Proving no real roots exist for a given expression

Show or prove that a specific polynomial (with no free parameter) has no real roots, typically by bounding or rewriting the expression to demonstrate it is always positive or negative.

jee-main 2012 Q61 View
Let $p , q , r \in R$ and $r > p > 0$. If the quadratic equation $p x ^ { 2 } + q x + r = 0$ has two complex roots $\alpha$ and $\beta$, then $| \alpha | + | \beta |$ is
(1) equal to 1
(2) less than 2 but not equal to 1
(3) greater than 2
(4) equal to 2
taiwan-gsat 2020 Q4 8 marks View
Let $f ( x )$ be a quadratic polynomial function with real coefficients such that $f ( x ) = 0$ has no real roots. Select the correct options.
(1) $f ( 0 ) > 0$
(2) $f ( 1 ) f ( 2 ) > 0$
(3) If $f ( x ) - 1 = 0$ has real roots, then $f ( x ) - 2 = 0$ has real roots
(4) If $f ( x ) - 1 = 0$ has a double root, then $f ( x ) - \frac { 1 } { 2 } = 0$ has no real roots
(5) If $f ( x ) - 1 = 0$ has two distinct real roots, then $f ( x ) - \frac { 1 } { 2 } = 0$ has real roots
taiwan-gsat 2022 Q12 5 marks View
Let $f ( x ) , g ( x )$ both be real-coefficient polynomials, where $g ( x )$ is a quadratic with positive leading coefficient. It is known that the remainder when $( g ( x ) ) ^ { 2 }$ is divided by $f ( x )$ is $g ( x )$ , and the graph of $y = f ( x )$ has no intersection with the $x$-axis. Select the option that cannot be the $y$-coordinate of the vertex of the graph of $y = g ( x )$.
(1) $\frac { \sqrt { 2 } } { 2 }$
(2) 1
(3) $\sqrt { 2 }$
(4) 2
(5) $\pi$