Question

Find the values of a which satisfies the equation$\sqrt{6a}\times \frac{2}{3}{a}^2-a^2\sqrt{12}+\frac{8a^2\pi }{3}=0$

Answer 1

$\begin{array}{c}\sqrt{6a}\times \frac{2}{3}{a}^2-a^2-a^2\sqrt{2}+\frac{8a^2\pi }{3}=0\\ \Rightarrow a^2\left(\frac{2\sqrt{6a}}{3}-\sqrt{12}+\frac{8\pi }{3}\right)=0\\ \Rightarrow a=0or\frac{2}{3}\sqrt{6a}=\sqrt{12}+\frac{8\pi }{3}\\ \Rightarrow \sqrt{6a}=\frac{3\sqrt{12}+8\pi }{2}\\ \Rightarrow \sqrt{6a}=3\sqrt{12}+4\pi \\ \Rightarrow 6a={\left(3\sqrt{3}+4\pi \right)}^2\\ \therefore a=\frac{{\left(3\sqrt{3}+4\pi \right)}^2}{6}\end{array}$

hence, roots are a=0, or $=\frac{{\left(3\sqrt{3}+4\pi \right)}^2}{6}$