Question

what's wrong with this: 4x³-2x=0 <=> 4x³=2x <=> 4x²=2 <=> x²=1/2 <=> x=sqrt(1/2) ?

Answer 1

forgot +/- from the last step

Answer 2

well, how do you know x is not 0?

Answer 3

the assignment would say x=1/sqrt(2)

Answer 4

the equivalence from step 2 to step 3 holds only if x is not 0

Answer 5

forgot +/- again

Answer 6

x is not 0

Answer 7

well, lol, sqrt(1/2) = 1/sqrt(2)

Answer 8

hmm

Answer 9

because \[\frac{1}{\sqrt(2)} * \frac{1}{\sqrt(2)} = \frac{1}{2}\]

Answer 10

thanks @Stiwan

Answer 11

\[4x ^{3}-2x=0,4x \left( x ^{2}-\frac{ 1 }{ 2 } \right)=0\]
now you can get all the three solutions.

Answer 12

When you divide both sides by x, you lose the root at zero. You are better off factoring.\[4x³-2x=0 \implies 2x^3-x=0 \implies x(2x^2-1)=x(\sqrt2x+1)(\sqrt2x-1)=0\]\[\implies x=0~or~x= \pm \frac{\sqrt2}{2}\]