Find the values of m for which the equation has 2 roots

[Math Tutor]

Find the values of m for which the equation has 2 different roots.
[tex](x - 1)(mx^2 - x + 2m - 1) = 0[/tex]

[Guest]

It is analyzed the equation: mx^2-x+2m-1

1). f(x_0) = f`(x_0) = 0 => m = (1+/- sqrt(3))/4 => x_1 = 1, x_2 = +/-sqrt(3) - 1

2). Delta = 0 => m = (1+/- sqrt(3))/4 => x_1 = 1, x_2 = +/-sqrt(3) - 1


y0urshad0w

[Math Tutor]

I do not understand why do you do these calculations:

1). f(x_0) = f`(x_0) = 0 => m = (1+/- sqrt(3))/4 => x_1 = 1, x_2 = +/-sqrt(3) - 1

[Guest]

[tex]m=\frac{1-\sqrt{3}}{4}[/tex] and [tex]m=\frac{1+\sqrt{3}}{4}[/tex]
its what i get ;]

[Math Tutor]

Yes last answers are correct.
Let's explain the solution:
x-1= 0 -> the first root
so [tex]mx^2-x+2m-1 = 0[/tex] must have exactly one root so we have to find for which m the discriminant=0

[Guest]

Mrs. Teacher, you must master mathematics in order to understand it. There are still some awesome ways to
work it out. YS