if $x^3 + px^2+qx+r = 0$ has three real roots, show that $p^2 \ge 3q$

Question

If $x^3 + px^2+qx+r = 0$ has three real roots, show that $p^2 \ge 3q$

Can I get help on this problem, thanks in advance!

See https://math.stackexchange.com/questions/1393869/conditions-for-distinct-real-roots-of-cubic-polynomials — lab bhattacharjee, Jan 21 '18 at 10:02

score 2 · Answer 1 · answered Jan 21 '18 at 10:03

2

Hint: The existence of $3$ real roots of $P(x)=x^3+px^2+qx+r$ implies the existence of how many real roots of $P'(x)=3x^2+2px+q$?

answered Jan 21 '18 at 10:03

José Carlos Santos

427,504

score 0 · Answer 2 · answered Jan 21 '18 at 10:07

0

Alternative hint: Express $p$ and $q$ in terms of the roots using Vieta's relations and see what transpires.

answered Jan 21 '18 at 10:07

Mark Bennet

100,194

score 0 · Answer 3 · edited Jan 21 '18 at 11:54

0

$f(x) = x^3 + px^2 +qx+r$
$f'(x) = 3x^2 + 2px +q$
Since $\operatorname{discr}{f(x)} > 0$
Therefore $\operatorname{discr}{f'(x)} > 0$
Thus
$\Rightarrow 4p^2 - 12q > 0$
$\Rightarrow p^2 > 3q$

edited Jan 21 '18 at 11:54

Bernard

175,478

answered Jan 21 '18 at 10:41

SDA OFFICIAL

31

Welcome to MSE. Please read the MathJax tutorial. – José Carlos Santos Jan 21 '18 at 10:47
Are you sure the O.P. knows about the discriminant of a cubic polynomial? Especially with a term in $x^2$? – Bernard Jan 21 '18 at 11:56

if $x^3 + px^2+qx+r = 0$ has three real roots, show that $p^2 \ge 3q$

3 Answers3