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Question

If all values of $$p$$ so that $$6$$ lies between roots of the equation $$ { x }^{ 2 }+2\left( p-3 \right) x+9=0$$, belong to the interval $$\left( -\infty ,-\dfrac{ a }{a+1 } \right) $$, then find $$a$$.


Solution

Let $$f(x)= { x }^{ 2 }+2\left( p-3 \right) x+9$$.
As 6 lies between the roots of $$f(x)=0$$, we have $$D>0$$ and $$af(6) <0$$
(I) Consider $$D>0$$ 
$$ (2\left( p-3 \right))^{ 2 }-4.1.9>0$$
$$ \Rightarrow \quad { \left( p-3 \right)  }^{ 2 }-9>0\quad \Rightarrow p(p-6)>0$$
$$ \Rightarrow \quad p\in \left( -\infty ,0 \right) \cup \left( 6,\infty  \right) $$
(II) Consider $$af(6) < 0$$ 
$$(36+12(p-3)+9) <0$$
$$\Rightarrow 12p+9<0 \Rightarrow p+\dfrac { 3 }{ 4 } <0$$ 
$$\Rightarrow p\in\left( -\infty ,-\dfrac{ 3 }{ 4 } \right) $$ 
Hence, the value of $$p$$ satisfying both inequalities at the same time are $$p\in\left( -\infty ,-\dfrac{ 3 }{ 4 } \right) $$.

Mathematics

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