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

Suggest Corrections

0

Similar questions
View More