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?
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 ??
x^2/3 - 7x^1/3 + 12 = 0
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
To solve the given equation x^(2/3) - 7x^(1/3) + 12 = 0, we can make a substitution to simplify the equation.
Let's substitute x^(1/3) with a. So, a = x^(1/3).
Now, rewrite the equation using the substitution:
(a^2) - 7a + 12 = 0
This equation is now in quadratic form, which we can easily solve. Factoring the quadratic:
(a-4)(a-3) = 0
Now, set each factor equal to zero and solve for a:
a - 4 = 0 --> a = 4
a - 3 = 0 --> a = 3
But remember, a represents x^(1/3), so we need to solve for x:
If a = 4, then x^(1/3) = 4. Cubing both sides, we get:
(x^(1/3))^3 = 4^3
x = 64
If a = 3, then x^(1/3) = 3. Cubing both sides:
(x^(1/3))^3 = 3^3
x = 27
So, the values of x that satisfy the equation are x = 64 and x = 27.
In response to your question, the values you provided x = 27 and x = 64 are correct.