How do you find the real zeros of $y = \frac{3}{4} {(x - \frac{2}{3})}^{3} - \frac{16}{9}$ $y=3/4(x-2/3)^3-16/9$ ?

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How do you find the real zeros of $y = \frac{3}{4} {(x - \frac{2}{3})}^{3} - \frac{16}{9}$ $y=3/4(x-2/3)^3-16/9$ ?

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Answer 1 · 2018-03-07T00:51:33

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Explanation:

You rewrite $\frac{3}{4} {(x - \frac{2}{3})}^{3} - \frac{16}{9} = 0$ $3/4(x-2/3)^3-16/9 = 0$ in the form :

$\frac{3}{4} {(x - \frac{2}{3})}^{3} = \frac{16}{9}$ $3/4(x-2/3)^3=16/9$

so that

${(x - \frac{2}{3})}^{3} = \frac{16}{9} \times \frac{4}{3} = {(\frac{4}{3})}^{3}$ $(x-2/3)^3=16/9 times 4/3 = (4/3)^3$

The only real solution to $x^{3} = a^{3}$ $x^3=a^3$ is $x = a$ $x=a$ (the other two are $x = a ω$ $x=a omega$ and $x = a ω^{2}$ $x=a omega^2$ , where $ω = exp (\frac{2 π}{3} i)$ $omega=exp({2pi}/3i)$ and $ω^{2}$ $omega^2$ are the complex cube roots of unity), so

$x - \frac{2}{3} = \frac{4}{3}$ $x-2/3 = 4/3$
and thus
$x = \frac{2}{3} + \frac{4}{3} = 2$ $x = 2/3+4/3=2$

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How do you find the real zeros of $y = \frac{3}{4} {(x - \frac{2}{3})}^{3} - \frac{16}{9}$ $y=3/4(x-2/3)^3-16/9$ ?

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Explanation:

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How do you find the real zeros of $y = \frac{3}{4} {(x - \frac{2}{3})}^{3} - \frac{16}{9}$ $y=3/4(x-2/3)^3-16/9$ ?