Question 31742: Solve:
(x-2)/(x^2 -2x-3)>=0
the answer to this is (-1,3)U(3 infinity) right?
Solve 5/(x+4)<=1
The answer to this is (negative infinity, -4)U[1, infinity)
Thank you,
Alexus Bandona
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! (x-2)/(x^2 -2x-3)>=0
=> (x-2)(x^2-3x+x-3) >= 0
=> (x-2)(x-3)(x+1) >= 0
The boundary points are x=-1, x=2, and x=3;
Let us consider some test values based on the boundary points.
for x=-2(x<-1), (x-2)(x-3)(x+1) = (-2-2)(-2-3)(-2+1) = (-4)(-5)(-1) = -20 (<0)
=> x < -1 does not satisfy the inequality
for x=0 (-10)
=> -1
for x=2.5 (2
(x-2)(x-3)(x+1)=(2.5-2)(2.5-3)(2.5+1)=(.5)(-.5)(3.5)=a negative number (<0)
=> does not satisfy the inequality
for x = 5 (x>3),
(x-2)(x-3)(x+1)=(5-2)(5-3)(5+1)=(3)(2)(6)=36 (>0) ... satisfies the inequality
So, all x-values for which -13 satisfy the given inequality; the given inequality has equal sign, meaning the boundary points are inclusive
Final result: the inequality is satisfies for all real x's meeting:
x >= -1 and x <= 2;
x >= 3;
In interval notation: x belongs to [-1,2]U[3,infinity)
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