How do you find the zeros, real and imaginary, of $y = x^{2} - 31 x - 9$ $y=x^2-31x-9$ using the quadratic formula?

Question

How do you find the zeros, real and imaginary, of $y = x^{2} - 31 x - 9$ $y=x^2-31x-9$ using the quadratic formula?

1 Answer

Impact of this question

1458 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-09T07:14:57

$x_{1} = \frac{+ 31 + \sqrt{(961 + 36)}}{2} = 31.2877$ $x_1=(+31+sqrt((961+36)))/(2)=31.2877$
and $x_{2} = \frac{+ 31 - \sqrt{(961 + 36)}}{2} = - 0.287653$ $x_2=(+31-sqrt((961+36)))/(2)=-0.287653$

Explanation:

from the given: $y = x^{2} - 31 x - 9$ $y=x^2-31x-9$

$a = 1$ $a=1$ , $b = - 31$ $b=-31$ , $c = - 9$ $c=-9$

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

$x = \frac{- - 31 \pm \sqrt{{(- 31)}^{2} - 4 (1) (- 9)}}{2 \cdot 1}$ $x=(--31+-sqrt((-31)^2-4(1)(-9)))/(2*1)$

$x = \frac{+ 31 \pm \sqrt{(961 + 36)}}{2}$ $x=(+31+-sqrt((961+36)))/(2)$

$x_{1} = \frac{+ 31 + \sqrt{(961 + 36)}}{2} = 31.2877$ $x_1=(+31+sqrt((961+36)))/(2)=31.2877$
and $x_{2} = \frac{+ 31 - \sqrt{(961 + 36)}}{2} = - 0.287653$ $x_2=(+31-sqrt((961+36)))/(2)=-0.287653$

have a nice day from the Philippines!

Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of $y = x^{2} - 31 x - 9$ $y=x^2-31x-9$ 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=x2−31x−9y=x^2-31x-9 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the zeros, real and imaginary, of $y = x^{2} - 31 x - 9$ $y=x^2-31x-9$ using the quadratic formula?