SEARCH HOME
Math CentralQuandaries & Queries

search

Question from Jyotishman, a student:

What can be the value of a for the equation to have 3 roots?
x^3-13x+a=0

Hi Jyotishman,

This problem involves geometry and calculus. The geometry tells me what it is I need to find and the calculus gives me a method to find it. I can get you started.

First I graphed y = f(x) = x3 - 13x + a with a = 0. The axes are in red and I didn't put the scale on the axes since I am interested in the shape not the scale. This graph crosses the x-axis in 3 places marked by the red dots.

y = x^3 - 13x

As a increases he graph moves upwards relative to the axes maintaining three distinct zeros until

y = x^3 - 13x

until a attains a value for which the polynomial f(x) = x3 - 13x + a two distinct roots, one of them I am going to call p as in the diagram. (One root is not on my diagram.) O

y = x^3 - 13x

If a increases beyond this value the resulting graph only crosses the x-axis at one point (not on the diagram).

y = x^3 - 13x

The task is thus to find p and then the value of a so that f(p) = p3 - 13p + a = 0.

p is a critical value for the polynomial f(x) so it can be found by solving f '(x) = 0.

Do you now see how to complete the problem? If you need further assistance write back.

Harley

About Math Central
 

 


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS