Question

If $x^2-4x+p=0$ has real roots, then
(a) p ≥ 4
(b) p ≤ 4
(c) p ≥ 5
(d) p ≤ 5

Answer 1

(b) p ≤ 4
$$\begin{gathered} \text{It is given that the equation (}{x}^2+4x+p=0)\text{has real roots.} \\ \therefore D\ge 0 \\ \Rightarrow b^2-4ac\ge 0 \\ \Rightarrow 16-4p\ge 0 \\ \Rightarrow 16\ge 4p \\ \Rightarrow p\le 4 \end{gathered}$$