Please Explain... (1 Viewer)

[Estel]

x^2+2bx+3c=0
What is the condition for b and c for the roots of the equation to be of opposite sign.

Obviously the discriminant will be >0, then b^2>3c.
But how does one obtain the other condition to ensure the opposite sign part?

My workin:
(x+b)^2 -b^2 + 3c=0
therefore sym: x=-b
intercepts:
-b+rt(b^2-3c)
-b-rt(b^2-3c)

then
-b+rt(b^2-3c) >0
-b-rt(b^2-3c) < 0

BUT the answer in the back is
c<0

 

[Heinz]

This is what I was thinking. Since its a quadractic . only 2 roots) you have to make the sum of the roots (-b/a) = 0 So.... -2b = 0 :. b = 0 is one condition. Because b^2 - 4ac has to be greater than 0 for real roots and b = 0 then -4ac (-12c) has to be greater than zero. :. c < 0

Edit: minor typo

 

[Estel]

Thanks.
Why is -4ac = -12c though... isn't a 1?

 

[Heinz]

Maybe I shouldve used different pronumerals
Ax^2 + Bx + C = 0
B^2 - 4AC > 0

but the equation is x^2+2bx+3c=0 so
A = 1, B = 2b and C = 3c
:. -4AC = -4*1*3c = -12c

 

[CM_Tutor]

Originally posted by Heinz
This is what I was thinking. Since its a quadractic . only 2 roots) you have to make the sum of the roots (-b/a) = 0 So.... -2b = 0 :. b = 0 is one condition. Because b^2 - 4ac has to be greater than 0 for real roots and b = 0 then -4ac (-12c) has to be greater than zero. :. c < 0

The answer is c < 0, but not for this reason. The sum of the roosts needs to be zero for the roots to be symmetric about the origin, but that is not a requirement for this question. All we need is one root positive and one negative. Hence, product of the roots must be negative.

ie alpha * beta = C / A < 0
(3c) / 1 < 0
c < 0

 

[Heinz]

Originally posted by CM_Tutor
The answer is c < 0, but not for this reason. The sum of the roosts needs to be zero for the roots to be symmetric about the origin, but that is not a requirement for this question. All we need is one root positive and one negative. Hence, product of the roots must be negative.

ie alpha * beta = C / A < 0
(3c) / 1 < 0
c < 0

I thought something wasnt right with my working since the question never mentioned anything about <b>equal</b> and opposite roots...

 

[velox]

ic....so when question asks
just opposite sign roots: c/a<0
Equal and opposite roots: ????

 

[nike33]

you dont need to memeroise this stuff..just think...let roots be @ and -@...therefore sum = 0 and -ve time +ve is -ve

 

[velox]

ah ic thanks so equal and opposite would be : -b/a = 0