How do you find the zeros, real and imaginary, of $y = 9 x^{2} - 8 x - 4$ $y=9x^2-8x-4$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = 9 x^{2} - 8 x - 4$ $y=9x^2-8x-4$ using the quadratic formula?

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Answer 1 · 2017-08-21T01:20:22

See a solution process below:

Explanation:

The quadratic formula states:

For $a x^{2} + b x + c = 0$ $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$9$ $color(red)(9)$ for $a$ $color(red)(a)$

$- 8$ $color(blue)(-8)$ for $b$ $color(blue)(b)$

$- 4$ $color(green)(-4)$ for $c$ $color(green)(c)$ gives:

$x = \frac{- (- 8) \pm \sqrt{{(- 8)}^{2} - (4 \cdot 9 \cdot - 4)}}{2 \cdot 9}$ $x = (-color(blue)((-8)) +- sqrt(color(blue)((-8))^2 - (4 * color(red)(9) * color(green)(-4))))/(2 * color(red)(9))$

$x = \frac{8 \pm \sqrt{64 - (- 144)}}{18}$ $x = (8 +- sqrt(64 - (-144)))/18$

$x = \frac{8 \pm \sqrt{64 + (144)}}{18}$ $x = (8 +- sqrt(64 + (144)))/18$

$x = \frac{8 \pm \sqrt{208}}{18}$ $x = (8 +- sqrt(208))/18$

$x = \frac{8 \pm \sqrt{16 \cdot 13}}{18}$ $x = (8 +- sqrt(16 * 13))/18$

$x = \frac{8 \pm \sqrt{16} \sqrt{13}}{18}$ $x = (8 +- sqrt(16)sqrt(13))/18$

$x = \frac{8 \pm 4 \sqrt{13}}{18}$ $x = (8 +- 4sqrt(13))/18$

$x = \frac{8 - 4 \sqrt{13}}{18}$ $x = (8 - 4sqrt(13))/18$ and $x = \frac{8 + 4 \sqrt{13}}{18}$ $x = (8 + 4sqrt(13))/18$

$x = \frac{4 - 2 \sqrt{13}}{9}$ $x = (4 - 2sqrt(13))/9$ and $x = \frac{4 + 2 \sqrt{13}}{9}$ $x = (4 + 2sqrt(13))/9$

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How do you find the zeros, real and imaginary, of $y = 9 x^{2} - 8 x - 4$ $y=9x^2-8x-4$ using the quadratic formula?

1 Answer

Explanation:

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How do you find the zeros, real and imaginary, of y=9x2−8x−4y=9x^2-8x-4 using the quadratic formula?

1 Answer

Explanation:

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How do you find the zeros, real and imaginary, of $y = 9 x^{2} - 8 x - 4$ $y=9x^2-8x-4$ using the quadratic formula?