√x - 6 = x^(1/4)
let √x = t
then we have
t - 6 = √t
square both sides
t^2 - 12t + 36 = t
t^2 - 13t + 36 = 0
(t-9)(t-3) = 0
t = 9 or t = 3
if t = 9, then x = 81
if t = 3, then x = 9
since we squared our equation, all answers must be verifies
if x = 81
LS = √81 - 6 = 9-6 = 3
RS = 81^(1/4) = 3
so x = 81 is valid
if x = 9
LS = √9 - 6 = 3-6 = -3
RS = 9^1/4) ≠ -3
so the only solution is x = 81
x^1/2-6=x^1/4
please explain steps to solving!
2 answers
if we let t = ∜x, we have
t^2-6 = t
t^2-t-6 = 0
(t-3)(t+2) = 0
t = 3 or -2
so, x = 81 or 16
But, since ∜x is positive, only x=81 satisfies the original equation.
So, the extraneous solution can arise even when the original equation is not squared. Trying to use x=16 gives us
√16 - 6 = ∜16
But
√16 - 6 = 4-6 = -2
and -2 is not ∜16
t^2-6 = t
t^2-t-6 = 0
(t-3)(t+2) = 0
t = 3 or -2
so, x = 81 or 16
But, since ∜x is positive, only x=81 satisfies the original equation.
So, the extraneous solution can arise even when the original equation is not squared. Trying to use x=16 gives us
√16 - 6 = ∜16
But
√16 - 6 = 4-6 = -2
and -2 is not ∜16