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.

Just line them up and shoot them.
The roots of $x^2 +ax -3b=0$ will also be roots of $(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$.  So if $x^2 +a -3b = 0$ has two real roots then $(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$ will have at least two real root.
The roots of $x^2 +ax -3b=0$ will, by the quadratic formula be, $\frac {-a\pm \sqrt{a^2 + 12b}}{2}$.  These will be two distinct roots if $a^2 +12b > 0$.  This will be one double root if $a^2 =12b$.  And this will be two complex if $a^2 + 12b < 0$.
The same holds holds true for the roots of $x^2 -cx+b = 0$.  This will have two distinct real root is $c^2 -4b >0$; will have one real root if $c^2 = 4b$; will have two complex roots if $c^2 -4b < 0$.
And $x^2 -dx + 2b=0$ will have two distinct real roots if $d^2 + 8b > 0$; one if $d^2 +8b = 0$; and none if $d^2 +8b < 0$.
So that's lining them up.  Now shoot them.
Suppose $a^2 + 12b \le 0$ and $c^2 - 4b \le 0$ and $d^2 + 8b \le 0$.
Then $b \le -\frac {a^2}{12}\le 0$ and $b \ge \frac {c^2}4 \ge 0$ and $b \le \frac {-d^2}8$.  So $b =0$ and $a=0$ and $c=0 and $d=0$.
So either $a=b=c=d=0$ and there is one real root $x = 0$.
Or at least one of $a^2 + 12b$ or $c^2 - 4b$ or $c^2 + 8b$ is positive.
And if any $v^2 + kb > 0$ then $\frac {- v\pm \sqrt{v^2 + kb}}2$ will be two distinct real roots.
