How do you solve #x^3=4x#?

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Answer 1 · 2016-06-09T10:33:15

#x = 0, " or "x= -2," or " x= 2#

the tempting method is just to divide by #x# which gives
#x^2 = 4#

However in dividing by #x# we have made an assumption that #x# is not 0, in which case it would not be allowed.

Make the equation = 0: #" " x^3 - 4x = 0#

Factorise fully: #" " x(x^2 -4) = 0#

So, #x(x + 2)(x - 2) = 0#

Each factor can be equal to 0, so
#x = 0, " or "x= -2," or " x= 2#

An equation with #x^3# should have 3 solutions, which we have.