# cubic equation

• Jul 20th 2009, 05:53 AM
furor celtica
cubic equation
the equation 3x^2 - x^3 = k has three real solutions for k
write down the set of possible values for k

i dont understand: can't k = anything as long as x remains real? and what is meant by '3x^2 - x^3 = k has three real solutions for k'?
• Jul 20th 2009, 06:53 AM
red_dog
$3x^2-x^3=k\Leftrightarrow x^3-3x^2+k=0$

Let $f(x)=x^3-3x^2+k, \ f:\mathbb{R}\to\mathbb{R}$

$f'(x)=3x^2-6x$

$f'(x)=0\Rightarrow x_1=0, \ x_2=2$

$f(0)=k, \ f(2)=k-4, \ \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty$

The Rolle's sequence is: $-\infty \ \ k \ \ k-4 \ \ \infty$

To have 3 real solutions the following inequalities must hold:

$\left\{\begin{array}{ll}k\geq 0\\k-4\leq 0\end{array}\right.\Rightarrow k\in[0,4]$
• Jul 20th 2009, 06:55 AM
bananaxxx
Draw the graph of the left hand side of the equation. Where you can draw a horizontal line such that the line intersects the graph three times is a value of k that you want - find all such values.

(drawing is meant for illustration only - it will need to be worked out mathematically. hint: consider turning points)
• Jul 20th 2009, 06:56 AM
bananaxxx
Quote:

Originally Posted by red_dog
$3x^2-x^3=k\Leftrightarrow x^3-3x^2+k=0$

Let $f(x)=x^3-3x^2+k, \ f:\mathbb{R}\to\mathbb{R}$

$f'(x)=3x^2-6x$

$f'(x)=0\Rightarrow x_1=0, \ x_2=2$

$f(0)=k, \ f(2)=k-4, \ \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty$

The Rolle's sequence is: $-\infty \ \ k \ \ k-4 \ \ \infty$

To have 3 real solutions the following inequalities must hold:

$\left\{\begin{array}{ll}k\geq 0\\k-4\leq 0\end{array}\right.\Rightarrow k\in[0,4]$

or red_dog will just give you the answer...
• Jul 20th 2009, 08:34 AM
furor celtica
thank you both