If the roots of the equation (m - n)x^2 + (n | vedclass

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(B) For the quadratic equation $Ax^2 + Bx + C = 0$,if the roots are equal,the discriminant $D = B^2 - 4AC = 0$.Here,$A = (m - n)$,$B = (n - l)$,and $C

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$2m = n + l$

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(B) For the quadratic equation $Ax^2 + Bx + C = 0$,if the roots are equal,the discriminant $D = B^2 - 4AC = 0$.Here,$A = (m - n)$,$B = (n - l)$,and $C = (l - m)$.Substituting these into $B^2 - 4AC = 0$:$(n - l)^2 - 4(m - n)(l - m) = 0$$n^2 + l^2 - 2nl - 4(ml - m^2 - nl + nm) = 0$$n^2 + l^2 - 2nl - 4ml + 4m^2 + 4nl - 4nm = 0$$l^2 + n^2 + 4m^2 + 2nl - 4ml - 4nm = 0$This expression is the expansion of $(l + n - 2m)^2 = 0$.Therefore,$l + n - 2m = 0$,which implies $2m = n + l$.

If the roots of the equation $(m - n)x^2 + (n - l)x + (l - m) = 0$ are equal,then which of the following is true for $l, m,$ and $n$?

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