Asked by mandy
Given that the quadratic equation is (2m+3)x^2+(4m-2)x-(m+1)=0.Find the value of m,if
•one of the root of the equation is negative
•one of the root of the equation is reciprocal of the other
•one of the root of the equation is negative
•one of the root of the equation is reciprocal of the other
Answers
Answered by
Steve
the roots are
x = -[(2m-1)±√(6m^2+m+4)]/(2m+3)
The product of the roots is
-(m+1)/(2m+3)
If that is negative, then
(m+1)>0 and (2m+3)>0, so m > -1
(m+1)<0 and (2m+3)<0, so m < -3/2
If the roots are reciprocals, then
-(m+1) = 2m+3
3m = -4
m = -4/3
x = -[(2m-1)±√(6m^2+m+4)]/(2m+3)
The product of the roots is
-(m+1)/(2m+3)
If that is negative, then
(m+1)>0 and (2m+3)>0, so m > -1
(m+1)<0 and (2m+3)<0, so m < -3/2
If the roots are reciprocals, then
-(m+1) = 2m+3
3m = -4
m = -4/3