Solving inequality with different power variables

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  • #1
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Homework Statement


Solve for k:
k2 - 16k < 0

In the answer it has 0 < k < 16, I do not know how they get there from the original question.
 

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  • #2
Ray Vickson
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Homework Statement


Solve for k:
k2 - 16k < 0

In the answer it has 0 < k < 16, I do not know how they get there from the original question.

Show us what you have done so far.
 
  • #3
BMW
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Show us what you have done so far.

Well I factorise it to this k(k - 16) < 0 then what? I tried dividing both sides by k, then I get k < 16 but how can I divide both sides by k as I don't know if its positive/negative? And how do I get the 0 < k?
 
  • #4
Ray Vickson
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Well I factorise it to this k(k - 16) < 0 then what? I tried dividing both sides by k, then I get k < 16 but how can I divide both sides by k as I don't know if its positive/negative? And how do I get the 0 < k?

There are two possible cases (since k = 0 is not allowed). These are
Case (1): k > 0
Case (2): k < 0
Just look at what you get in each case. Of course, you have already dealt with case (1), so now go ahead and look at case (2).
 
  • #5
you just solve it like it were an equality. when you have k(k-16)=0 you know that you have k=0 and k-16=0
 
  • #6
BMW
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you just solve it like it were an equality. when you have k(k-16)=0 you know that you have k=0 and k-16=0

Ok, then I get k < 0 and k < 16, but it should be 0 < k < 16
 
  • #7
BMW
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There are two possible cases (since k = 0 is not allowed). These are
Case (1): k > 0
Case (2): k < 0
Just look at what you get in each case. Of course, you have already dealt with case (1), so now go ahead and look at case (2).

Ok, so if I say k < 0 (case 2), then when I divide both sides by k I flip the sign. Then I get k > 16 ?????
 
  • #8
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I would take a different tack. The equation k(k - 16) = 0 is different from your inequality, but there is a connection. The two solutions are k = 0 and k = 16.

The two solutions divide the number line into three regions: (-∞, 0), (0, 16), (16, ∞). Note that these are all open intervals: they don't include any endpoints.

Since the only values for which k(k - 16) = 0 are 0 or 16, any other values make this expression positive or negative.

Pick any number from the first interval (-∞, 0). Determine the sign of x(x - 16). If it's negative, this interval is part of your solution.

Do the same for the other two intervals.
 
  • #9
CompuChip
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It's always tricky to deal with inequalities in this way, as you need to keep track of when you divide by negative numbers so you can flip the sign. Once you know that the equality k² - 16k = 0 holds for k = 0 and k = 16, I always prefer to draw them on the number line:

Code:
          0                    0                         <-- k² - 16k
----------|--------------------|-----------------
         k=0                 k=16

Now you know that on each of the three pieces, an inequality will hold. It cannot flip signs because if you go from k² - 16k > 0 to k² - 16k < 0 there should be a point in between where k² - 16k = 0, but you have just found all of these. So you just have to put either < or > over the three intervals, and the easiest way to do that is pick a k inside of them:

Code:
  </>?    0        </>?        0         </>?            <-- k² - 16k
----.-----|-----.--------------|------.----------
 k=-1    k=0   k=1           k=16     k=20

Plugging these in, you find
(-1)² - 16 (-1) = 1 + 16 = 17 > 0
1² - 16 (1) = 1 - 16 = -15 < 0
(20)² - 16(20) = 400 - 320 = 80 > 0

so you can complete the diagram as
Code:
   >0     0       <0           0         >0              <-- k² - 16k
----------|--------------------|-----------------
         k=0                 k=16

so the inequality k² - 16k < 0 is solved by 0 < k < 16.
 
  • #10
Ray Vickson
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Ok, so if I say k < 0 (case 2), then when I divide both sides by k I flip the sign. Then I get k > 16 ?????

Right: you have both k < 0 and k > 16 in case (2). What does that tell you?
 
  • #11
LCKurtz
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It's always tricky to deal with inequalities in this way, as you need to keep track of when you divide by negative numbers so you can flip the sign. Once you know that the equality k² - 16k = 0 holds for k = 0 and k = 16, I always prefer to draw them on the number line:

Code:
          0                    0                         <-- k² - 16k
----------|--------------------|-----------------
         k=0                 k=16

Now you know that on each of the three pieces, an inequality will hold. It cannot flip signs because if you go from k² - 16k > 0 to k² - 16k < 0 there should be a point in between where k² - 16k = 0, but you have just found all of these. So you just have to put either < or > over the three intervals, and the easiest way to do that is pick a k inside of them:

Code:
  </>?    0        </>?        0         </>?            <-- k² - 16k
----.-----|-----.--------------|------.----------
 k=-1    k=0   k=1           k=16     k=20

Plugging these in, you find
(-1)² - 16 (-1) = 1 + 16 = 17 > 0
1² - 16 (1) = 1 - 16 = -15 < 0
(20)² - 16(20) = 400 - 320 = 80 > 0

so you can complete the diagram as
Code:
   >0     0       <0           0         >0              <-- k² - 16k
----------|--------------------|-----------------
         k=0                 k=16

so the inequality k² - 16k < 0 is solved by 0 < k < 16.


I would like to add to this. Once you have the number lines with the places where the factors of ##k(k-16)## are ##0##, and you know the only two places that function can change sign are at the roots of the factors (##k=0,~k=16##), you can find out the sign of the expression by analyzing the signs of the factors. You don't need to plug in any values to do this.

For example, it is clear that if ##k>16##, both ##k## and ##k-16## are positive. So above the line to the right of ##16## put ##+~+##. Now think of ##k## sliding to the left across ##k=16##. That changes the sign of the factor ##k-16## to negative, so above that section of the line put ##+~-##. Now as ##k## moves left across ##k=0## the ##k## factor changes sign so on the left part of the line put ##-~-##. The sign of the product depends on how many minus signs there are and it obviously negative on ##(0,16)##. You just have to look at which factor changes sign. One advantage of doing it this way is it doesn't matter how many factors there are or whether they are in the numerator or denominator. Just remember if a factor is squared, it doesn't change sign when the variable crosses its root.
 

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