Find a quadratic equation with roots (4+i) and (4-i).

Question

anonymous · Answer

@terenzreignz

terenzreignz · Answer

Whenever you need to find an equation with roots r and s, simply do this:
$$\Large f(x) = (x-r)(x-s)$$
where r = 4+i
and s = 4-i

anonymous · Answer

okay I'm stuck at the point where I have to multiply... [x-(4+i)][x-(4-i)].... I don't know how to use the foil method for that .

terenzreignz · Answer

Just foil (x-r) and (x-s) first, and the replace afterwards.

anonymous · Answer

okay hold on

anonymous · Answer

okay I get $$x^2-8x+16-i^2$$

terenzreignz · Answer

Yup... you're almost done ^_^

anonymous · Answer

$$-i^2 $$  ^^^ is equal to 1...?

terenzreignz · Answer

That's correct :D

anonymous · Answer

so then the equation is $$x^2 -8x+17=0$$

terenzreignz · Answer

Bravo ^_^

anonymous · Answer

okay thank you! it was really that foiling part that had me stuck, cause I knew the method for finding equations given the roots. Didn't realize I could just replace the (x-r)(x-s) with the correct terms... so thanks for pointing that out!

terenzreignz · Answer

No problem.
:)

anonymous · Answer

ax^2 + bx + c = 0 with 2 complex roots: (4 + i) and (4 - i)
Product of  2 roots x1*x2 = (4 + i)(4 - i) = 16 + 1 = 17 = ac
Suppose a = 1 and c = 17.
Sum of 2 roots: (4 + i) + (4 -i) = 8 = -b. Then b = -8.
Back to the equation: x^2 - 8x + 17 = 0
D = b^2 - 4ac = -4 = 4i^2.
The 2 roots are:(4 + i) and (4 -i). Correct.