Question

If the equation $a{x}^2+bx+c=0$ does not have $2$ distinct real
roots and $a+c>b,$ then prove that $\int \left(x\right)\ge 0,\forall x\in R.$

Answer 1

$f\left(x\right)=a{x}^2+bx+c$

But, since $f\left(x\right)$ does not have $2$ distinct real root, the conditions can be either

$f\left(x\right)\ge 0$ or $f\left(x\right)\le 0$ $\forall x\in R$

graphs possible OR for $f\left(x\right)\ge 0$

graphs possible for $f\left(x\right)\le 0$

Now, $f(-1)=a(-1{)}^2+b(-1)+c=a-b+c$

Since $f\left(x\right)$ does not have $2$ distinct real roots.

$f(-1)\ge 0$ or $f(-1)\le 0$

Case $1$: when $f\left(x\right)\ge 0$

$f(-1)\ge 0$

$a-b+c\ge 0$

$a+c\ge b$

Also, $a+c>b$ [given]

Hence, case $1$ is satisfied.

Hence, we can say that $f\left(x\right)\ge 0\forall x\in R$.