May 25th, 2019, 07:58 AM  #1 
Member Joined: Jul 2017 From: KOLKATA Posts: 43 Thanks: 2  Number of Integral solutions
if x2 x+1 =p has exactly one solution then the no. of integral values of p is

May 25th, 2019, 08:26 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra 
$$x > 2 \implies x2x+1=(x2)(x+1)=3=p$$ Thus, $p=3$ gives us an infinite number of solutions. Since we require only one solution, we must have $p\ne3$. $$x < 1 \implies x2x+1=(2x)+(x+1)=03=p$$ Thus, $p=3$ gives us an infinite number of solutions. Since we require only one solution, we must have $p\ne 3$. \begin{align}1 \le x \le 2 \implies x2x+1&=p \\ (2x)(x+1)&=p \\12x&=p \\ x &= \frac12(1p)\end{align} This holds as long as \begin{alignat}{2}1 &\le \frac12(1p) &&\le 2 \\ 2 &\le 1p &&\le 4 \\ 3 &\le p &&\le 3 \end{alignat} But we already eliminated the limiting values, so there are five possible integer values of $p$. Specifically $$p \in \{2,1,0,1,2\}$$ 

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integral, number, solutions 
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