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.Com
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)   (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(39616)   (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.



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:
.
Raising this to the power, you find that or .


NEXT, set f(x) to 0, and SOLVE FOR x, using Completing the Square.

, the full expression shown so it will be less of a mystery
Answer by MathTherapy(10551)   (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.




Using the QUADRATIC EQUATION formula: , and with:

a being 1; b being 2; and c being 3,  becomes: 









 ------  and 







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

RELATED QUESTIONS


I am trying to figure out how to factor the difference of two cubes of 
n^6 - 343=... (answered by Earlsdon,scott8148)

How do i find the complex zeros of quadratic Functions, such as this one.... (answered by ReadingBoosters)

I am trying to help my child with her homework.  So, if you can explain it to me, maybe I  (answered by drk)

Please help me with the following problem, the directions read:

Graph the equation.... (answered by scott8148)

what is the answer to #30 & 40 on page 186- Holt mcdougal larson Algebra1... (answered by ReadingBoosters)

Two verticies of a rectangle are on the x-axis and the other two vertices on the curve... (answered by ikleyn)

I've been trying and trying to find the answer to this. Please help

Match the... (answered by Alan3354)

Please help me figure out how to find the complex zeros of the equation:... (answered by Fombitz)

If m<1 = x.  find:

M<2
M<3
M<4

Trying to figure out how to find the measurement... (answered by jim_thompson5910)