quad eq. 2

Asked by varunsinghal2011 | 9th May, 2010, 01:55: PM

Expert Answer:

Dear Student,

The critical points are where the quantity inside the modulus (in the given expression) becomes zero. Hence the critical points here are:-2 and -1.

Case1:

-2:

2 -(x+2) - [-(2 x+1 - 1)] = 2 x+1 +1................(because for X  -2, |x+2| and |2 x+1 - 1| are both negative)

2 -(x+2) +(2 x+1 - 1) = 2 x+1 +1

2 -(x+2)=21

-x-2 = 1

x=-3

which is in the domain considered, i.e.,X  -2, Hence x=-3 is a solution.

Similarly;

Case 2:

-2-1:

2 (x+2) - [-(2 x+1 - 1)] = 2 x+1 +1

2 (x+2) +(2 x+1 - 1) = 2 x+1 +1

2 (x+2)=21

x+2 = 1

x=-1

which is in the considered domain; -2-1. Hence x=-1 is a solution.

Case 3:

x>-1:

2 (x+2) - (2 x+1 - 1) = 2 x+1 +1

2 (x+2) - 2 x+1 + 1 = 2 x+1 +1

2 (x+2) = 2.2 x+1

2 (x+2) = 2 x+2

LHS=RHS (for all) values of x. Hence, the entire considered interval is the solution.

So, x-1; x=-3.

Regards Topperlearning.

Answered by  | 29th May, 2010, 11:43: AM

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