3x^2 kx-k=3x prove that this equation has rational roots for all ration values of K thanks
5 answers
You are missing a + or - sign in there.
Its 3x^2 kx-k=3x prove that this equation has rational roots for all rational values of K.
Sorry.
3x^2 kx-k
makes no sense. Is it
3x^2+kx-k
or
3x^2-kx-k
?
As it stands, since a b means a*b, you have
3x^2*kx-k = 3k
3kx^3-k = 3x
3kx^3-3x-k = 0
which clearly does not work.
3x^2 kx-k
makes no sense. Is it
3x^2+kx-k
or
3x^2-kx-k
?
As it stands, since a b means a*b, you have
3x^2*kx-k = 3k
3kx^3-k = 3x
3kx^3-3x-k = 0
which clearly does not work.
3x^2 kx-k=3x
It's 3x^2+kx-k=3k
1 answer