How do you find the zeros, real and imaginary, of $y = x^{2} - x + 81$ $y=x^2 -x+81$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = x^{2} - x + 81$ $y=x^2 -x+81$ using the quadratic formula?

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Answer 1 · 2018-06-26T05:46:16

The quadratic formula for a function of the form

$y = a x^{2} + b x + c$ $y=ax^2+bx+c$

is

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

In your case a=1, b=-1 and c=81

Substitute.

$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (81)}}{2 (1)}$ $x=(-(-1)+-sqrt((-1)^2-4(1)(81)))/(2(1))$

Simplify

$x = \frac{1 \pm \sqrt{- 323}}{2} \to \frac{1 \pm \sqrt{323 i^{2}}}{2} \to \frac{1 \pm i \sqrt{323}}{2}$ $x=(1+-sqrt(-323))/2 rarr (1+-sqrt(323i^2))/2 rarr (1+-isqrt(323))/2$

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How do you find the zeros, real and imaginary, of $y = x^{2} - x + 81$ $y=x^2 -x+81$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = x^{2} - x + 81$ $y=x^2 -x+81$ using the quadratic formula?