# Finding variable for fraction = 0

• May 17th 2009, 03:06 PM
anon_404
Finding variable for fraction = 0
What approach should i take to finding the values of x for this equation

$\frac{x^3-3x^2+1}{x^3}=0$
• May 17th 2009, 03:08 PM
derfleurer
Find all x for which the the numerator equals 0 and the denominator doesn't.
• May 17th 2009, 03:18 PM
anon_404
Quote:

Originally Posted by derfleurer
Find all x for which the the numerator equals 0 and the denominator doesn't.

Could you give some hints on how i could do that :(
• May 17th 2009, 03:35 PM
pickslides
$
\frac{x^3-3x^2+1}{x^3}=0
$

multiply both side by $x^3$

$
x^3-3x^2+1=0
$

now use the factor theorem

if $f(a)=0$ then $(x-a)$ is a factor
• May 17th 2009, 03:39 PM
VonNemo19
First multiply both sides by $x^3$ getting

$x^3-3x^2+1=0$

Then recall that the zeros of any polynomial equation of the form $a_nx^n+a_{n-1}x^{n-1}......a_1x+a_0=0$ can be expressed as a ratio of $\frac{p}{q}$ where $p$ are all possible factors of $a_0$ and $q$ are the possible factors of the leading coefficient $a_n$.

Obviously some of the zeros are of a repeating nature, so you can narrow the field down a little bit. In more complicated expressions, you would normally use synthetic division to test the roots, but here it should be evident which ones are the right choice.(Wink)

This could help.

Box 5.3a: The Cubic Formula

--------------------------------------------------------------------------------

Solving a Cubic
Four centuries before European mathematicians solved the problems of the cubic and quartic equations, Indian mathematician and astronomer, Bhaskara, had solved these problems. You can see a biography of Bhaskara by clicking here.

There are also pdf converters available online if you need to have any pdf document in another format.)

Note that this material is not at all likely to be on the exam. The formula is not even all that useful. It's given here merely to satisfy your curiosity. So you may skip this box if you like.

If you are given a cubic equation in the form of

x3 + px2 + qx + r = 0 eq. 5.3a-1

and need to solve for x, then the first thing you do is substitute variables. Everywhere you see an x in the cubic, replace it with
p
x = u - eq. 5.3a-2
3

When you get done squaring and cubing this expression, then substituting stuff back in and gathering like terms, you will get
u3 + au + b = 0 eq. 5.3a-3a

where
p2
a = q - eq. 5.3a-3b
3

and
pq 2p3
b = r - + eq. 5.3a-3c
3 27

Now compute A and B by

A =

eq. 5.3a-4a

and

B =

eq. 5.3a-4b

Then you have the following solutions for u

u = A + B eq. 5.3a-5a
____
u = -(1/2)(A + B) + √-3/4 (A - B) eq. 5.3a-5b
____
u = -(1/2)(A + B) - √-3/4 (A - B) eq. 5.3a-5c

Of course, you know how convert from the u solutions to the x solutions (hint: look at 5.3a-2).

You may be troubled by the expression, √-3/4, which is not a real number, but is a complex number. Note that A and B will also nonreal complex numbers whenever

b2 a3
+ < 0 eq. 5.3a-6
4 27

So the cubic formula does require that you understand the arithmetic of complex numbers.

And if you are not up on complex numbers, don't worry. We will be reviewing them in a later section.
• May 17th 2009, 04:06 PM
anon_404
It confuses me that on the graph of this function, where y = 0, x does not = $\frac{1}{3}$