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i counter checked already... it had to be -3 < m <2.

then it is x^2 + 6.

Explain My last two steps?

then it is x^2 + 6.

Explain My last two steps?

You had not learned this method of solving inequality? using number line...

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

oh its less than 0

wait why isit less than 0 how do you know if its all positive

wait why isit less than 0 how do you know if its all positive

read my solution x^2 - 2mx - m + 6 > 0

Do you get this?

Do you get this?

yeah

so if eqn is >0 D is < 0?

so if my eqn is <0 D is > 0??

ohhh okay thank you i get it now

the expression now always greater than 0 means it never meet the x-axis... means D < 0.

You slowly think...

while when eqn <0 it means it cuts x axis so discriminant is > 0

expression less than zero, it also never cuts the x-axis,,,,, D is still < 0

but if eqn <0 it cuts x axis?

so only an eqn = 0 then can d >0 or = 0 and have point(s) of intersection

so eqn < 0 just means negative curve graph and does not cut x axis

(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

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