The real roots of the equation `x^(2//3)+x^(1//3)-2=0` are

Options

A. 1,8
B. `-1,-8`
C. `-1,8`
D. `1,-8`

Answer

To solve the equation \( x^{2/3} + x^{1/3} - 2 = 0 \) for its real roots, we can follow these steps:

### Step 1: Substitute \( y = x^{1/3} \)

Let \( y = x^{1/3} \). Then, we can express \( x^{2/3} \) as \( y^2 \). The equation becomes:

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

### Step 2: Factor the quadratic equation

Next, we need to factor the quadratic equation \( y^2 + y - 2 = 0 \). We look for two numbers that multiply to \(-2\) and add to \(1\). The numbers \(2\) and \(-1\) fit this requirement. Thus, we can factor the equation as:

\[ (y - 1)(y + 2) = 0 \]

### Step 3: Solve for \( y \)

Setting each factor to zero gives us:

1. \( y - 1 = 0 \) → \( y = 1 \)
2. \( y + 2 = 0 \) → \( y = -2 \)

### Step 4: Substitute back to find \( x \)

Now, we substitute back for \( y \):

1. For \( y = 1 \):
\[ x^{1/3} = 1 \implies x = 1^3 = 1 \]

2. For \( y = -2 \):
\[ x^{1/3} = -2 \implies x = (-2)^3 = -8 \]

### Step 5: List the real roots

The real roots of the original equation \( x^{2/3} + x^{1/3} - 2 = 0 \) are:

\[ x = 1 \quad \text{and} \quad x = -8 \]

### Final Answer:

The real roots of the equation are \( x = 1 \) and \( x = -8 \).

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