• Maths Problem.
    18 replies, posted
Hey. As you might know, Israel's education level sucks. And I came across something my mid-high never taught me in my transfer to high-school project: functions with x^3. The question I was asked it "Suggest a way to find the solution of the inequation: (x²-4)(x-2)<0" How would I do this? I don't want the answer I just want to know how. By the way, I'm going to year 10. Unlike most high schools, many Israeli high schools start with year 10 instead of the usual 9.
Just multiply everything in the first bracket (x^2-4) by x, and then by -2. So (x²-4)(x-2) = (x^3-4x)+(-2x^2+8) = x^3 - 2x^2 - 4x + 8
Yeah, I know that much. But how does it help me get to the solution?
That would expand to: x^2 + 0x - 4 X^2 + 2x - 2x - 4 [x(x + 2) -2 (x+2)] Which means that: (x-2)(x+2)(x-2) < 0 So: x < 2 or x < -2
then you factor it using pasqel's triangle and once that's been done you find the zeros of the equation [editline]12:21PM[/editline] lucaio's got it
hahaha, I'm so stupid. Lucaio's post is right. I'm a prime example of how to fail your a-level exams.
That was "fun"! Haven't done maths in a while!
But wait, the zone -2<x<2 is positive, so how is it possible that the two answers are x<2 and x<-2? THAT'S where I got stuck, I actually already did what you all posted except for pasqel's triangle which I have never heard about.
wow i havent done this stuff in a long while :/ alright, you need to plug the values back into the equation
Poop you got me there. :/ That would just mean the answer would be x<2 then, as x<2 includes x<-2 Actually, ignore that. My mathematical skills seem to have gone now that I have done my exams.
:psyduck:
do your OWN FUCKIN homework
I seem to have confused the fast threads section of Facepunch as well as myself. InsaneInThe, are you dumb? I'm not asking for the answer, I'm asking to learn how to do it so I CAN do my own homework.
pay attention in class then
[QUOTE=InsaneInThe;16352214]pay attention in class then[/QUOTE] I said that my stupid school didn't teach us th... Did you even read the OP?!
no
I'll just be leaving now.....
I guess the suggestion should include two stages: The first is to solve that inequation and get the two results, x<2 and x<-2, and the second is to find the 0 points to conclude if one of these answers are impossible. For example, in this function one of the 0 points is (-2,0), and since x<2 includes (-2,0), it is not lesser than 0, therefore the real answer is x<-2.
[QUOTE=Lucaio;16351826]That would expand to: x^2 + 0x - 4 X^2 + 2x - 2x - 4 [x(x + 2) -2 (x+2)] Which means that: (x-2)(x+2)(x-2) < 0 So: x < 2 or x < -2[/QUOTE] You almost got it. Now notice that this simplifies to (x+2)(x-2)² < 0. Since a real number squared is always positive (or zero, but that doesn't satisfy the inequality), you can divide both sides by (x-2)² and get (x+2) < 0, so the answer is x < -2. Note that this method wouldn't work if the terms didn't factor so nicely, a more general way would be to look at the terms (x²-4) and (x-2) separately and find regions where one was positive and the other negative so that their product was < 0.
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