Find roots of function

**gk99** · March 27th 2010, 04:02 AM

Firstly this question relates to an assignment so please do not provide answer.

I am asked to find the roots of a function which appears to be a cubic function.

Function

$\frac{x^3}{3} - x^2 - 3x$

I am given a graph which clearly shows 3 roots, which are approx -2, 0 and 5. However I need to work out the exact points.

I have read the The "Cubic Formula" which talks about the cubic formula however I honestly cannot get my head around it. Does anyone know of a better link which explains it better or can explain it better ?

Unfortunately the book I have been given doesn't touch on this formula.

**skeeter** · March 27th 2010, 04:44 AM

Originally Posted by gk99

Firstly this question relates to an assignment so please do not provide answer.

I am asked to find the roots of a function which appears to be a cubic function.

Function

$\frac{x^3}{3} - x^2 - 3x$

I am given a graph which clearly shows 3 roots, which are approx -2, 0 and 5. However I need to work out the exact points.

I have read the The "Cubic Formula" which talks about the cubic formula however I honestly cannot get my head around it. Does anyone know of a better link which explains it better or can explain it better ?

Unfortunately the book I have been given doesn't touch on this formula.

no need for the cubic formula ...

$\frac{x^3}{3} - x^2 - 3x = 0$

factor out an $x$ from each term ...

$x \left(\frac{x^2}{3} - x - 3\right) = 0$

now you know one root is $x = 0$ ... solve the quadratic factor for the other two roots.

**Prove It** · March 27th 2010, 04:47 AM

Originally Posted by gk99

Firstly this question relates to an assignment so please do not provide answer.

I am asked to find the roots of a function which appears to be a cubic function.

Function

$\frac{x^3}{3} - x^2 - 3x$

I am given a graph which clearly shows 3 roots, which are approx -2, 0 and 5. However I need to work out the exact points.

I have read the The "Cubic Formula" which talks about the cubic formula however I honestly cannot get my head around it. Does anyone know of a better link which explains it better or can explain it better ?

Unfortunately the book I have been given doesn't touch on this formula.

If you are finding the roots, you are finding the values of $x$ that make the equation

$\frac{x^3}{3} - x^2 - 3x = 0$ true.

Multiply both sides by $3$ to get rid of the fraction...

$x^3 - 3x^2 - 9x = 0$

Now factorise and use the Null Factor Law to solve for $x$ .

$x^3 - 3x^2 - 9x = 0$

$x(x^2 - 3x - 9) = 0$

So $x = 0$ or $x^2 - 3x - 9 = 0$ .

The first root is obvious, the other 2 require either completing the square or using the Quadratic Formula...

$x^2 - 3x - 9 = 0$

$x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 9 = 0$

$\left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{36}{4} = 0$

$\left(x - \frac{3}{2}\right)^2 - \frac{45}{4} = 0$

$\left(x - \frac{3}{2}\right)^2 = \frac{45}{4}$

$x - \frac{3}{2} = \pm \sqrt{\frac{45}{4}}$

$x - \frac{3}{2} = \pm \frac{3\sqrt{5}}{2}$

$x = \frac{3}{2} \pm \frac{3\sqrt{5}}{2}$ .

So the three roots are

$x = \frac{3 - 3\sqrt{5}}{2}, x = 0, x = \frac{3 + 3\sqrt{5}}{2}$ .

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