Question

Can someone show me how I can find the zeros of this equation?
x^4 - 8x^2 - 9 = 0

Answer 1

let x^2=a

so substitute a for x^2 (and for x^4 you would substitute a^2)

can you do it now? if not say so and I'll help you more.

Answer 2

Sorry, I still don't get it...
Can someone help me?

Answer 3

Yes. x^4 - 8x^2 - 9 = 0

(x^2)^2 - 8 (x^2) - 9 = 0 now we will substitute the "a"
a^2 - 8a - 9 = 0

Now, which method do you prefer ? completing the square OR quadratic formula ?

Answer 4

Oh, okay, I understand the substituting for "a" now. I'm not sure, I have only heard of the quadratic formula, so that I guess?

Answer 5

So the quad. form is as follows.
\[\huge\color{blue}{ \frac{-b±\sqrt{b^2-4ac} }{2a} } \]

now, lets plug in your values .
\[\huge\color{blue}{ \frac{-(-8)±\sqrt{(-8)^2-4(1)(-9)} }{2(1)} } \]
\[\huge\color{blue}{ \frac{+8±\sqrt{64-(-36)} }{2(1)} } \]
\[\huge\color{blue}{ \frac{+8±\sqrt{100} }{2(1)} } \]
\[\huge\color{blue}{ \frac{+8±10 }{2} } \]
\[\huge\color{blue}{ +4±5 } \]

so a = 9 OR -1


So far so good?

Answer 6

Also, just to put this there, that polynomial WAS factorable @feelgoodinc in case you feel better using that method.

But Zelman's method (quadratic formula) works all the time, so you'd do well to master this method too.

Answer 7

Hello ? This is NOT your final answer (unless you want to get it wrong) !!