# Quadratics - why do roots remain the same when dividing an expression?

• Jun 1st 2011, 05:11 PM
jaffs
Quadratics - why do roots remain the same when dividing an expression?
Hi,

Could anybody explain why if I divide a quadratic, the roots remain the same? Or point me in the direction of an article or tutorial that explains this.

Take for example: -3x^2 - 15x + 9

I can divide this by 2, by (x^3 - 9x + 23), by (9x^4 - 5x^3 + 6x^2 9x + 9000)^(97/2) and so on. I see the peaks changing, yet if I'm asked to find the roots, apparently I can just discount the denominator.. for.. some reason. I've tried to find out why, but many places seem to just take it for granted without explanation. Maybe it's just really obvious (Thinking)

Thank you,
J.K
• Jun 1st 2011, 05:55 PM
Wilmer
Got no idea what you're doing or asking...
• Jun 1st 2011, 06:04 PM
jaffs
Heh sorry, I'll put it another way.

If I take, say (-2x^2 +10x + 4) and find the roots, they're exactly the same as for (-2x^2 +10x + 4) / (x^3 - 9x + 23).

So, as long as the top stays the same, the roots don't change regardless of what I put at the bottom.

I'm trying to understand why, but not solve any specific problem.

I'm not sure if that's any clearer :)
• Jun 1st 2011, 06:20 PM
TheEmptySet
Quote:

Originally Posted by jaffs
Heh sorry, I'll put it another way.

If I take, say (-2x^2 +10x + 4) and find the roots, they're exactly the same as for (-2x^2 +10x + 4) / (x^3 - 9x + 23).

So, as long as the top stays the same, the roots don't change regardless of what I put at the bottom.

I'm trying to understand why, but not solve any specific problem.

I'm not sure if that's any clearer :)

I don't think that is true from example consider the function

$f(x)=\frac{x^2-1}{(x-1)}$

$x^2-1$ has two roots but $f(x)$ only has 1.

plot &#40;x&#94;2-1&#41;&#47;&#40;x-1&#41; - Wolfram|Alpha
• Jun 1st 2011, 07:09 PM
TheChaz
Quote:

Originally Posted by jaffs
Heh sorry, I'll put it another way.

If I take, say (-2x^2 +10x + 4) and find the roots, they're exactly the same as for (-2x^2 +10x + 4) / (x^3 - 9x + 23).

So, as long as the top stays the same, the roots don't change regardless of what I put at the bottom.

I'm trying to understand why, but not solve any specific problem.

I'm not sure if that's any clearer :)

You give an example of a rational function. The roots (i.e. x-intercepts) will be where the numerator = 0 BUT where the denominator is NOT = 0.

As { } has demonstrated, when there is a common factor (e.g. x - 1), there is a hole rather than a zero/root/x-intercept.

So to answer your question, the roots are unchanged because you are not dividing by the corresponding factors.
• Jun 1st 2011, 07:27 PM
jaffs