How do you find the zeros, real and imaginary, of $y = - x^{2} - 22 x + 67$ $y=- x^2-22x+67$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 22 x + 67$ $y=- x^2-22x+67$ using the quadratic formula?

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Answer 1 · 2016-04-28T09:32:14

Zeros of $y = - x^{2} - 22 x + 47$ $y=-x^2-22x+47$ are $- 11 + 2 \sqrt{67}$ $-11+2sqrt67$ and $- 11 - 2 \sqrt{47}$ $-11-2sqrt47$

Explanation:

To find zeros of $y = - x^{2} - 22 x + 67$ $y=-x^2-22x+67$ , one needs to find values of $x$ $x$ for which $y = f (x) = 0$ $y=f(x)=0$ .

For $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , solution is given by quadratic formula $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$ . Hence for $- x^{2} - 22 x + 67 = 0$ $-x^2-22x+67=0$ or $x^{2} + 22 x - 67 = 0$ $x^2+22x-67=0$ ,

$x = \frac{- 22 \pm \sqrt{22^{2} - 4 \times 1 \times (- 67)}}{2}$ $x=(-22+-sqrt(22^2-4xx1xx(-67)))/2$

= $\frac{- 22 \pm \sqrt{484 + 268}}{2}$ $(-22+-sqrt(484+268))/2$

= $\frac{- 22 \pm \sqrt{752}}{2}$ $(-22+-sqrt(752))/2$

= $\frac{- 22 \pm 4 \sqrt{47}}{2}$ $(-22+-4sqrt47)/2$

= $- 11 \pm 2 \sqrt{47}$ $-11+-2sqrt47$

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 22 x + 67$ $y=- x^2-22x+67$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y=−x2−22x+67y=- x^2-22x+67 using the quadratic formula?

1 Answer

Explanation:

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Impact of this question

How do you find the zeros, real and imaginary, of $y = - x^{2} - 22 x + 67$ $y=- x^2-22x+67$ using the quadratic formula?