Question

f(x)=-x^2+px-(3p+4) if f(x) is possible for at least one real x then the value(s) of p is/are

Answer 1

Given,

$\[f\left( x \right) = {x^2} + px - \left( {3p + 4} \right)\]$

Solution,

Know that for at least one real x, the value of the determinant should be greater than or equal to zero.

Therefore,

$\[\begin{array}{l} \Rightarrow {\left( p \right)^2} + 4\left( 1 \right)\left( {3p + 4} \right) \ge 0\\ \Rightarrow {p^2} + 12p + 16 \ge 0\\ \Rightarrow \left( {p + 1.52} \right)\left( {p + 10} \right) \ge 0\end{array}\]$

So,

$\[ \Rightarrow p \in \left( {1.2,1.52} \right)\]$