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The sum of squares of roots of the equation $x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0$ is
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Verified Answer
The correct answer is:
65
Given, equation is
$$
\begin{array}{rlrl}
& & x^{2 / 3}+x^{1 / 3}-2 & =0 \\
\Rightarrow & & x^{2 / 3}+x^{1 / 3} & =2 \\
& \Rightarrow & \left(x^{2 / 3}+x^{1 / 3}\right)^3 & =2^3 \\
& & & \\
{\left[\therefore(a+b)^3\right.} & \left.=a^3+b^3+3 a b(a+b)\right]
\end{array}
$$
$\begin{aligned} & \Rightarrow \quad x^2+x+3 x \times 2=8 \Rightarrow \\ & x^2+7 x-8=0 \\ & \Rightarrow \quad(x+8)(x-1)=0 \Rightarrow x=1 ;-8 \\ & \text { Sum of square roots }=(1)^2+(-8)^2 \\ & \quad=1+64=65\end{aligned}$
$$
\begin{array}{rlrl}
& & x^{2 / 3}+x^{1 / 3}-2 & =0 \\
\Rightarrow & & x^{2 / 3}+x^{1 / 3} & =2 \\
& \Rightarrow & \left(x^{2 / 3}+x^{1 / 3}\right)^3 & =2^3 \\
& & & \\
{\left[\therefore(a+b)^3\right.} & \left.=a^3+b^3+3 a b(a+b)\right]
\end{array}
$$
$\begin{aligned} & \Rightarrow \quad x^2+x+3 x \times 2=8 \Rightarrow \\ & x^2+7 x-8=0 \\ & \Rightarrow \quad(x+8)(x-1)=0 \Rightarrow x=1 ;-8 \\ & \text { Sum of square roots }=(1)^2+(-8)^2 \\ & \quad=1+64=65\end{aligned}$
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