How do you solve $|x^2 - 3x| = 18$ ?

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Answer 1 · 2016-04-04T10:00:25

Only real roots are $x=6$ and $x=-3$ .

Complex roots are $x=3/2+isqrt(63)/2$ and $x=3/2-isqrt(63)/2$

As $|x^2-3x|=18$ , we have either $x^2-3x=18$ i.e. $x^2-3x-18=0$ or

$x^2-3x=-18$ i.e. $x^2-3x+18=0$

For solving $x^2-3x-18=0$ by factorizing, we split middle term as

$x^2+3x-6x-18=0$ or $x(x+3)-6(x+3)=0$ or $(x-6)(x+3)=0$ i.e. $x=6$ or $x=-3$ .

As determinant given by $b^2-4ac$ for $x^2-3x+18=0$

is $(-3)^2-4*1*18=9-72=-63$ ,

Hence, we do not have any real roots of $x^2-3x+18=0$

Complex roots for this are given by $x=(3+-sqrt(-63))/2$ or

$x=3/2+isqrt(63)/2$ or $x=3/2-isqrt(63)/2$

How do you solve |x^2 - 3x| = 18|x^2 - 3x| = 18?