Show that the cubic equation x^3 + 3x -15 = 0 has only one real root.
Step-by-step guide is appreciated. Thanks
$\displaystyle \Delta=-(4p^3+27q^2)$ 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.
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 .