Numbers of roots in Quadratic Equations If $a,b,c,d$ are real numbers, then show that the equation $$(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$$ has at least two real roots.

Adding the discriminants of the factors we get $(a^2+12b)+(c^2-4b)+(d^2-8b)=a^2+c^2+d^2.$
If the sum is positive then at least one of discriminants is positive and so at least one of  three factors has two distinct roots.
Let $a^2+c^2+d^2=0$ then the equation is $(x^2-3b)(x^2+b)(x^2+2b)=0$ and evidantly has 2 or 4 distinct roots for $b\ne 0$
Thus the equation has less than 2 distinct real roots only when $a=b=c=d=0$
