## Let x2+kx=0;kx^2+kx=0;k is a real number .The set of values of kk for which the equation f(x)=0f(x)=0 and f(f(x))=0f(f(x))=0 have same real solution set.

Let f(x)=x2+kx;k is a real number.The set of values of k for which the equation f(x)=0 and f(f(x))=0 have same real solution set. The equation x2+kx=0 has solutions x=0,−k. So the solutions of the equation f(f(x))=0 should be x=0,−k. f(f(x))=(x2+kx)2+(x2+kx)k=0 gives x4+k2x2+2kx3+x2k+xk2=0 x4+2kx3+(k2+k)x2+xk2=0 x(x3+2kx2+(k2+k)x+k2)=0 has the solutions x=0,x=−k Therefore x3+2kx2+(k2+k)x+k2=0 has the solution x=−k. But … Read more