How do you solve #-x ^ { 4} + 3x ^ { 3} - 6x ^ { 2} = 0#?

1 Answer
Jim G.
Sep 27, 2017

#x=0" multiplicity 2 or "x=3/2+-15/2i#

Explanation:

#"multiply through by "-1#

#rArrx^4-3x^3+6x^2=0#

#"factor out "x^2#

#rArrx^2(x^2-3x+6)=0#

#"equate each factor to zero and solve for x"#

#x^2=0rArrx=0" (multiplicity 2)"larr" 2 real roots"#

#x^2-3x+6=0toa=1,b=-3,c=6#

#"check the value of the "color(blue)"discriminant"#

#Delta=b^2-4ac=(-3)^2-24=-15#

#Delta<0rArr" there are complex roots"#

#"solve using the "color(blue)"quadratic formula"#

#x=(3+-sqrt(-15))/2=3/2+-15/2i#

#rArrx=3/2+-15/2ilarr" 2 complex conjugate roots"#

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