How do you find the zeros, real and imaginary, of $y= 3x^2-6x+2$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y= 3x^2-6x+2$ using the quadratic formula?

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Answer 1 · 2016-04-04T05:20:23

$1 +- sqrt3/3$

Explanation:

Use the improved quadratic formula (Google, Yahoo Search)
$D = d^2 = b^2 - 4ac = 36 - 24 = 12 --> d = +- 2sqrt3$
There are 2 real roots:
$x = -b/(2a) +- d/(2a) = 6/6 +- (2sqrt3)/6 = 1 +- sqrt3/3$

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How do you find the zeros, real and imaginary, of $y= 3x^2-6x+2$ using the quadratic formula?

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Humanities

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How do you find the zeros, real and imaginary, of y= 3x^2-6x+2 y= 3x^2-6x+2 using the quadratic formula?

1 Answer

Explanation:

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How do you find the zeros, real and imaginary, of $y= 3x^2-6x+2$ using the quadratic formula?