C-05-Q098mediumsingle_mcqSolution set of y2=3 yy^2 = \sqrt{3}\, yy2=3y is—a{0,3}\{0, \sqrt{3}\}{0,3}b{3}\{\sqrt{3}\}{3}c{0}\{0\}{0}d{0,3}\{0, 3\}{0,3}ব্যাখ্যাRearranging gives y2−3 y=0y^2-\sqrt{3}\,y=0y2−3y=0, so y(y−3)=0y(y-\sqrt{3})=0y(y−3)=0. Thus y=0y=0y=0 or y=3y=\sqrt{3}y=3, giving the solution set {0,3}\{0,\sqrt{3}\}{0,3}.