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
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.
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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
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!!