Questions: Find all the real solutions to the following equation. x^2 / 3+2 x^1 / 3-8=0 Choose ALL that apply. x=-6 x=64 x=-2 x=4 x=-8 x=-12

Solution Steps

To solve the equation \(x^{2/3} + 2x^{1/3} - 8 = 0\), we can perform a substitution to simplify it. Let \(y = x^{1/3}\). Then the equation becomes \(y^2 + 2y - 8 = 0\), which is a quadratic equation in terms of \(y\). We can solve this quadratic equation using the quadratic formula. Once we find the values of \(y\), we can substitute back to find the corresponding values of \(x\).

We start with the equation

\[ x^{2/3} + 2x^{1/3} - 8 = 0. \]

To simplify, we let

\[ y = x^{1/3}. \]

This transforms our equation into

\[ y^2 + 2y - 8 = 0. \]

Next, we apply the quadratic formula

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 1\), \(b = 2\), and \(c = -8\). Plugging in these values, we get:

\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}. \]

This results in two potential solutions for \(y\):

\[ y_1 = \frac{4}{2} = 2 \quad \text{and} \quad y_2 = \frac{-8}{2} = -4. \]

Now we substitute back to find \(x\):

1. For \(y_1 = 2\):

\[ x^{1/3} = 2 \implies x = 2^3 = 8. \]

2. For \(y_2 = -4\):

\[ x^{1/3} = -4 \implies x = (-4)^3 = -64. \]

Final Answer