Range of $z,$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$

Question

If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$. Then value of $z$ lie in the interval.

My Try:: Let $x,y,z$ be the roots of the quadratic equation

$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$

Let $xyz = p, $ Then let $f(t) = t^3-3t^2-9t-p.$

Now we will check the extermes value of $f(t)$

$f^{'}(t) = 3t^2-6t-9 = 3(t^2-2t-3)$

$f^{''}(t) = 3(2t-3)$

Now for Max. or Min. , $f^{'}(t)=0$ or $t=-1\;,t=3$

So $f^{''}(-1) = -15=-$(ve)

so Local Max. at $t=-1$

and $f^{''}(3) = 9 = +$(ve)

So Local Min. at $t=3$

Now My question is How can I check Range of $z$ using Graph of cubic equatio using Derivatives

Thanks

Answer 1

We have $x+y+z = 3$ and $xy+yz+zx = -9$

$xy+z(3-z)=-9$

$\Rightarrow xy+3z-z^2=-9$

$\Rightarrow xy+3z-z^2+9=0$

$\Rightarrow -xy-3z+z^2-9=0$

$\Rightarrow z^2-3z-9=xy$.....(1)

By A.M G.M inequality we have,

$\displaystyle |xy|\le(\frac{x+y}{2})^2=(\frac{3-z}{2})^2$

Using this in 1 we have,

$-(\frac{3-z}{2})^2\le z^2-3z-9\le (\frac{3-z}{2})^2$

$4(z^2-3z-9)\le (z^2-6z+9)$

$3z^2-6z-45\le 0$

$z^2-2z-15\le0$

Evaluate this and the left part and take the intersection of the two answers as your answer.

Answer 2

Equating the values of $x$

$$3-y-z=-\frac{9+yz}{y+z}\implies y^2+y(z-3)+z^2-3z-9=0$$

As $y$ is real, the discriminant $(z-3)^2-4\cdot1\cdot(z^2-3z-9)$ must be $\ge0$

$$\implies z^2-2z-15\le0\implies (z-5)(z+3)\le0$$

As $(x-a)(x-b)\le 0$ for $a\le b\implies a\le x\le b$

So,$-3\le z\le 5$

Observe that the given conditions are symmetric wrt $x,y,z$

So, $x,y,z$ will have the same range.