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
+33616 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.
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+33616 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.
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