g(x)=(-m-1)x+3<f(x)=(m=1)x2+(m-3)x+3m+1

ibanez1608

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Jul 12, 2013
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24
when does
g(x)=(-m-1)x+3<f(x)=(m=1)x2+(m-3)x+3m+1
i got the equation:
(1-m)x2+(2-2m)x-3m+2<0
the answer is m>=1
how can it be???? i dont know if X is +/-
 
when does
g(x)=(-m-1)x+3<f(x)=(m=1)x2+(m-3)x+3m+1
i got the equation:
(1-m)x2+(2-2m)x-3m+2<0
the answer is m>=1
how can it be???? i dont know if X is +/-

Got a headache trying to decipher this. Typos in there too.

Is this what you are trying to do?

When is g(x) < f(x)? With \(\displaystyle g(x)=(-m-1)x+3\ \ and\ \ f(x)=(m-1)x^2+(m-3)x+3m+1\) ??

Plus, don't mix lower case x with upper case X.
 
i got the equation:
(1-m)x2+(2-2m)x-3m+2<0
the answer is m>=1
how can it be???? i dont know if X is +/-

the equation \(\displaystyle ax^2+bx+c \) has always the same sign when there are no roots, and then it has the same sign as \(\displaystyle a\)
So, you want that \(\displaystyle 1-m =<0 => m>=1 \)
 
Got a headache trying to decipher this. Typos in there too.

Is this what you are trying to do?

When is g(x) < f(x)? With \(\displaystyle g(x)=(-m-1)x+3\ \ and\ \ f(x)=(m-1)x^2+(m-3)x+3m+1\) ??

Plus, don't mix lower case x with upper case X.
i didnt mix i solove it and this is what i got, its the same thing
 
the equation \(\displaystyle ax^2+bx+c \) has always the same sign when there are no roots, and then it has the same sign as \(\displaystyle a\)
So, you want that \(\displaystyle 1-m =<0 => m>=1 \)
i didnt understand u
 
i didnt understand u

If you take the discriminant, you have Δ\(\displaystyle =(2-2m)^2-4(1-m)(2-3m)=-8m^2+12m-4 \), with roots \(\displaystyle \frac{1}{2} \) and 1.
The sign of an equation does not change if Δ<0.
You want that your equation (1-m)x2+(2-2m)x-3m+2 is always <0.
So, Δ<0 => \(\displaystyle m< \frac{1}{2} \) & \(\displaystyle m>1 \). (1)
Then, the equation (1-m)x2+(2-2m)x-3m+2 is also always <0 when the term of \(\displaystyle x^2 \) is <0,
so \(\displaystyle 1-m<0 => m>1 \) (2)

From the relations (1) and (2) you have that \(\displaystyle m>1 \).
 
If you take the discriminant, you have Δ\(\displaystyle =(2-2m)^2-4(1-m)(2-3m)=-8m^2+12m-4 \), with roots \(\displaystyle \frac{1}{2} \) and 1.
The sign of an equation does not change if Δ<0.
You want that your equation (1-m)x2+(2-2m)x-3m+2 is always <0.
So, Δ<0 => \(\displaystyle m< \frac{1}{2} \) & \(\displaystyle m>1 \). (1)
Then, the equation (1-m)x2+(2-2m)x-3m+2 is also always <0 when the term of \(\displaystyle x^2 \) is <0,
so \(\displaystyle 1-m<0 => m>1 \) (2)

From the relations (1) and (2) you have that \(\displaystyle m>1 \).

now i got you vary reasonable thanks man
 
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