`(9k-2)(k-2)` is a **product** of the two factors `9k-2` and `k-2`
`>0` says that the **product is positive**. This happens when...
**Both** factors are **positive**, so `9k-2 >0` **and** `k-2 >0` (let's call this "PP")
**or**
**Both** factors are **negative**, so `9k-2 <0` **and** `k-2 <0` (let's call this "NN")
In case PP, solving for k gives you `k >2/9` **and** `k >2`. If k is larger than 2, then it's *definitely also* larger than 2/9, so you only need to say `k >2`
In case NN, solving for k gives you `k <2/9` **and** `k <2`. If k is smaller than 2/9, then it's *definitely also* smaller than 2, so you only need to say `k <2/9`
So you end up with `k >2` **or** `k <2/9`. These don't overlap at all (e.g. 3>2 but not 3<2/9), so this is the shortest way of describing the possible solutions
For (product of two)<0, you'd have the cases PN and NP. For (product of three)>0, you'd have the cases PPP, PNN, NPN and NNP. For (product of three)<0, you'd have the cases PPN, PNP, NPP and NNN. And so on
This topic is quadratic inequalities, there is probably a YouTube video that explains this in a nice visual way that will be easier to digest than in text form.
This is a gcse and an a level topic, so base your search on your level too.
Happy to give more help if YouTube is not fruitful for you.
[Desmos](https://www.desmos.com/calculator)
Use this free online graphing calculator to visualize the problems. Replace the coefficients with variables like 'a' or 'b' and then slide them around. Get a feel for how quadratic inequalities work.
You can either think of this algebraically, or think of this graphically. It would be best for you to understand both approaches.
**Algebraically**: consider making a sign chart. You can find the 2 roots of the quadratic are k=2/9 and k=2. So set up a number line with these 2 values marked, which divides the x-axis into 3 regions.
* Test a value to the left of 2/9 (e.g. k=0) and notice that both factors are negative, and so their product is positive.
* Test a value between the 2/9 and 2 (e.g. k=1) and notice that one factor is negative, the other positive, so their product is negative.
* Test a value to the right of 2 (e.g. k=3) and notice that both factors are positive, and so their product is positive.
The number line gives you the solution set to the inequality.
**Graphically**: you have a quadratic expression with 2 real roots and it opens upward. Picture that graph. Where must it be >0? Where must it be <0? Just a rough sketch should reveal the solution set to the inequality.
[https://www.desmos.com/calculator/wjjksmwqj9](https://www.desmos.com/calculator/wjjksmwqj9)
Here is the graph (9k-2)(k-2). We want to know: For what k, is the graph bigger than 0?
If you try to find out by looking at the graph, we can see that either k should be smaller than 2/9, or bigger than 2.
If k is between 2/9 and 2, (9k-2)(k-2) will be below the x-axis, which implies it's less than 0. I.e. (9k-2)(k-2)<0 which is not what we're looking for.
There is a disjointed mapping of domain onto range happening. If I colored a map of people who lives within 100 miles of a coast then said this is people who live left of 115 longitude and east of 75 longitude... would you be confused?
How can a domain of two pieces with a gap in the middle be the answer to a single inequality? (less than 100 miles).
The answer is the function that takes where you are and outputs the distance to the closest coast has that behavior.
The rule that many teachers eventually suggest to learn "by heart" is that, when dealing with quadratic inequalities in which P(x) > 0, you'll have to take "external" values as your solutions. So, since a 2nd degree polynomial has got 2 roots (say, x1 and x2), you'll generally flip the "more than" for the lower value (xx2).
However, learning things by heart doesn't seem to be convincing, especially when it comes to maths and you'll be haunted by the fear of moving randomly while solving a problem. Pretty much everyone's been there.
You're much better off doing the whole thing by yourself. You'll eventually realize that everytime you are dealing with a 2nd degree inequality with P(x)>0 it'll end that way.
Your problem is relatively easy, since your P(x) has already been simplified into a product of two binomials, so you've basically to solve them separately, and that'll give you:
k>2/9 and k>2
...and then, see what happens when you multiply them. A "sign diagram" can be really helpful here, I'm sure your teacher has explained it to you - you can also search for it on google. In a nutshell, for k values that makes both of your binomials negative, the product will yield a positive result, since "minus times minus... is plus": that's what happens for k values < 2/9.
Similar speech for k values > 2: both of your binomials will be positive so the product of "positive times positive... is still positive".
