Hi cleanncrazi!!
1) x^2 - 5x + 6 > 0
First of all solve
x^2 - 5x + 6 = 0
x1 = [5 + sqrt(25 - 4*1*6)] / (2*1) =
= [5 + 1] / 2 =
= 6 / 2 =
= 3
x2 = [5 - sqrt(25 - 4*1*6)] / (2*1) =
= [5 - 1] / 2 =
= 4 / 2 =
= 2
Then we have that:
x^2 - 5x + 6 = (x-3)*(x-2) > 0
Then (x-3) and (x-2) must have the same sign and not be equal to zero:
Case 1:
x-3 > 0 and x-2 > 0 ,
then
x>3 and x>2 ,
the first inequality implies the second, so it is suffice to say:
x > 3
Case 2
x-3 < 0 and x-2 < 0 ,
then
x<3 and x<2 ,
now the second inequality implies the first, so it is suffice to say:
x < 2
The answer to this question is:
x > 3
or
x < 2
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2) (x-3)/(x+4) > 2
The first condition here is (x+4) =/= 0
in other words x =/= -4 (=/= means different)
Now we can work:
If (x+4)>0 then:
x > -4
and
(x-3)/(x+4) > 2 <==> (x-3) > 2*(x+4) = 2x + 8 <==>
x-3 > 2x+8 <==> x-2x > 8+3 = 11 <==>
-x > 11 <==> x < -11
But remember that we have started this considering that x > -4 and
finished it finding that x < -11, this is absurd. Then x cannot be
greater than -4 to satisfy this inequality.
If (x+4)< 0 then:
x < -4
and
(x-3)/(x+4) > 2 <==> (x-3) < 2*(x+4) = 2x + 8 <==>
x-3 < 2x+8 <==> x-2x < 8+3 = 11 <==>
-x < 11 <==> x > -11
But remember that we have started this considering that x < -4 and
finished it finding that x > -11 , summing up we have:
-11 < x < -4
The answer to this question is:
-11 < x < -4
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3) 1/(x+1) < 1/(x-1)
First of all it must be:
x =/= 1 and x =/= -1
-Case x+1 > 0 and x-1 > 0:
In this case x > -1 and x > 1, so it is suffice to say x>1.
Then:
1/(x+1) < 1/(x-1) <==> (x-1) < (x+1) <==>
x-1 < x+1 <==> x-x < 1+1 <==> 0 < 2
x > 1 satisfies this inequation.
-Case x+1 > 0 and x-1 < 0:
In this case x > -1 and x < 1, so we can say -1 < x < 1
Then:
1/(x+1) < 1/(x-1) <==> (x-1) > (x+1) <==>
x-1 > x+1 <==> x-x > 1+1 <==> 0 > 2
ABSURD
This case is not possible for this inequation.
-Case x+1 < 0 and x-1 < 0:
In this case x < -1 and x < 1, so it is suffice to say x<-1.
Then:
1/(x+1) < 1/(x-1) <==> (x-1) < (x+1) <==>
x-1 < x+1 <==> x-x < 1+1 <==> 0 < 2
x < -1 satisfies this inequation.
-Case x+1 < 0 and x-1 > 0:
In this case x < -1 and x > 1, ABSURD
This case is not possible for this inequation.
The answer to this question is:
x > 1
or
x < -1
----------------------------------------------------------
4) w^2 + 3w < 18
w^2 + 3w < 18 <==> w^2 + 3w -18 < 0
Now we solve
w^2 + 3w -18 = 0
w1 = [-3 + sqrt(9 - 4*1*(-18))] / (2*1) =
= [-3 + sqrt(9+72)] / 2 =
= [-3 + sqrt(81)] / 2 =
= [-3 + 9] / 2 =
= 6 / 2 =
= 3
w2 = [-3 - sqrt(9 - 4*1*(-18))] / (2*1) =
= [-3 - sqrt(9+72)] / 2 =
= [-3 - sqrt(81)] / 2 =
= [-3 - 9] / 2 =
= -12 / 2 =
= -6
Now we have that:
w^2 + 3w < 18 <==> w^2 + 3w -18 < 0 <==>
(w-3)*(w+6) < 0
This is the same to say that (w-3) and (w+6) must have different sign
at the same time and both must be different to zero:
-Case w-3>0 and w+6<0:
In this case is w>3 and w<-6 . ABSURD
This case cannot satisfy this inequation.
-Case w-3<0 and w+6>0:
In this case is w<3 and w>-6 , then we have that
-6 < w < 3
satisfies the inequation.
The answer to this question is:
-6 < w < 3
-----------------------------------------------------------
5) 2/(x-2) < 3/(x+1)
First of all it must be:
x =/= 2 and x =/= -1
-Case x+1 > 0 and x-2 > 0:
In this case x > -1 and x > 2, so it is suffice to say x>2.
Then:
2/(x-2) < 3/(x+1) <==> 2*(x+1) < 3*(x-2) <==>
2x+2 < 3x-6 <==> 2x-3x < -6-2 <==> -x < -8 <==>
x > 8
We started considering x > 2 to find that also must be x > 8, then:
x > 8 satisfies this inequation.
-Case x+1 > 0 and x-2 < 0:
In this case x > -1 and x < 2, so we can say -1 < x < 2
Then:
2/(x-2) < 3/(x+1) <==> 2*(x+1) > 3*(x-2) <==>
2x+2 > 3x-6 <==> 2x-3x > -6-2 <==> -x > -8 <==>
x < 8
We started considering -1 < x < 2, to find that also must be x < 8,
this is obviously true, then:
-1 < x < 2 satisfies the inequation.
-Case x+1 < 0 and x-2 < 0:
In this case x < -1 and x < 2, so it is suffice to say x < -1.
Then:
2/(x-2) < 3/(x+1) <==> 2*(x+1) < 3*(x-2) <==>
2x+2 < 3x-6 <==> 2x-3x < -6-2 <==> -x < -8 <==>
x > 8
We started considering x < -1 to find that also must be x > 8, this is
ABSURD, then x < -1 cannot satisfies this inequation.
-Case x+1 < 0 and x-2 > 0:
In this case x < -1 and x > 2, ABSURD
This case is not possible for this inequation.
The answer to this question is:
x > 8
or
-1 < x < 2
-----------------------------------------------------------
I hope that this helps you. If you find something wrong (like a typo)
or unclear feel free to request for a clarification. I will gladly
give you further assistance on this if it is needed.
Best regards.
livioflores-ga |
