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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

 Mar 23, 2015

Best Answer 

 #1
avatar+27566 
+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.

.

 Mar 23, 2015
 #1
avatar+27566 
+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
 #2
avatar+99377 
0

That is because   1-2m    would have to equal 0.

 Mar 23, 2015

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