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I have a problem on my textbook:

$x,y \in\mathbb{R}$ and $x^2+xy+y^2+ \sqrt{3}y + 1=0$ is given. Find the value of $xy$.

I couldn't find.

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    If you don't mind, which textbook is this from? – 2017-01-19
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    I took the question from fdd's (a Turkish publisher) tests for ygs preperation. – 2017-01-19

5 Answers 5

1

Lets group the terms so we get a quadratic in $y$ $$y^2+(x+\sqrt{3})y+x^2+1=0$$ Now for a solution to exist we must have that the discriminant $\Delta\geq 0$ $$\Delta=(x+\sqrt{3})^2-4(x^2+1)=-3x^2+2\sqrt{3}-1=-(3x^2-2\sqrt{3}+1)=-(\sqrt{3}x-1)^2$$ Since a square is always $\geq 0$ we have that $\Delta\leq0$ (because of the minus) which implies $\sqrt{3}x-1=0$,$x=\frac{1}{\sqrt{3}}$ From the quadratic formula we have(since $\Delta=0$) $$y_{1,2}=\frac{-(x+\sqrt{3})}{2}=-\frac{2}{\sqrt{3}}$$

3

Multiply both sides on 4 $$x^2+xy+y^2+ \sqrt{3}y + 1=0$$ $$4x^2+4xy+y^2+3y^2+ 4\sqrt{3}y + 4=0$$ $$(2x+y)^2+(\sqrt3y+2)^2=0$$ Then $y=-\frac2{\sqrt3}, x=\frac1{\sqrt3}$

$$xy=-\frac2{3}$$

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    You multiply everything by 4, except the last two terms? Edit: fixed now; (y). – 2017-01-19
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    Did you know this equation was going to have a unique solution (or at least finitely many) before you started working on it? – 2017-01-19
2

You can rewrite this equation as :

$$\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+y\sqrt{3}+1=0$$

which can be transformed further into :

$$\left(x+\frac{y}{2}\right)^2+\frac{3}{4}\left(y^2+\frac{4y}{\sqrt3}+\frac{4}{3}\right)=0$$

Finally, you should notice that :

$$y^2+\frac{4y}{\sqrt3}+\frac{4}{3}=\left(y+\frac{2}{\sqrt3}\right)^2$$

When the sum of two positive numbers is zero, each of them must be $0$.

Hence :

$$x+\frac{y}{2}=0\qquad\mathrm{and}\qquad y+\frac{2}{\sqrt3}=0$$

which leads to :

$$x=\frac{1}{\sqrt3}\qquad\mathrm\qquad y=-\frac{2}{\sqrt3}$$

Finally :

$$\boxed{xy=-\frac{2}{3}}$$

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    How did you know to factor it as the sum of two squares? Did you know from the beginning that the equation was going to have one (or finitely many) solutions? – 2017-01-19
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    The phrasing of the question ("find the value of $xy$") made me think that there would be only one answer. Putting a quadratic expression in one or several variable(s) into canonical form is a ... canonical method worth to know ! – 2017-01-19
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    Haha thanks I'll try to remember that. Rather than factoring it into squares, I was trying to factor it as $(ax + by + c)(dx + e y + d)$. But then I realized this must be impossible, because substituting $y=0$ into the original equation yields no solution for $x$. Anyway I asked a related question to this here if you're interested: http://math.stackexchange.com/questions/2104331/when-substituting-how-do-i-know-if-ill-get-a-new-equation-or-just-an-identity – 2017-01-19
2

Setting $t=xy\in\mathbb R$, the equation becomes (knowing that $y\ne0$):

$$\frac{t^2}{y^2}+t+y^2+\sqrt3y+1=0.$$

The discriminant of this quadratic equation is $$\Delta=b^2-4ac=\frac{-3y^2-4\sqrt3y-4}{y^2}=-3\left(1-\dfrac2{\sqrt3y}\right)^2$$ and for a real solution to exist, it must be zero.

Then

$$xy=t=-\frac b{2a}=-\frac{y^2}2=-\frac23.$$

1

Here we have two variables. Looks impossible. So our approach is pretty obvious. Its one of those kinds of problem where there equation holds only for a particular value of x and y.
$ (x + (y/2))^2 + ((√3/2)y +1)^2 =0 $ .
Two non negative terms (squares) adding to zero implies both 0.
Therefore $ x=-y/2$ and $ y= -(2/√3) $