How can I show b^2 > 24c in a cubic with one maximum and one minimum?

[camrocker]

Hello, been thinking on this one for a little while, and can't seem to figure it out. Problem statement is:

The cubic curve $$y = 8x^3 + bx^2 + cx + d$$ has two distinct points P and Q, where the gradient is zero.

Show that $$b^2 > 24c$$.

It seems simple enough, but I can't logic it out. This equation has two distinct points where gradient is zero, so one maximum and one minimum, right? I played around with an online graphing calculator and saw that $$b^2 > 24c$$ for two points of zero gradient is in fact true, but don't see how to mathematically prove/show this.

What direction should I be taking to show that $$b^2 > 24c$$? Thanks for any tips!

 

[mathmari]

The gradient of the curve $$y = 8x^3 + bx^2 + cx + d$$ is $$y' = 24x^2 + 2bx + c$$

Since it is a second order polynomial it has two roots when the discriminant $\Delta$ is $>0$.

$$\Delta>0 \Rightarrow 4b^2-4 \cdot 24 c>0 \Rightarrow 4b^2>4 \cdot 24c \Rightarrow b^2>24c$$

 

[camrocker]

The gradient of the curve $$y = 8x^3 + bx^2 + cx + d$$ is $$y' = 24x^2 + 2bx + c$$

Since it is a second order polynomial it has two roots when the discriminant $\Delta$ is $>0$.

$$\Delta>0 \Rightarrow 4b^2-4 \cdot 24 c>0 \Rightarrow 4b^2>4 \cdot 24c \Rightarrow b^2>24c$$

Thank you so much! I finally got to the discriminant point some time after posting this, but I tossed it aside because I forgot that b was 2b not just b, so I got $$b^2 > 96c$$. Thanks for the help!