If the quadratic equation kx^2+kx+1=0 has no real roots, what values of k are possible?

The solutions to quadratic equations, which are of the form ax^2+bx+c=0 with a≠0, are given by the Quadratic Formula: x= [-b+-sqrt(b^2-4ac)]/(2a). However, all real numbers, whether negative or not, square to give a non-negative real number; thus, the equation will have no real roots if b^2-4ac<0 because it is impossible for real numbers to satisfy this inequality. From comparing the general form of a quadratic equation with the quadratic equation in the question, kx^2+kx+1=0, we can see that a=k, b=k and c=1 and that a≠0 implies k≠0. Therefore, since there are no real roots to this equation then k^2-4k<0 and since both terms contain a factor of k, we can factorise this equation to give k(k-4)=k(k-4)<0. Finally, by plotting the graph of k(k-4), remembering to include the two roots of k=0 and k=4 and that for large positive/negative k k^2-4k increases, we can see that k^2-4k<0 is only satisfied when 0<k<4, which does not contradict the condition k≠0.