Question

Let x, y, z be real varibles which satisfy the equation xy+yz+zx=7 and x+y+z=6. Find the range in which the varibles lie.

Answer 1

Given equations

$xy+yz+zx=7$

$\Rightarrow xy+(x+y)z=7$ .....(1)

$x+y+z=6$

$\Rightarrow z=6-(x+y)$ .....(2)

Substituting the value from eq(2) in eq(1), we get

$xy+(x+y)\{6-(x+y\left)\right\}=7$

$\Rightarrow xy+6x+6y-x^2-y^2-2xy=7$

$\Rightarrow x^2+y^2+xy-6x-6y+7=0$

$\Rightarrow x^2+x(y-6)+(y^2-6y+7)=0$

Since, $x$ is real

$\Rightarrow D\ge 0$

$\Rightarrow (y-6{)}^2-4(y^2-6y+7)\ge 0$

$\Rightarrow y^2+36-12y-4y^2+24y-28\ge 0$

$\Rightarrow -3y^2+12y+8\ge 0$

$\Rightarrow 3y^2-12y-8\le 0$ ....(3)

Now, taking $3y^2-12y-8=0$, we will find the roots

$y=\frac{12\pm \sqrt{144+96}}{6}$

$y=\frac{12\pm \sqrt{240}}{6}$

$y=\frac{12\pm 4\sqrt{15}}{6}$

$y=\frac{6\pm 2\sqrt{15}}{3}$

So, the inequality (3) becomes

$[y-\left(\frac{6-2\sqrt{15}}{3}\right)][y-\left(\frac{6+2\sqrt{15}}{3}\right)]\le 0$

$\Rightarrow \frac{6-2\sqrt{15}}{3}\le y\le \frac{6+2\sqrt{15}}{3}$

Since, $x,y,z$ are symmetrical placed . So, $y,z$ also lies in thesame interval.