- #1
BMW
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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.
BMW said: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.
Ray Vickson said:Show us what you have done so far.
BMW said: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?
Eagle's Wings said: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
Ray Vickson said: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).
0 0 <-- k² - 16k
----------|--------------------|-----------------
k=0 k=16
</>? 0 </>? 0 </>? <-- k² - 16k
----.-----|-----.--------------|------.----------
k=-1 k=0 k=1 k=16 k=20
>0 0 <0 0 >0 <-- k² - 16k
----------|--------------------|-----------------
k=0 k=16
BMW said: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 ?
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.CompuChip said: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.
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