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

How many integers satisfy the inequality

|10(x+1)/x^2+2x+3|=1?

Grade:11

4 Answers

Akash Kumar Dutta
98 Points
9 years ago

Dear Vivek,

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.

Regards.

Veer Kashyap
18 Points
9 years ago

I am sorry.

The question is 

|10(x+1)/x^2+2x+3|1

shiddhant bhattacharya
25 Points
9 years ago

4 integers satisfy the equation which u hav given. take out the tha mode and u get 2 cases on the rite hand side of the equation +1 and -1, solve them both u get 2 answers each. so total 4

Abhishekh kumar sharma
34 Points
9 years ago

only 1.

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