How do you find the zeros, real and imaginary, of $y = 8 x^{2} - 4 x - 11$ $y=8x^2-4x-11$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = 8 x^{2} - 4 x - 11$ $y=8x^2-4x-11$ using the quadratic formula?

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Answer 1 · 2016-02-19T12:24:33

$x = \frac{1 \pm \sqrt{23}}{4}$ $x = (1+-sqrt(23))/4$

Explanation:

For $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ :

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

Therefore $a = 8, b = - 4, c = - 11$ $a=8, b=-4, c=-11$ and:

$x = \frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot 8 \cdot - 11}}{2 \cdot 8}$ $x = (4+-sqrt((-4)^2-4*8*-11))/(2*8)$
$x = \frac{4 \pm \sqrt{16 + 352}}{16}$ $x = (4+-sqrt(16+352))/16$
$x = \frac{4 \pm \sqrt{368}}{16}$ $x = (4+-sqrt(368))/16$
$x = \frac{4 \pm 4 \sqrt{23}}{16}$ $x = (4+-4sqrt(23))/16$
$x = \frac{1 \pm \sqrt{23}}{4}$ $x = (1+-sqrt(23))/4$

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How do you find the zeros, real and imaginary, of $y = 8 x^{2} - 4 x - 11$ $y=8x^2-4x-11$ using the quadratic formula?

1 Answer

Explanation:

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How do you find the zeros, real and imaginary, of y=8x2−4x−11y=8x^2-4x-11 using the quadratic formula?

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Explanation:

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How do you find the zeros, real and imaginary, of $y = 8 x^{2} - 4 x - 11$ $y=8x^2-4x-11$ using the quadratic formula?