How do you find the zeros, real and imaginary, of $y = - x^{2} - 22 x + 6$ $y=- x^2-22x+6$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 22 x + 6$ $y=- x^2-22x+6$ using the quadratic formula?

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Answer 1 · 2016-04-10T10:59:06

Zeros are $- 11 + \sqrt{127}$ $-11+sqrt127$ and $- 11 - \sqrt{127}$ $-11-sqrt127$

Explanation:

The roots of general form of equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ are zeros of the general form of equation $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ .

In the given equation $y = - x^{2} - 22 x + 6$ $y=-x^2-22x+6$ , we see that $a = - 1$ $a=-1$ , $b = - 22$ $b=-22$ and $c = 6$ $c=6$ .

As discriminant $b^{2} - 4 a c = {(- 22)}^{2} - 4 (- 1) (6) = 484 + 24 = 508$ $b^2-4ac=(-22)^2-4(-1)(6)=484+24=508$

As discriminant is positive but not a complete square, we have real but irrational roots.

Hence, using quadratic formula $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$ ,

the zeros are the given equation are $x = \frac{- (- 22) \pm \sqrt{508}}{2 \cdot (- 1)}$ $x=(-(-22)+-sqrt508)/(2*(-1))$

or $x = - \frac{22 \pm \sqrt{508}}{2} = - \frac{22 \pm 2 \sqrt{127}}{2}$ $x=-(22+-sqrt508)/2=-(22+-2sqrt127)/2$ i.e.

zeros are $- 11 + \sqrt{127}$ $-11+sqrt127$ and $- 11 - \sqrt{127}$ $-11-sqrt127$

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 22 x + 6$ $y=- x^2-22x+6$ using the quadratic formula?

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Explanation:

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Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y=−x2−22x+6y=- x^2-22x+6 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 $y = - x^{2} - 22 x + 6$ $y=- x^2-22x+6$ using the quadratic formula?