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# A bit confused by Quadratic Inequalities watch

1. So I've been doing these questions and getting most of them right, but I'm still not exactly sure what it is I'm doing.

e.g x^2-x-6=0

The answer is {x:x<-2} and {x:x>3}. I did this by drawing the graph and looking at the values for which the solution is bigger than X but I don't actually get what i'm doing, I'm just following what I learnt in the video. Why is the area I'm looking at outside the curves? I get it's bigger than 0 so its above 0 on the x axis (i think) but what exactly am I doing? I know it sounds stupid lol :/

Furthermore I got stuck with this question:

Use set notation to describe the set of vlaues for x for which:

x^2-x-6>0 and 10-2x<5

So I got x<-2 or x>3 and also 2.5<x.

I saw that the answer was {x:x>3} but I put {x:x<-2} union {x:x>3}.

Again, I'm not 100% as to what I'm doing so I'm not sure why I'm wrong. Is it because 2.5<x and x>3 overlap on the number line? And if so, why do you completely disregard x<-2?

Sorry this is a long post but im lowkey baffed
thanks
2. Plot
y =Q(x)
y< Q(x)
On desmos
3. (Original post by emperorCode)
Plot
y =Q(x)
y< Q(x)
On desmos
So all the blue is shaded outside the reigon of the graph? what does that mean? when the equation of the graph is bigger than y the values x must be outside the roots?
4. Sorry, made an error before

You're solving for when the blue curve is above the y-axis - the two red regions can't happen at the same time - so it has to be
x<-2. OR x> 3
Which is a union
5. (Original post by emperorCode)

You're solving for when the blue curve is above the y-axis - the two red regions can't happen at the same time - so it has to be
x<-2. OR x> 3
Which is a union
Ok, that makes slightly more sense.

For x^2-x-12>0 and x>-5 the answer is {x:-5<x<-3} U {x:x>4}

Is this because x>-5 overlaps with both inequalities so satisfies both values of x from x^2-x-12>0 ?

6. The regions satisfying the quadratic is the orange
The regions satisfying the linear is the purple
(The other orange section is under the origin)
The trick is to find the regions for each graph individually using the method from before and then combine them using an intersection.

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