# Thread: Show that the equation have one real root

1. ## 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

2. ## Re: Show that the equation has only one real root

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

3. ## Re: Show that the equation has only one real root

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?

4. ## 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.

5. ## 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 .

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