Question

Find a quadratic equation with roots -1+4i and -1-4i

Answer 1

Algebra 2 please help

Answer 2

I got the quadratic equation to be \[x ^{2}+2x-15\]

Answer 3

Let me check it, then I'll explain it, ok?

Answer 4

Ok thank you!!

Answer 5

Ok, I believe that that is correct; it's been a long time since I taught this!!! The roots that we are considering are x = -1+4i and x = -1-4i. But in order to go backwards from that to the original quadratic, we have to have (x +/- something)(x +/- something) to get an x^2 term, cuz quadratics ARE x^2 terms. So if x = -1+4i, then the root would be (x + 1 + 4i). And the other root would be (x + 1 - 4i). So here's what we have:
\[(x+1+4i)(x+1-4i)\]

Answer 6

Now we FOIL that out, starting with the first x in the first set of parenthesis:
\[x ^{2}+x-4ix\]

Answer 7

Now let's move to the 1 in the first set of parenthesis:
\[x+1-4i\]

Answer 8

@IMStuck hi! when your done please come back to my question :)

Answer 9

Now let's move to the 4i in the first set of parenthesis:
\[4ix+4i-16i ^{2}\]

Answer 10

Now let's put it all together:
\[x ^{2}+x-4ix+x+1-4i+4ix+4i-16i ^{2}\]

Answer 11

Combining like terms and canceling out the ones that eliminate each other, we are left with:
\[x ^{2}+2x-16i ^{2}+1\]

Answer 12

The thing you need to remember here is that i^2 = -1:
\[x ^{2}+2x-16(-1)+1=x ^{2}+2x+16+1=x ^{2}+2x+17\]

Answer 13

I did my negatives incorrectly when I gave you my initial equation. A 2x -(-16) is 2x + 16. My bad.

Answer 14

So the equation, as far as I can possibly remember, is \[x ^{2}+2x+17\]

Answer 15

I am going to check that now using the quadratic formula...will return in just a moment!

Answer 16

Yes, that is correct. Good luck with those! They are no different than FOILing out regular roots.

Answer 17

Another way of looking at it.

\(x=-1\pm4i\implies x+1=\pm4i\implies (x+1)^2=(4i)^2\)

Which gives \(x^2+2x+17=0\).