 ## Question

secondary 3 | A Maths

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##### Hehe

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pls help i cry so hard my braincells

Date Posted: 1 week ago
Views: 46
NotATutor
1 week ago
x^2 + 6 or x^2 - 6?
Hehe
1 week ago
SORRYYYYY ITS +6

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1st
Date Posted: 1 week ago
NotATutor
1 week ago
i counter checked already... it had to be -3 < m <2.
then it is x^2 + 6.
Explain My last two steps?
NotATutor
1 week ago
You had not learned this method of solving inequality? using number line...
Hehe
1 week ago
cause for part a) i got (m+3)(m-2)>0
so since its >0 i put
positive x positive >0
m+3>0 and m-2>0
m>-3 and m>2
or
negative x negative >0
m+3<0 and m-2<0
m<-3 and m<2
NotATutor
1 week ago
in the first place, why greater than 0? discriminant should be less than zero... Let us sort this out first before I tell you how to solve the inequality
Hehe
1 week ago
oh its less than 0
wait why isit less than 0 how do you know if its all positive
NotATutor
1 week ago
read my solution x^2 - 2mx - m + 6 > 0
Do you get this?
Hehe
1 week ago
yeah
Hehe
1 week ago
how do you know if your discriminant is less or more than 0?
so if eqn is >0 D is < 0?
so if my eqn is <0 D is > 0??
NotATutor
1 week ago
Then the discriminant should be less than 0. D less tham 0 means the quadratic never meet the x axis.
Hehe
1 week ago
ohhh okay thank you i get it now
NotATutor
1 week ago
discriminat > 0 means the curve cuts x-axis at two points. D = 0 means the curve touches the x-axis once. D <0 means the curve never meet the x-axis.....
the expression now always greater than 0 means it never meet the x-axis... means D < 0.
You slowly think...
Hehe
1 week ago
so for eg when the eqn is >0 it means it did not cut x axis so discriminant is < 0
while when eqn <0 it means it cuts x axis so discriminant is > 0
NotATutor
1 week ago
expression less than zero, it also never cuts the x-axis,,,,, D is still < 0
Hehe
1 week ago
but if eqn <0 it cuts x axis?
NotATutor
1 week ago
expression > 0 or < 0 , both cases also the curve never meet the x-axis. .... Because the x-axis is the f(x) = 0 line. > 0 or < 0 ..... both also never = 0
Hehe
1 week ago
so only an eqn = 0 then can d >0 or = 0 and have point(s) of intersection
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Date Posted: 1 week ago
Hehe
1 week ago
so eqn < 0 just means negative curve graph and does not cut x axis
NotATutor
1 week ago
right..... if the y = ax^2 + b x + c and the question never say anything, then it could be any of the six cases... if a > 0 then it should be any of the U shapes. If a < 0 then it should be any of the n shapes
NotATutor
1 week ago
If clear, then we can discuss about the (m + 3)(m - 2).... we have different method of solving this type of inequality... What you teacher taught you?
Hehe
1 week ago
my cher say d= eg; (m+3)(m-2)>0 it is (m+3)>0 , (m-2)>0 or (m+3)<0 ,(m-2)<0
(positive x positive or neg x neg)
then for (m+3)(m-2)<0,
m+3>0, m-2<0 or m+3<0 m-2>0
(pos x neg or neg x pos)
so for this qn
(m+3)(m-2)>0
so since its >0 i put
positive x positive >0
m+3>0 and m-2>0
m>-3 and m>2
or
negative x negative >0
m+3<0 and m-2<0
m<-3 and m<2
NotATutor
1 week ago
this method is very difficult to learn anyway, in this case is should be (m + 3)(m- 2) <0
1. There are two factors: (m + 3) and (m -2)
2. to have a negative product (< 0), one of them > 0 then the other one must be < 0.
3. two cases
4. case 1 (m + 3) > 0 and (m - 2) < 0
m > -3 and m < 2
which part of the number line satisfy both?
-3 < m < 2
5. case 2 (m + 2) < 0 and (m - 2) > 0
m < -3 and m > 2
which part of the number line satisfy both?
none....
6. final set of solution: -3 < m < 2 