How do you find the zeros, real and imaginary, of $y = - x^{2} - x + 5$ $y=-x^2- x +5$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = - x^{2} - x + 5$ $y=-x^2- x +5$ using the quadratic formula?

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Answer 1 · 2016-06-10T13:09:44

Zeros of $y = - x^{2} - x + 5$ $y=-x^2-x+5$ are

$\frac{- 1 - \sqrt{21}}{2}$ $(-1-sqrt21)/2$ and $\frac{- 1 + \sqrt{21}}{2}$ $(-1+sqrt21)/2$

Explanation:

Quadratic formula gives the roots or zeros of general form of quadratic equation $a x^{2} + b x + c$ $ax^2+bx+c$ as $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

Hence zeros of $y = - x^{2} - x + 5$ $y=-x^2-x+5$ are

$\frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \times (- 1) \times 5}}{2 \cdot (- 1)}$ $(-(-1)+-sqrt((-1)^2-4xx(-1)xx5))/(2*(-1))$ or

$\frac{1 \pm \sqrt{1 + 20}}{- 2}$ $(1+-sqrt(1+20))/(-2)$ or

$\frac{- 1 \pm \sqrt{21}}{2}$ $(-1+-sqrt21)/2$ or

i.e. $\frac{- 1 - \sqrt{21}}{2}$ $(-1-sqrt21)/2$ and $\frac{- 1 + \sqrt{21}}{2}$ $(-1+sqrt21)/2$

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How do you find the zeros, real and imaginary, of $y = - x^{2} - x + 5$ $y=-x^2- x +5$ using the quadratic formula?

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Explanation:

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How do you find the zeros, real and imaginary, of $y = - x^{2} - x + 5$ $y=-x^2- x +5$ using the quadratic formula?