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My question is:

Find the square root of the polynomial-

$$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$

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Let $t=\dfrac xy +\dfrac yx$.

Then $t^2=\dfrac {x^2}{y^2} +\dfrac{y^2}{x^2}+2$.

Your "polynomial" becomes finally an actual polynomial:

$$t^2-2-2t+3=t^2-2t+1=(t-1)^2=\left(\frac xy+\frac yx -1\right)^2.$$

So, the two square roots are $$\pm \left(\frac xy+\frac yx -1\right).$$

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As a matter of correct terminology, what you have there isn’t a polynomial: the variables in the denominator prevent that.

I suggest letting $u=\dfrac{x}y$, so that $\dfrac{y}x=\dfrac1u$; then you can rewrite the expression as $$u^2+\frac1{u^2}-2\left(u+\frac1u\right)+3\;,$$ getting it in terms of a single variable, and put it all over a common denominator:

$$\frac{u^4+1-2u^3-2u+3u^2}{u^2}\tag{1}\;.$$

The denominator of $(1)$ is a perfect square, so you’re almost done if you can write the numerator as a perfect square. The numerator is $u^4-2u^3+3u^2-2u+1$. If it’s the perfect square of anything, it must be of some expression of the form $u^2+au+1$: that’s the only way to get the $u^4$ and $1$ terms by squaring. Now $$(u^2+au+1)^2=u^4+2au^3+(a^2+2)u^2+2au+1$$ (as you should check for yourself). You should be able to use this to figure out what $a$ has to be, and then all you have to do is substitute back to replace the $u$’s with $x$’s and $y$’s.

There are slicker solutions, but this one generalizes fairly easily to other problems in which a ratio of quantities appears repeatedly, as $x/y$ does here. It’s the repeated appearance of $x/y$ (sometimes squared and sometimes inverted, but still recognizable) that prompted my to make the substitution.

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$$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$
$$=\frac{{x^4} + {y^4} - 2\left({x^2} +{y^2}\right)xy + 3{x^2}{y^2}}{{x^2}{y^2}}$$ $$=\frac{{x^4} + {y^4} + 2{x^2}{y^2}+ {x^2}{y^2} - 2\left({x^2} +{y^2}\right)xy }{{x^2}{y^2}}$$ $$=\frac{({x^2}+{y^2})^2+ {(xy)^2} - 2\left({x^2} +{y^2}\right)xy }{{x^2}{y^2}}$$ $$=\frac{({x^2}+{y^2} - xy)^2}{{(xy)^2}}$$

the square root of the polynomial = $$\pm \frac{{x^2}+{y^2} - xy}{xy}$$

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The polynomial is

$ \frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left( \frac{x}{y} + \frac{y}{x} \right) + 3$

Denote $ z = \frac{x}{y}$ and have

$ z^2 + z^{-2} - 2(z+z^{-1}) + 3 $

Assume $y \ne 0$ and multiply by $z^2$ and have

$ z^4 + 1 - 2z^3 - 2z + 3z^2 = 0$

Instructions how to solve a general quartic equation can be found in Wikipedia.

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  • $\begingroup$ I just saw the answer of @Phira , in that case there is a trick (which Phira shown), in the general case you need to work harder as I wrote. (e.g. $(x/y)^2 + 3(y/x)^2 + 5(x/y) + 7(y/x)$. $\endgroup$ – Zachi Evenor Jun 9 '12 at 10:59

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