Question

Find all the zeros of the equation.

-4x4 - 44x2 + 3600 = 0

A. 5, -5, 6i, -6i
B. 5, 6i
C. 5, -5, 6i, 0
D. -5, -6i

Answer 1

The highest exponent is 4, so we know there must be 4 zeros. That eliminates B and D. We also know that complex roots must come in complex conjugate pairs in a polynomial with only real coefficients. That rules out another choice.

In case you've forgotten, a complex conjugate pair is \(a \pm bi\) where \(i=\sqrt{-1} \)

Answer 2

You could also factor out a 4, then do a substitution \(u = x^2\) and solve the resulting quadratic by your favorite means. Undo the substitution to give two quadratics, solve them, and find your zeros.

Answer 3

it is C.

Answer 4

No!

Answer 5

A then..

Answer 6

Come on, reread what I said in my first post...no guessing required.

Answer 7

ill never get this

Answer 8

Sure you will. It's nowhere near as complicated as it might seem.

You always have as many zeros as the the exponent of the highest power of the variable. We have \(x^4\) here, so we must have 4 zeros. Some of them may be identical, but they get counted individually.

Okay so far?

Answer 9

okay.

Answer 10

Now, we have a polynomial with only real numbers as coefficients — \(i\) doesn't appear anywhere, does it?

Answer 11

We don't have anything like \[P(x) = 3x^2 + (2+i)x + 7\]

Answer 12

right

Answer 13

Okay, now let's look at what happens when you multiply \((a+bi)(a-bi)\)
\[(a+bi)(a-bi) = a^2 -abi + abi -b^2i^2=a^2-b^2i^2\]But \[i=\sqrt{-1} \rightarrow i^2=-1\]so we can remove \(i^2\) and replace it with \(-1\) giving us\[a^2-b^2(-1) = a^2+b^2\]
Notice that there's no \(i\) anywhere to be seen!

Answer 14

but where did u get 6i from?

Answer 15

Now let's try multiplying something without a conjugate pair:
\[(a+1)(a+i) = a^2+ai + 1a + 1i\]Nothing we can do here will make that pesky \(i\) go away. The bottom line is that if we have a complex root, we must also have its conjugate or we have to have an imaginary number in our coefficients. Do you see that?

Answer 16

the only way we can completely get rid of a term with \(i\) in it is by multiplying by the same term with an opposite sign. That means our complex roots (if we have any) come in conjugate pairs.

I got the 6i from the answers — though I can easily find them on my own in this case. But we already had two real zeros, and there are 4 zeros altogether, so if the answer choices proposed that one of the two remaining zeros was a complex number, I had to then choose the answer choice that had its conjugate pair. Does that make sense?

Answer 17

Let's see what happens if we choose the one with 0 instead. Because the zeros of a polynomial are just the spots where it equals 0, and you can write the polynomial as the product of \((x-r_1)(x-r_2)...(x-r_n)\) where \(r_1,r_2,...r_n\) are the zeros or roots of the polynomial, we can construct the polynomial from the list of zeros:
\[(x-5)(x+5)(x-6i)(x-0) = 150 i x-25 x^2-6 i x^3+x^4\]

Well, that doesn't look anything like our polynomial, does it? It has two imaginary coefficients, and the other numbers are wrong, too.

However, if we instead expanded\[(x-5)(x+5)(x-6i)(x+6i) \]we get\[-900+11 x^2+x^4\]which is our polynomial divided by 4. Because it = 0, we can divide off that factor of 4 without changing the zeros.

Answer 18

The takeaways here are
1) always have as many roots/zeros as the highest exponent
2) complex roots come in conjugate pairs
3) a little bit of thinking can save doing a lot of algebra!

Answer 19

oh okay thank you!! I got it now

Answer 20

Good!

Answer 21

it's sort of like doing little logic puzzles sometimes.