Expert: Ahmed Salami - 7/21/2004

Question
Hi I am a high school student and I wanted to solve these problems. I am having trouble with the one's below.....

The question was -

Solve each of the following inequalities. Write solutions in interval notation.

1)-7<(2x-5)/2<1
2)2x^3-18x<=0
3)|x/x-1|<1

I also wanted to know is there was an answer to this.
the question was -

factor as completly as possible

4) k^15p-8

I could'nt get this one either.
the question was-

Simplify the following.

5)(7-i)/(3+i)+(4-3i)/(3i)

waiting for your reply,
Avinash

Answer
Hi Avinash,
For -7<(2x-5)/2<1 , we solve separately
-7<(2x-5)/2
-14< 2x - 5
-9 < 2x
-9/2 < x
also
(2x-5)/2<1
2x -5 < 2
2x < 7
x < 7/2
so the complete solution is
-9/2 < x < 7/2

For 2x^3-18x<=0
2x (x + 3)(x - 3)<= 0
We do this by testing, take one number each in the intervals
x < -3 , -3 < x < 0 , 0 < x < 3 , x > 3
Insert each one into the inequality 2x (x + 3)(x - 3)<= 0
and select only the ones that satisfy the inequality. The range in which the numbers lie is our solution. Try it out and see what you can make of it. The first and third intervals should be the solution.

For |x/x-1|<1 , we have
-1 < x/x-1 < 1 and solving separately gives
-1 < x/x-1 now we have to make sure we multiply by a positive quantity to maintain the inequality sign and we then use (x-1)^2 , therefore
-(x-1)^2 < x(x-1)
-(x^2 - 2x + 1) < x^2 - x
-x^2 + 2x -1 < x^2 - x
0 < 2x^2 - 3x + 1
(2x - 1)(x - 1)> 0 and we employ the usual testing procedure to get x < 1/2 and x > 1
For the other part,
x/x-1 < 1 we do the same thing and arrive at
x(x-1)< (x-1)^2
x^2 - x < x^2 - 2x + 1
x - 1 < 0
x < 1
So combining all the intervals and testing shows that
x < 1/2 is the only solution to the problem.

I don't think i get the fourth question.
As for the last one,
(7-i)/(3+i)= (7-i)/(3+i) . (3-i)/(3-i)
          = (20 - 10i)/10 = 2 - i
(4-3i)/(3i)= (4-3i)/(3i) . (-3i)(-3i)
          = (-9 - 12i)/ 9 = -1 - 4/3i
So (7-i)/(3+i)+(4-3i)/(3i)
= (2 - i) + (-1 - 4/3i)
= 1 - 7/3i

Really hope all these helps. But remember that the method of solution is paramount.
Good luck.
Regards.