## How many integers satisfy the inequality, Physics

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How many integers satisfy the inequality |10(x+1)/x^2+2x+3|=1?

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#### Shrikvin

3/29/2013 6:27:51 AM

first thing thats not an inequality, and second thing its very easy if thats the question. the LHS = |10/x^2 + 10/x + 2x + 3| = 1, 10/x^2 will also be an integer only if x^2<10.....i.e. x<3.
hence x= 0,+1,-1,+2,-2,+3,-3. but the only value that holds the equality is -1.
hence the no. of INTEGERS is only 1.  