If the equation (p^2 −4)(p^2 −9)x^3 +[((p−2)/2)]x^2 +(p−4)(p−3)(p−2)x+{2p−1}=0. is satisfied by all values of x in (0,3] then sum of all possible integral values of ′p′ is ? {.} = fractional part function. [.]= greatest integer function.


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Question Number 33944 by rahul 19 last updated on 28/Apr/18
$If the equation (p^2 −4)(p^2 −9)x^3 +[((p−2)/2)]x^2 +(p−4)(p−3)(p−2)x+{2p−1}=0. is satisfied by all values of x in (0,3] then sum of all possible integral values of ′p′ is ? {.} = fractional part function. [.]= greatest integer function.$
$I f the equation$ $(p^{2} - 4) (p^{2} - 9) x^{3} + [\frac{p - 2}{2}] x^{2} + (p - 4) (p - 3) (p - 2) x + {2 p - 1} = 0 .$ $is satisfied by all values of x in (0, 3] t h e n$ $s u m o f a l l p o s s i b l e i n t e g r a l v a l u e s o f$ $^{'} p^{'} i s ?$ ${.} = f r a c t i o n a l p a r t f u n c t i o n .$ $[.] = g r e a t e s t i n t e g e r f u n c t i o n .$
	Answered by MJS last updated on 28/Apr/18
	$p∈Z ⇒ {2p−1}=0 now it′s easy (p^2 −4)(p^2 −9)=0 ⇒ p∈{−3; −2; 2; 3} [((p−2)/2)]=0 ⇒ [(p/2)]=1 ⇒ p∈{2; 3} (p−4)(p−3)(p−2)=0 ⇒ {2; 3; 4} p∈{2; 3} sum({2; 3})=5$
	$p \in Z \Rightarrow {2 p - 1} = 0$ $now {it}^{'} s easy$ $(p^{2} - 4) (p^{2} - 9) = 0 \Rightarrow p \in {- 3; - 2; 2; 3}$ $[\frac{p - 2}{2}] = 0 \Rightarrow [\frac{p}{2}] = 1 \Rightarrow p \in {2; 3}$ $(p - 4) (p - 3) (p - 2) = 0 \Rightarrow {2; 3; 4}$ $p \in {2; 3}$ $s u m ({2; 3}) = 5$
	Commented by rahul 19 last updated on 28/Apr/18

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