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x^2(x-3)(x-9)^5>0

I need help solving this in interval notation

Comments

I am not sure if the ^5 covers both (x-3) AND (x-9) or if it just covers the (x-9).  If it encompasses both, you need to put another pair of parenthesis around the two like this ((x-3)(x-9))^5  That information would be needed in order to answer the question correctly.

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3 Answers

To solve x^2(x-3)(x-9)^5>0, we need to look at each factor.

X^2 will always be positive (even exponents give positive results).

(x-3) can be positive or negative.

(x-9)^5 can be positive or negative.

For the entire expression to be positive, either:

        x^2 is positive AND (x-3) is positive AND (x-9)^5 is positive (+++)

    or x^2 is positive AND (x-3) is negative AND (x-9)^5 is negative (+--).

This give us two scenarios:

 A)         x^2>0; x>0

           & x-3>0; x>3

           & (x-9)^5>0; x-9>0; x>9

              x>0 & x>3 & x>9 : x>9 : (9,infinity)

B)           x^2>0; x>0

           & x-3<0; x<3

          & (x-9)^5<0; x-9<0; x<9

             x>0 & x<3 & x<9 : 0<x<3 (0,3)

Combining A and B; x>9 or 0<x<3; (9,infinity)U(0,3)

 

Comments

x> 0 is true for all real numbers except when x = 0.  This solution is incomplete.

The solution does not include x=0.  It specifies x>0, not x≥0.

x>9 or 0<x<3 says x is either greater than (but not including) 9, or between 0 and 3 (but not including 0 or 3).

I'm sure, that Kenneth W's phrase "This solution is incomplete." was not referred to x=0 as a solution, but to your final answer, ma'am. You missed set of negative numbers, which satisfies the given inequality too.

Thanks for clarifying, Natilya.  I did mean to imply that x<0 also qualifies.

Now I see what you meant, Kenneth.  Thank you for pointing this out.  The full solution therefore is:

(-infinity,0)U(0,3)U(9,infinity)

Comment

First, thank you for using proper notation.  This will help me organize the answer a lot better.

x2 * (x-3) * (x-9)5 > 0

I would begin solving this by finding all the places where the equation equals 0 first.

x2 * (x-3) * (x-9)5 = 0

x = 0

x - 3 = 0  => (x= 3)

(x - 9)5 = 0  =>(x = 9)

So your three main points worth looking at are {0, 3, 9}. This means there are four regions that need to be tested to see whether or not any of them are positive.  (x<0 , 0<x<3, 3<x<9, 9<x)  We can use any four numbers that lie within those regions.  I suggest using {-1, 1, 4, 10} as your four numbers.

x = 10

102*(10-3)*(10-9)5 > 0

100* 7 *1>0

700 > 0  (TRUE)

So we definitely know that the interval (9, infinity) is part of the solution for this problem.  If 10 is true, 9.1 is true, and so is 348724, and so is any other number greater than 9.  I leave it to you to find out if the remaining number choices {-1, 1, 4} are true or not.

HINT #1: 0, 3 and 9 are not part of your solution because they solve to 0.  Even if -1, 1, and 4 are all true, you must seperate them into separate intervals.  If all four numbers were true, your solution would look like {(-∞, 0), (0, 3), (3, 9), (9, ∞)}

HINT #2: Not all four numbers are true.

Zeros of expression x^2(x-3)(x-9)^5 are 0, 3 and 9.

Those three numbers divide the numeric line into four intervals
(-∞, 0), (0, 3), (3, 9), (9, +∞)
Let's take any number from each interval, plug in and analyze the sign of expression.
<--------|-----o----|----o-------|------o------|------>
- ∞      - 1      0     1     3       6      9       10   + ∞
(-1)^2(-1-3)(-1-9)^5 > 0
(+)  •  ( - ) •  ( - ) 

(1)^2(1-3)(1-9)^5 > 0
(+) • ( - ) • ( - )

(6)^2(6-3)(6-9)^5 < 0
(+) • (+) • ( - ) 

(10)^2(10-3)(10-9)^5 > 0
 (+) •  ( + ) •  ( + )

So, the answer is
( - ∞, 0)U(0, 3)U(9, + ∞)