Quadratic EquationHard
Question
The complete set of real values of $p$ for whcih both roots of the equation $x^{2} + 2(p - 3)x + 9 = 0$ lie in $( - 6,1)$ is
Options
A.$\left\lbrack 6,\frac{27}{4} \right)$
B.$\left( \frac{27}{4},9 \right)$
C.$\lbrack 6,9)$
D.$( - 2,0\rbrack \cup (2,9)$
Solution
Let $f(x) = x^{2} + 2(P - 3)x + 9$
$$\begin{matrix} & \ \begin{matrix} P \geq 0 \Rightarrow P \in ( - \infty,0\rbrack \cup \lbrack 6,\infty) \\ - 6 < - \frac{b}{2a} < 1\ \Rightarrow \ P \in (2,9) \\ f( - 6) > 0\ \Rightarrow \ P < \frac{27}{4} \\ f(1) > 0\ \Rightarrow \ P > - 2 \\ \text{~Hence,~}\ P \in \left\lbrack 6,\frac{27}{4} \right) \end{matrix} \end{matrix}$$
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