another finding zeros problem

[PaulKraemer]

Hi,

I just asked a similar question, but I am having trouble finding the zeros on another function:

f(x) = x^4 - 8x^2 + 1

I used the rational zero theorem to find that the possible zeros should be +/- 1.

I tried using synthetic division to test both +1 and -1 to see if either of these really are zeros, and I got a remainder in both cases.

If anyone could tell me what I'm missing, I'd really appreciate it.

Thanks in advance,
Paul

 

[galactus]

\(\displaystyle x^{4}-8x^{2}+1=0\)

Let \(\displaystyle u=x^{2}\)

\(\displaystyle u^{2}-8u+1=0\).

Per the quad formula:

\(\displaystyle u=-(\sqrt{15}-4), \;\ u=\sqrt{15}+4\)

Thus, upon resubbing, \(\displaystyle x=\pm \sqrt{-(\sqrt{15}-4)}, \;\ x=\pm\sqrt{\sqrt{15}+4}\)

 

[Mrspi]

Hi,

I just asked a similar question, but I am having trouble finding the zeros on another function:

f(x) = x^4 - 8x^2 + 1

I used the rational zero theorem to find that the possible zeros should be +/- 1.

I tried using synthetic division to test both +1 and -1 to see if either of these really are zeros, and I got a remainder in both cases.

If anyone could tell me what I'm missing, I'd really appreciate it.

Thanks in advance,
Paul

This function doesn't have any RATIONAL zeroes, but it does have four real zeros. If you use a graphing calculator to graph this function, you should see that this is true. Now, to FIND those values for x which will make the function have a value of 0, you might try this:

We want to know when f(x) = 0, or, when
0 = x[sup:v0fp1keq]4[/sup:v0fp1keq] - 8 x[sup:v0fp1keq]2[/sup:v0fp1keq] + 1
let u = x[sup:v0fp1keq]2[/sup:v0fp1keq]

Now substitute.....

0 = u[sup:v0fp1keq]2[/sup:v0fp1keq] - 8u + 1

Do you see a quadratic equation here? You can use the quadratic formula to find the values of u which satisfy this equation. Then, remember that you're really looking for x, and that

u = x[sup:v0fp1keq]2[/sup:v0fp1keq]

One value you'll get for u is (4 - sqrt 15), which is approximately 0.1207167. So,

0.1207167 = x[sup:v0fp1keq]2[/sup:v0fp1keq]
Take the square root of both sides and you'll have two of the values of x which will make f(x) = 0.

You'll also have another value for u which will produce the remaining two values for x.

I hope this helps you.

OOPS...I see Galactus beat me to it.