# Show that the equation have one real root

• Mar 28th 2013, 08:23 AM
alexander9408
Show that the equation have one real root
Show that the cubic equation x^3 + 3x -15 = 0 has only one real root.
Step-by-step guide is appreciated. Thanks :D
• Mar 28th 2013, 08:48 AM
Nehushtan
Re: Show that the equation has only one real root
$\displaystyle \Delta=-\left[4(3)^3+27(-15)^2\right]<0$.
• Mar 28th 2013, 08:50 AM
alexander9408
Re: Show that the equation has only one real root
Quote:

Originally Posted by Nehushtan
$\displaystyle \Delta=-\left[4(3)^3+27(-15)^2\right]<0$.

Sorry, I don't understand at all, can you explain, please?
• Mar 28th 2013, 09:00 AM
Nehushtan
Re: Show that the equation has one real root
$\displaystyle \Delta=-(4p^3+27q^2)$ is the discriminant of the cubic equation $\displaystyle x^3+px+q=0$.

• If $\displaystyle \Delta > 0$ the equation has three distinct real roots.

• If $\displaystyle \Delta=0$ the equation has three real roots, not all distinct.

• If $\displaystyle \Delta<0$ the equation has one real root and two complex roots.
• Mar 28th 2013, 01:09 PM
MINOANMAN
Re: Show that the equation have one real root
Alexander
get the function f(x)=x^3+3x-15
the first derivative is f'(x) = 3x^2+3 which is > 0 for all real numbers x .this means that the function f(x) =x^3+3x -15 is a monotone increasing function and it will meet once and only once the x-axis ..Thus the equation x^3+3x-15 = 0 has one and only one real root.
the other two roots wil be conjugate complex numbers .