How do you find the zeros, real and imaginary, of $y = - x^{2} - 2 x - 4$ $y= -x^2-2x-4$ using the quadratic formula?

Question

How do you find the zeros, real and imaginary, of $y = - x^{2} - 2 x - 4$ $y= -x^2-2x-4$ using the quadratic formula?

1 Answer

Impact of this question

1572 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-08T02:21:03

$x = = 1 + \sqrt{3} i$ $x==1+sqrt3i$
$x = - 1 - \sqrt{3} i$ $x=-1-sqrt3i$

Explanation:

Identify the values of $a, b & c$ $a,b & c$ and substitute in using the quadratic formula: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b \pm sqrt(b^2-4ac)) / (2a)$

$a = - 1$ $a=-1$

$b = - 2$ $b=-2$

$c = - 4$ $c=-4$

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

$x = \frac{2 \pm \sqrt{4 - 16}}{- 2}$ $x = (2 \pm sqrt(4-16)) / (-2)$

$x = \frac{2 \pm \sqrt{- 12}}{- 2}$ $x = (2 \pm sqrt(-12)) / (-2)$

$x = \frac{2 \pm 2 \sqrt{3} i}{- 2}$ $x = (2 \pm 2 sqrt(3)i) / (-2)$

$x = - 1 \pm \sqrt{3} i$ $x = -1 \pm sqrt(3)i$

$x = 1 + \sqrt{3} i$ $x=1+sqrt3i$ , $x = - 1 - \sqrt{3} i \leftarrow$ $x=-1-sqrt3ilarr$ Final solutions

Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of $y = - x^{2} - 2 x - 4$ $y= -x^2-2x-4$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y= -x^2-2x-4y=−x2−2x−4y= -x^2-2x-4 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the zeros, real and imaginary, of $y = - x^{2} - 2 x - 4$ $y= -x^2-2x-4$ using the quadratic formula?