Question

The sum of squares of roots of the equation \[ x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0 \] is

• A. \(82\)
• B. \(65\)
• C. \(50\)
• D. \(37\)

Hint:

For equations involving fractional powers such as \(x^{1/3}\) and \(x^{2/3}\), substitute \(t=x^{1/3}\) to convert the equation into a quadratic equation in \(t\).

Answer 1

The Correct Option is B

Step 1: Substitute \(t=x^{\frac{1}{3}}\).
Let \[ t=x^{\frac{1}{3}} \] Then \[ x^{\frac{2}{3}}=t^2 \] The given equation becomes \[ t^2+t-2=0 \] Factorizing, \[ (t+2)(t-1)=0 \] Thus, \[ t=1 \quad \text{or} \quad t=-2 \]

Step 2: Find the corresponding values of \(x\).
Since \[ t=x^{\frac{1}{3}}, \] we get For \(t=1\), \[ x=1^3=1 \] For \(t=-2\), \[ x=(-2)^3=-8 \] Hence, the roots of the given equation are \[ 1,\,-8 \]

Step 3: Calculate the sum of squares of the roots.
The sum of squares of the roots is \[ 1^2+(-8)^2 \] \[ =1+64 \] \[ =65 \]

Step 4: Final conclusion.
Therefore, the required sum of squares of the roots is \[ \boxed{65} \]