deduce values of k for which cubic has three distinct real roots

• Apr 10th 2010, 03:03 AM
differentiate
deduce values of k for which cubic has three distinct real roots
Find the equation at the tangent to $y = x^3$ at the point $(t,t^3)$.

Hence deduce the values of k for which the equation $x^3 = kx -2$ will have 3 real and distinct roots

The equation of the tangent is $y = 3t^2x - 2t^3$

what do I do next?

Thank you
• Apr 10th 2010, 03:50 AM
HallsofIvy
For what values of x is the tangent line to $x^3- kx$ horizontal? If there are two such values and the value of y at one of those points is greater than 2 and at the other less than two, then the equation will have three distince real roots. Do you see why?
• Apr 12th 2010, 08:32 PM
differentiate
sorry. I don't see why
• Apr 13th 2010, 12:37 AM
HallsofIvy
Well, do you know what the graph of a general cubic looks like? In general it looks like an "S" on its side- that is, the graph rises to some maximum value, then back down to a minimum value, then back up again (or the other way, first a minimum, then up to a maximu, then down again).

In either case, the maximum value and minimum values will be where the derivative is 0. If the maximum is above y= -2 and the minimum below it, the graph must cross y= -2 going up to the maximum, then cross it again going down to the minimum, and cross it a third time going up again.

In particular, $y= x^3- kx$ has derivative $y'= 3x^2- k$ which will be 0 at $x= \pm\sqrt{\frac{k}{3}$. At $x= \sqrt{k}{3}$ $y= \left(\frac{k}{2}\right)^{3/2}- \frac{k^{3/2}}{\sqrt{3}}= k^{3/2}\left(\frac{1}{2^{3/2}- \frac{1}{\sqrt{3}}\right)$. $\frac{1}{2^{3/2}- \frac{1}{\sqrt{3}}$ is about -.224. That value of y will be greater than 2 for $-.224k^{3/2}> 2$ or $k^{3/2}< -8.9$ or $k< - 4.3$