How do you find the roots of $x^3-18x+27=0$ ?

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How do you find the roots of $x^3-18x+27=0$ ?

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Answer 1 · 2016-10-31T18:19:57

The roots are $(3, -3/2+3sqrt5/2, -3/2-3sqrt5/2)$

Explanation:

Let $f(x)=x^3-18x+27$
Then by trial and error, $f(3)=27-54+27=0$
so $x=3$ is a root
Then we do a long division
$x^3$ $color(white)(aaaaa)$ $-18x+27$ $∣$ $x-3$
$x^3-3x^2$ $color(white)(aaaaaaaaa)$ $∣$ $x^2+3x-9$
$0+3x^2-18x$
$color(white)(aaaa)$ $0-9x$
$color(white)(aaaaaa)$ $-9x+27$
$color(white)(aaaaaa)$ $-9x+27$
$color(white)(aaaaaaaa)$ $0+0$
To find the other roots, we must solve
$x^x+3x-9=x^2+3x+9/4-9-9/4$
$(x+3/2)^2-45/4$
So $(x+3/2)^2=45/4$ $=>$ $x+3/2=+-sqrt45/2$
$x=-3/2+-3sqrt5/2$

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How do you find the roots of $x^3-18x+27=0$ ?

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How do you find the roots of x^3-18x+27=0x^3-18x+27=0?

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How do you find the roots of $x^3-18x+27=0$ ?