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 negatif

Guest Mar 23, 2015

#1**+5 **

The solutions to this quadratic may be expressed as:

$$\\x_+=\frac{1-2m+\sqrt{6m^2+m+4}}{2m+3}\\\\x_-=\frac{1-2m-\sqrt{6m^2+m+4}}{2m+3}$$

There are an infinite number of values of m that have one root negative and one root positive. However, if you are looking for the value of m that makes one root the negative** of the other**, then that is clearly m = 1/2.

.

Alan
Mar 23, 2015

#1**+5 **

Best Answer

The solutions to this quadratic may be expressed as:

$$\\x_+=\frac{1-2m+\sqrt{6m^2+m+4}}{2m+3}\\\\x_-=\frac{1-2m-\sqrt{6m^2+m+4}}{2m+3}$$

There are an infinite number of values of m that have one root negative and one root positive. However, if you are looking for the value of m that makes one root the negative** of the other**, then that is clearly m = 1/2.

.

Alan
Mar 23, 2015