\sqrt{-2}=2\sqrt{2i}\not=0$$

So please go back and scour your book for what it says about its interpretation of the square root symbol.

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    $\begingroup$ I feel that this answer is the only one so far that really addresses the question at hand, +1. $\endgroup$ – Ennar Mar 25 '16 at 22:21
  • $\begingroup$ @Ennar, thanks for the comment (and the +1). $\endgroup$ – Barry Cipra Mar 25 '16 at 22:35
  • $\begingroup$ Does there exist any other meaning beside the one, you call conventional? $\endgroup$ – Sufyan Naeem Mar 25 '16 at 23:01
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    $\begingroup$ @SufyanNaeem, we could invent a convention that $\sqrt{x}=-\sqrt{|x|}i$ if $-1\lt x\lt0$ (and keep the rest of the usual convention for $x\le-1$). That would make the equation $\sqrt{-1/2}+\sqrt{-1/2}+\sqrt{-2}=0$ correct, but it would probably run afoul on other examples. Plus, I rather doubt that's what the OP's book has in mind.... $\endgroup$ – Barry Cipra Mar 25 '16 at 23:18
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    $\begingroup$ @Man_Of_Wisdom, in that case, as my example shows, the book is contradicting itself. $\endgroup$ – Barry Cipra Mar 26 '16 at 13:50
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It is fine to use $\sqrt a \sqrt b = \sqrt {ab}$ as long as $a$ and $b$ are positive reals. If $a,b \lt 0$ the result does not hold true. The result might also not hold true if $a$ and $b$ are complex.

A very simple example:

Let $a = -4$ and $b = -9$

So, $$\begin{align} \sqrt a \sqrt b &=\sqrt{-4} \sqrt{-9} \\ & = 2i \times 3i \\ & = -6 \end{align}$$

Also, $$\begin{align} \sqrt {ab} &= \sqrt{-4 \times -9} \\ &= \sqrt {36} \\ &= 6 \end{align}$$

See, here the equation $\sqrt a \sqrt b = \sqrt {ab}$ does not hold true, as we get different answers.

In case of positive reals, the equation holds true. If you delve into complex numbers or negative reals, you run into trouble.

So, if you can prove that the two numbers $a, b \in \mathbb{R}$ are not both negative, you can use the equation without any trouble. But, if you can't, then be aware that you might get a wrong answer. Plus, complex numbers are a whole different ball game. Don't use this equation an any case involving complexes, or the chances of error become very high.

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  • $\begingroup$ The $\alpha , \beta$ refer to the roots in the question. Since OP is multiplying by $\sqrt {\frac \alpha \alpha}$ and $\sqrt {\frac \beta \beta}$, they can't be zero. $\endgroup$ – Pratyush Yadav Mar 25 '16 at 22:24
  • $\begingroup$ The two roots of the given quadratic cannot both be positive real numbers. $\endgroup$ – Barry Cipra Mar 25 '16 at 22:26
  • $\begingroup$ I think one of us misunderstood the other. The equation given in the first quotation can have both positive roots (they can also be negative or non-real for all I know), unless more information is provided. All we know is that $p,q \gt 0$. We don't know anything about the nature of roots. I don't exactly understand what OP is trying to say in the 2nd quotation. I'm only referring to the first equation. That equation can have both positive roots as long as no new information is provided. $\endgroup$ – Pratyush Yadav Mar 25 '16 at 22:32
  • $\begingroup$ I am referring to the equation $pt^2+qt+q$ as well. By Vieta, the sum of the roots is $-q/p$ and the product is $q/p$. If the two roots are both positive, both their sum and their product is positive. But $-q/p$ and $q/p$ have opposite sign. $\endgroup$ – Barry Cipra Mar 25 '16 at 23:32
  • $\begingroup$ Okay. Sorry. My bad. I didn't notice that. I'll rephrase then. Thanks for pointing it out. $\endgroup$ – Pratyush Yadav Mar 25 '16 at 23:42

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