this part is correct:
<< a^2 - 7a + 12 = 0
(a-4)(a-3) = 0
a = 3
2 = 4 >>
as for x2/3 - 7x1/3 + 12 = 0
I don't know what you mean
is it
(x^2)/3 - 7x/3 + 12 = 0 ??
Solve:
x2/3 - 7x1/3 + 12 = 0
a^2 - 7a + 12 = 0
(a-4)(a-3) = 0
a = 3
2 = 4
x = 27
x = 64
Is this correct?
3 answers
x^2/3 - 7x^1/3 + 12 = 0
the 2/3 and 1/3 are fractions
does this help?
the 2/3 and 1/3 are fractions
does this help?
ahh, ok,
how about let y = x^(1/3)
then your equation is
y^2 - 7y + 12 = 0
(y-3)(y-4) = 0
y = 3 or y = 4
then x^(1/3) = 3, x = 27
x^(1/3) = 4, then x = 64
so you were right
how about let y = x^(1/3)
then your equation is
y^2 - 7y + 12 = 0
(y-3)(y-4) = 0
y = 3 or y = 4
then x^(1/3) = 3, x = 27
x^(1/3) = 4, then x = 64
so you were right