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#