Find all Real solutions of the Equation:
(ENter as comma sparated answers, if there are no real solutions then just leave as NRS)
x^(4/3)-9x^(2/3)+14=0
x=?
set u = x^(2/3)
then you have
u^2-9u+14 = 0
(u-2)(u-7) = 0
so, that means
x^(2/3) = 2
x^(2/3) = 7
I expect you can handle that, now...
Thank you for the help Mr. Steve
To find the real solutions of the equation x^(4/3) - 9x^(2/3) + 14 = 0, we can employ a technique called substitution. This technique involves introducing a new variable in place of the given expression to simplify the equation.
Let's substitute a new variable, u, for the term x^(2/3). Therefore, u = x^(2/3).
By doing this substitution, the equation becomes:
u^2 - 9u + 14 = 0.
Now, we can solve this quadratic equation for u by factoring or by using the quadratic formula.
The factored form of the equation is: (u - 2)(u - 7) = 0.
Setting each factor equal to zero gives us two possible values for u:
u - 2 = 0 -> u = 2,
u - 7 = 0 -> u = 7.
However, we should remember that u represents x^(2/3). So, substituting back x^(2/3) for u, we have:
x^(2/3) = 2 -> x = (2)^(3/2) = 2^(3/2) = sqrt(2)^3 = 2.83 (approx),
x^(2/3) = 7 -> x = (7)^(3/2) = 7^(3/2) = sqrt(7)^3 = 10.55 (approx).
Therefore, the real solutions for the given equation are approximately x = 2.83 and x = 10.55.
Thus, the answer is x = 2.83, 10.55.