Official solution
y \[ \ge \] x2 \[ \Rightarrow \] upper region of y = x2
y \[ \le \] 2x \[ \Rightarrow \] lower region of y = 2x
According to question, area of OABC = 2 \[ \times \] area of OAC
\[ \Rightarrow \] \[\int\limits^{4}_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy\] = 2\[\int\limits^{\alpha }_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy\]
\[\Rightarrow \left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^4 = 2\left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^\alpha \]
\[ \Rightarrow {{16} \over 3} - 4 = 2\left[ {{2 \over 3}{{\left( \alpha \right)}^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]\]
\[ \Rightarrow {4 \over 3} = 2\left[ {{2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]\]
\[ \Rightarrow {2 \over 3} = {2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}\]
\[ \Rightarrow 8 = 8{\alpha ^{{3 \over 2}}} - 3{\alpha ^2}\]
\[ \Rightarrow 3{\alpha ^2} - 8{\alpha ^{{3 \over 2}}} + 8 = 0\]