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