Question

Prove that for any value of $a$, the inequation $(a^2+3)x^2+(a+2)x-6<0$ is true for atleast one negative $x$.

Answer 1

Consider a quadratic polynomial $(a^2+3)x^2+(a+2)x-6$

Now coefficient of $x^2$ is always positive $(\because a^2+3>0)$

So for this polynomial to be negative for atleast one $x$ discriminant need to be greater than zero$(D>0)$

$D=(a+2{)}^2-4(a^2+3\left)\right(-6)=25a^2+4a+76=(5a+\frac{2}{5}{)}^2+76-\frac{4}{25}>0$

So $D>0$

Hence, coefficient of $x^2$ is positive and $D>0$ means this polynomial will have atleast one negative value for sure.