quadratic help (1 Viewer)

[kaha167]

find the values of m for which 9x^2 + 6x + m^2 is always positive.
How do i solve this?
Thanks for the help,

 

[ohexploitable]

The coefficient of x² is already +ve so you just need the discriminant to be +ve.

So...

b² - 4ac > 0
36 - 36m² > 0
-36m² > -36
m² < 1
m² - 1 < 0
(m-1)(m+1) < 0

-1 < m < 1

 

[hscishard]

Positive definite. A>0 and discriminant is <0
All values are above the x axis(therefore positive)

b^2-4ac<0
36-36m^2<0
Divide by 36
1-m^2<0
Bring to right hand side
m^2-1>0
(m+1)(m-1)>0
m>1, m<-1

 

[hscishard]

The coefficient of x² is already +ve so you just need the discriminant to be +ve.

So...

b² - 4ac > 0
36 - 36m² > 0
-36m² > -36
m² < 1
m² - 1 < 0
(m-1)(m+1) < 0

-1 < m < 1

:S If the discriminant is positive it's indefinite.

 

[ohexploitable]

Oh lol... yeah... hscishard is correct... hehe

 

[adomad]

or what you can do is draw 9x^2 + 6x . and then shift it up by a value to make it positive definite. then square root the value of the shifting value and then add a plus minus sign in the front.