SOLUTION: I'm trying to figure out how to find "Complex Zeros of Quadratic Functions" but the Holt McDougal video lesson isn't helping whatsoever and I've researched as much as I could. I'm

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I'm trying to figure out how to find "Complex Zeros of Quadratic Functions" but the Holt McDougal video lesson isn't helping whatsoever and I've researched as much as I could. I'm       Log On


   

Question 937988: I'm trying to figure out how to find "Complex Zeros of Quadratic Functions" but the Holt McDougal video lesson isn't helping whatsoever and I've researched as much as I could. I'm trying to find the zeros of "f(x)=x²+2x+3" and the video says the answer is x=6+2i√3. This involves imaginary numbers.
Found 3 solutions by nerdybill, josgarithmetic, MathTherapy:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
By definition, "complex zeros" involves imaginary numbers.
.
f(x)=x²+2x+3
since we can't factor the right side we apply the "quadratic equation":
x = (-b +- sqrt(b^2 - 4ac))/(2a)
x = (-2 +- sqrt(2^2 - 4(1)(3)))/(2(1))
x = (-2 +- sqrt(4 - 12))/2
x = (-2 +- sqrt(-8))/2
x = (-2 +- sqrt(-2*2*3))/2
x = (-2 +- 2sqrt(-3))/2
x = (-2 +- 2isqrt(3))/2
x = -1 +- sqrt(3)i
.
there are two complex roots:
x = -1-sqrt(3)i
and
x = -1+sqrt(3)i

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
What happens if you put the function into a graph?

You can try the old slow way of making a data table for getting several points and then plotting them, and try to guess the ROOTS or ZEROS.

graph%28300%2C300%2C-6%2C6%2C-6%2C6%2Cx%5E2%2B2x%2B3%29

This parabola does not cross nor touch the x-axis, and therefore has NO REAL ZEROS.



But the function DOES have Complex zeros.
Accept a number, i such that:
highlight_green%28i%5E2=-1%29.
Raising this to the 1%2F2 power, you find that i=sqrt%28-1%29 or i=-sqrt%28-1%29.


NEXT, set f(x) to 0, and SOLVE FOR x, using Completing the Square.
x%5E2%2B2x%2B3=0
x%5E2%2B2x%2B%282%2F2%29%5E2%2B3=%282%2F2%29%5E2, the full expression shown so it will be less of a mystery
x%5E2%2B2x%2B1%2B3=1
%28x%5E2%2B2x%2B1%29%2B3=1
%28x%2B1%29%5E2%2B3=1
%28x%2B1%29%5E2=-2
x%2B1=0%2B-+sqrt%28-2%29
x=-1%2B-+sqrt%28-2%29
x=-1%2B-+sqrt%28-1%2A2%29
x=-1%2B-+sqrt%28-1%29sqrt%282%29
highlight%28highlight%28x=-1%2B-+i%2Asqrt%282%29%29%29

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

I'm trying to figure out how to find "Complex Zeros of Quadratic Functions" but the Holt McDougal video lesson isn't helping whatsoever and I've researched as much as I could. I'm trying to find the zeros of "f(x)=x²+2x+3" and the video says the answer is x=6+2i√3. This involves imaginary numbers.


f%28x%29+=+x%5E2+%2B+2x+%2B+3

Using the QUADRATIC EQUATION formula: x+=+%28-+b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29, and with:

a being 1; b being 2; and c being 3, x+=+%28-+b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29 becomes: 

x+=+%28-+2+%2B-+sqrt%28%282%29%5E2+-+4%281%29%283%29%29%29%2F%282%281%29%29

x+=+%28-+2+%2B-+sqrt%284+-+12%29%29%2F2

x+=+%28-+2+%2B-+sqrt%28-+8%29%29%2F2

x+=+%28-+2+%2B-+sqrt%28-+1+%2A+2%5E2+%2A+2%29%29%2F2

x+=+%28-+2+%2B-+2i+%2A+sqrt%282%29%29%2F2 ------ sqrt%28-+1%29+=+i and sqrt%282%5E2%29+=+2

x+=+%282%28-+1+%2B-+i%2Asqrt%282%29%29%29%2F2

x+=+%28cross%282%29%28-+1+%2B-+i%2Asqrt%282%29%29%29%2Fcross%282%29

highlight_green%28x+=+-+1+%2B-+sqrt%282%29%2Ai%29

This is the correct answer, but is nowhere close to what you claim the answer is. You need to re-check the problem!!