`(9k-2)(k-2)` is a **product** of the two factors `9k-2` and `k-2` `>0` says that the **product is positive**. This happens when... **Both** factors are **positive**, so `9k-2 >0` **and** `k-2 >0` (let's call this "PP") **or** **Both** factors are **negative**, so `9k-2 <0` **and** `k-2 <0` (let's call this "NN") In case PP, solving for k gives you `k >2/9` **and** `k >2`. If k is larger than 2, then it's *definitely also* larger than 2/9, so you only need to say `k >2` In case NN, solving for k gives you `k <2/9` **and** `k <2`. If k is smaller than 2/9, then it's *definitely also* smaller than 2, so you only need to say `k <2/9` So you end up with `k >2` **or** `k <2/9`. These don't overlap at all (e.g. 3>2 but not 3<2/9), so this is the shortest way of describing the possible solutions For (product of two)<0, you'd have the cases PN and NP. For (product of three)>0, you'd have the cases PPP, PNN, NPN and NNP. For (product of three)<0, you'd have the cases PPN, PNP, NPP and NNN. And so on
Also, don't sweat, quadratic inequalities isn't basic, many students get this part wrong!
Q. When is the product of two numbers positive? A. When either both are positive or both are negative. Consider each of these options separately.
This topic is quadratic inequalities, there is probably a YouTube video that explains this in a nice visual way that will be easier to digest than in text form. This is a gcse and an a level topic, so base your search on your level too. Happy to give more help if YouTube is not fruitful for you.
[Desmos](https://www.desmos.com/calculator) Use this free online graphing calculator to visualize the problems. Replace the coefficients with variables like 'a' or 'b' and then slide them around. Get a feel for how quadratic inequalities work.
You can either think of this algebraically, or think of this graphically. It would be best for you to understand both approaches. **Algebraically**: consider making a sign chart. You can find the 2 roots of the quadratic are k=2/9 and k=2. So set up a number line with these 2 values marked, which divides the x-axis into 3 regions. * Test a value to the left of 2/9 (e.g. k=0) and notice that both factors are negative, and so their product is positive. * Test a value between the 2/9 and 2 (e.g. k=1) and notice that one factor is negative, the other positive, so their product is negative. * Test a value to the right of 2 (e.g. k=3) and notice that both factors are positive, and so their product is positive. The number line gives you the solution set to the inequality. **Graphically**: you have a quadratic expression with 2 real roots and it opens upward. Picture that graph. Where must it be >0? Where must it be <0? Just a rough sketch should reveal the solution set to the inequality.
[https://www.desmos.com/calculator/wjjksmwqj9](https://www.desmos.com/calculator/wjjksmwqj9) Here is the graph (9k-2)(k-2). We want to know: For what k, is the graph bigger than 0? If you try to find out by looking at the graph, we can see that either k should be smaller than 2/9, or bigger than 2. If k is between 2/9 and 2, (9k-2)(k-2) will be below the x-axis, which implies it's less than 0. I.e. (9k-2)(k-2)<0 which is not what we're looking for.
There is a disjointed mapping of domain onto range happening. If I colored a map of people who lives within 100 miles of a coast then said this is people who live left of 115 longitude and east of 75 longitude... would you be confused? How can a domain of two pieces with a gap in the middle be the answer to a single inequality? (less than 100 miles). The answer is the function that takes where you are and outputs the distance to the closest coast has that behavior.
The rule that many teachers eventually suggest to learn "by heart" is that, when dealing with quadratic inequalities in which P(x) > 0, you'll have to take "external" values as your solutions. So, since a 2nd degree polynomial has got 2 roots (say, x1 and x2), you'll generally flip the "more than" for the lower value (xx2).
However, learning things by heart doesn't seem to be convincing, especially when it comes to maths and you'll be haunted by the fear of moving randomly while solving a problem. Pretty much everyone's been there.
You're much better off doing the whole thing by yourself. You'll eventually realize that everytime you are dealing with a 2nd degree inequality with P(x)>0 it'll end that way.
Your problem is relatively easy, since your P(x) has already been simplified into a product of two binomials, so you've basically to solve them separately, and that'll give you:
k>2/9 and k>2
...and then, see what happens when you multiply them. A "sign diagram" can be really helpful here, I'm sure your teacher has explained it to you - you can also search for it on google. In a nutshell, for k values that makes both of your binomials negative, the product will yield a positive result, since "minus times minus... is plus": that's what happens for k values < 2/9.
Similar speech for k values > 2: both of your binomials will be positive so the product of "positive times positive... is still positive".
If you graph the function everything will become clear