Polynomial function

[TheLivingMartyr]

I'm doing my integration homework, and I can do the integration fine, but one of the functions i've got now is

 

[math]

4x^2 + \frac{1}{x^2}

[/math]

 

It asks me to find where this function crosses the x-axis, ie. Solve it. But seeing as there are positive and negative powers of x I don't know how to go about solving it. I thought maybe you could do the following:

 

[math]

4x^2 + \frac{1}{x^2} = 0

[/math]

[math]

4x^4 + 1 = 0

[/math]

[math]

(2x^2 + 1)^2 - 4x^2 = 0

[/math]

[math]

So -2x^2 + 1 = 0

[/math]

[math]

2x^2 - 1 = 0

[/math]

[math]

(\surd{2}x + 1)(\surd{2}x - 1) = 0

[/math]

[math]

x = \frac{\surd{2}}{2}... or... x = - \frac{\surd{2}}{2}

[/math]

 

I'm not sure if you can do it this way? but if you can could you tell me whether or not my answer is correct, and if you cant do it this way, could someone tell me how you are supposed to solve such an equation. Thanks

[Schrödinger's hat]

I'm not sure if you can do it this way? but if you can could you tell me whether or not my answer is correct, and if you cant do it this way, could someone tell me how you are supposed to solve such an equation. Thanks

 

1) If you want to know if your answer is correct, try putting it in the original function

 

2) Not quite sure what you're doing with this line:

[math] So -2x^2 + 1 = 0 [/math]

 

Maybe try from here:

[math] 4x^2 + \frac{1}{x^2} = 0 [/math]

[math] 4x^4 =-1 [/math]

 

Other things to think about that are good for checking your answers ahead of time:

 

If you put a real number in [math]x[/math], what is the sign of [math]x^2[/math].

 

When is [math]4x^2[/math] less than 4?

 

When is [math]\frac{1}{x^2}[/math] less than 1?

 

Are either of these ever negative (for real [math]x[/math])?

 

What does that mean if [math]4x^2 + \frac{1}{x^2}[/math] is less than 1?

 

If you ask questions like these you can usually figure out what your answer should be like without doing the rearranging.

Also a good idea is to plot it on a calculator or similar.

Edited October 22, 2011 by Schrödinger's hat

[TheLivingMartyr]

wait a minute.....oh dear, it seems the equation

[math]

4x^2 + \frac{1}{x^2}

[/math]

is never going to cross the x-axis. I see what I did wrong now, I forgot about the constant of integration earlier on in the question. I'll do that and see if i get any better results Thanks for your help, i realised my folly!

 

The constant turned out to be -4, so i'll rewrite the equation:

 

Solve:

[math]

4x^2 + \frac{1}{x^2} - 4 = 0

[/math]

 

So here, i can see that there would be values of x where y would be zero. So again, I'm wondering if it is mathematically acceptable to times by x² ie, change it to

[math]

4x^4 + 1 - 4x^2 = 0

[/math]

 

or would that be losing some of the solutions? sorry, I've just never had to solve things like this before.

[Cap'n Refsmmat]

When [math]x\neq 0[/math], I think that method works just fine. (When [math]x=0[/math], the first equation becomes undefined because of division by zero.)

[DrRocket]

wait a minute.....oh dear, it seems the equation

[math]4x^2 + \frac{1}{x^2}[/math]

is never going to cross the x-axis. I see what I did wrong now, I forgot about the constant of integration earlier on in the question. I'll do that and see if i get any better results Thanks for your help, i realised my folly!

 

The constant turned out to be -4, so i'll rewrite the equation:

 

Solve:

[math]4x^2 + \frac{1}{x^2} - 4 = 0[/math]

 

So here, i can see that there would be values of x where y would be zero. So again, I'm wondering if it is mathematically acceptable to times by x² ie, change it to

[math]4x^4 + 1 - 4x^2 = 0[/math]

 

or would that be losing some of the solutions? sorry, I've just never had to solve things like this before.

 

Since x=0 is not a zero of [math]4x^2 + \frac{1}{x^2} - 4 = 0[/math] or of [math]4x^4 + 1 - 4x^2 = 0[/math] you have not changed the zeros of the original rational function. In many cases multiplying by [math] x^2[/math]ould add a zero at [math] x=0[/math], but in this case you have simple eliminated the pole due to the

[math]\frac {1}{x^2}[/math] term.

 

But you have dramatically changed the general character.

 

[math]4x^2 + \frac{1}{x^2} - 4 = 0[/math] is very large near 0 because of the [math]\frac{1}{x^2} [/math] term and grows quadratically for large values of x, while [math]4x^4 + 1 - 4x^2 = 0[/math] is approximately 1 near zero and grows like a quartic..

 

Note that [math]4x^4 + 1 - 4x^2 = 0[/math] is quadratic in [math] x^2[/math] so you should be able to find the zero.

Edited October 22, 2011 by DrRocket