How to prove b^2 > 24c for a cubic with 1 max/1 min

[teetar]

Homework Statement

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

Show that $b^2 > 24c$

Homework Equations

None that I can think of.

The Attempt at a Solution

There's two distinct points where the gradient is zero, since it's third degree these must be the local maximum and minimums points. I graphed the equation using an online graphing tool and some sliders, and saw that it was in fact true that $b^2$ has to be greater than $24c$ for there to be these points with zero gradient, but I'm completely lost on how to show this mathematically.

What direction should I go into start myself off? Thanks for any tips!

 

[Svein]

How do you find the equation for a tangent to a curve?

 

[teetar]

How do you find the equation for a tangent to a curve?

I could get the gradient of the tangent from the first derivative $\frac{dy}{dx} = m = 24x^2 + 2bx + c$, and I believe the equation of the line would then be $y - y_1 = m(x - x_1)$, right? Sorry, I'm not picking up on where to go next.

 

[SteamKing]

I could get the gradient of the tangent from the first derivative $\frac{dy}{dx} = m = 24x^2 + 2bx + c$, and I believe the equation of the line would then be $y - y_1 = m(x - x_1)$, right? Sorry, I'm not picking up on where to go next.

If you want to find where the gradient (slope) is zero, and the slope of a tangent to the polynomial is m = 24x² + 2bx + c, then what must you do?

 

[teetar]

If you want to find where the gradient (slope) is zero, and the slope of a tangent to the polynomial is m = 24x² + 2bx + c, then what must you do?

You must solve for $0 = 24x^2 + 2bx + c$, right? How do I get an inequality out of that?

 

[SteamKing]

You must solve for $0 = 24x^2 + 2bx + c$, right? How do I get an inequality out of that?

Re-read the question from the OP.

 

[RUber]

In order for the quadratic to have 2 (real) solutions, something has to be true about (2b)^2 - 4(24)(c).

 

[Mark44]

Questions that involve derivatives do not belong in the Precalc subsection, so I have moved this thread to the Calculus subsection..

 

[teetar]

Sorry I'm not replying until now, been busy.

Re-read the question from the OP.

Thanks! Guess I just needed to clear my head, now it makes sense. I appreciate the help!

In order for the quadratic to have 2 (real) solutions, something has to be true about (2b)^2 - 4(24)(c).

Thanks a bunch! I don't know why I missed this the entire time, but it is indeed the solution. Of course, due to my organization on paper, I put b in instead of 2b and was faced with 10 more minutes of confusion.

Questions that involve derivatives do not belong in the Precalc subsection, so I have moved this thread to the Calculus subsection..

Thanks, sorry for posting in the wrong section.