Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
What interval? Also, your equation is cubic, not quadratic. – Kaster Mar 2 '13 at 9:01
Does this work? $x+y=3-z$, $xy\le(3-z)^2/4$, $-9=xy+(x+y)z\le(3-z)^2/4+(3-z)z$, and so on. – Gerry Myerson Mar 2 '13 at 9:16
Just an aside @juantheron , Your tag for the question is algebra-precalculus and you solution uses calculus. – Abhra Abir Kundu Mar 2 '13 at 9:22
up vote 2 down vote accepted

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


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

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

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

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

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

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

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


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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.