SOLUTION: Could someone please help me? Solve the following inequalities. Write your answer using interval notation. (x+4)(2x-3)>0 x^2-2x<8 Thank you

Question 61886: Could someone please help me?
Solve the following inequalities. Write your answer using interval notation.
(x+4)(2x-3)>0
x^2-2x<8
Thank you
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
Solve the following inequalities. Write your answer using interval notation.
(x+4)(2x-3)>0
x+4=0 and 2x-3=0
x+4-4=0-4 and 2x-3+3=0+3
x=-4 and 2x=3
x=-4 and 2x/2=3/2
x=-4 and x=3/2
These are your critical numbers, they tell you the possible intervals, (-infinity,-4),(-4,3/2), (3/2,infinity):
Check the interval (-infinity,-4) by using a test number in that interval like -5
(-5+4)(2(-5)-3)>0 ?
(-1)(-13)>0
13>0 This is true.
This tells us that the interval (-infinity, -4) is in the solution.
:
Now test the interval (-4,3/2) with a test number like 0.
(0+4)(2*0-3)>0 ?
(4)(-3)>0
-12>0 This is not true so the interval (-4,3/2) is not part of the solution.
:
Now test the interval (3/2,infinity) with a test number like 2.
(2+4)(2*2-3)>0 ?
(6)(1)>0
6>0 This is true
This tells us that the interval (3/2,infinity) is part of the solution.
The Solution is: (-infinity,-4)U(3/2,infinity)
:
x^2-2x<8
x^2-2x-8<8-8
x^2-2x-8<0 Factor
(x-4)(x+2)<0
x-4=0 and x+2=0
x-4+4=0+4 and x+2-2=0-2
x=4 and x=-2
These tell you that your possible intervals are:(-infinity,-2),(-2,4), and (4,infinty)
:
Test the interval of (-infinity,-2) with a test number like, -3.
(-3-4)(-3+2)<0 ?
(-7)(-1)<0
7<0 This is not true.
(-infinty,-2) is not part of the solution.
Test the interval (-2,4) with a test number like 0.
(0-4)(0+2)<0 ?
-4(2)<0
-8<0 this is true.
(-2,4) is part of the solution.
Test the interval (4,infinity) with a test number like 5.
(5-4)(5+2)<0 ?
1(7)<0
7<0 this is not part of the soltuion.
The solution therefore is: (-2,4)
Happy Calculating!!!
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