# EXPRESS the solution set for the inequality x^2 - 2x -3 <= 0 in set builder notation.

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The given quadratic equation is x^2-2x-3 < = 0.

To determine the solution set , we factorise the left side of the inequation:

x^2 -2x-3 < = 0

x^2- 3x+x -3 < = 0

x(x-3) -1(x-3) < = 0

(x-3)(x-1) < = 0

Therefore the expression in the form of product on the left is a negative quantity. This possible if x is in between the roots 1 and 3.

Therefore the solution set is ; S = {x : 1 < = x <= 3}.

We have the inequation x^2 - 2x -3 <= 0

x^2 - 2x -3 <= 0

=> x^2 - 3x + x - 3 <=0

=> x(x-3)+1(x-3) <=0

=> (x+1)(x-3)<=0

Now this is possible if either of (x+1) and (x-3) is negative or 0.

For x+1 <=0 and x-3 =>0

=> x<=-1 and x=> 3

This gives no valid values.

For x+1 =>0 and x-3 <=0

=> x =>-1 and x<=3

This is possible and gives us values of x ranging from x= -1 to x=3.

**The solution here is [-1 , 3]**