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

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Question 176614: Find a quadratic equation with roots (4+i) and (4-i)
Found 3 solutions by solver91311, josmiceli, Earlsdon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
If x+=+4+%2B+i then x+-+%284+%2B+i%29 is a factor of the quadratic. Similarly, x+-+%284+-+i%29 is a factor of the quadratic and we can say:

%28x+-+%284+%2B+i%29%29%28x+-+%284+-+i%29%29+=+0.

Just multiply the two factors to get back to the original quadratic. Hints: Treat these two factors as binomials with x being one term and 4+%2B-+i as the other term. Remember that multiplying a conjugate pair results in the difference of two squares, that is: %28a+%2B+b%29%28a+-+b%29+=+a%5E2+-+b%5E2. Therefore %284+%2B+i%29%284+-+i%29+=+16+-+%28-1%29+=+17

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I can write
%28x+-+%284%2Bi%29%29%28x+-+%284-i%29%29+=+0
This equation is true if either x+=+4+%2B+i
or x+=+4+-+i
The function y(x) is
y+=+%28x+-+%284%2Bi%29%29%28x+-+%284-i%29%29
y+=+x%5E2+-%284%2Bi%29x+-+%284-i%29x+%2B+%284%2Bi%29%284-i%29
y+=+x%5E2+-+4x+-+ix+-+4x+%2B+ix+%2B+16+%2B+1 note that i%5E2+=+-1
y+=+x%5E2+-+8x+%2B+17
check:
0+=+%284-i%29%5E2+-+8%2A%284-i%29+%2B+17
0+=+16+-+8i+-+1+-+32+%2B+8i+%2B+17
0+=+0
OK

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the quadratic equation whose roots are: %284%2Bi%29 and %284-i%29
Basically, we'll work in reverse of solving the quadratic equation, starting with:
x+=+%284%2Bi%29 and x+=+%284-i%29 rewrite these as:
x-%284%2Bi%29+=+0 and x-%284-i%29+=+0...and we know that these are factors of the original quadratic equation, so we can reconstitute the equation by multiplying these factors. First, simplify the factors.
%28x-%284%2Bi%29%29%2A%28x-%284-i%29%29+=+%28x-4-i%29%2A%28x-4%2Bi%29 Now we'll perform the multiplication.
%28x-4-i%29%2A%28x-4%2Bi%29+=+x%5E2-4x%2Bxi-4x%2B16-4i-xi%2B4i-i%5E2 Now we'll simplify this.
x%5E2-8x%2B16-i%5E2+=+x%5E2-8x%2B17
The original quadratic equation is:
highlight%28x%5E2-8x%2B17+=+0%29
Let's check using the quadratic formula:x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x+=+-%28-8%29%2B-sqrt%28%28-8%29%5E2-4%281%29%2817%29%29%29%2F2%281%29
x+=+%288%2B-sqrt%2864-68%29%29%2F2
x+=+%288%2B-sqrt%28-4%29%29%2F2
x+=+%288%2F2%29%2B2i%2F2 or x+=+%288%2F2%29-2i%2F2 Simplifying, we get:
highlight%28x+=+4%2Bi%29 or highlight%28x+=+4-i%29