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

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(B) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = (m - n)$,$b = (n

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$2m = n + l$

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(B) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = (m - n)$,$b = (n - l)$,and $c = (l - m)$.Substituting these into the condition $b^2 - 4ac = 0$:$(n - l)^2 - 4(m - n)(l - m) = 0$Expanding the terms:$(n^2 + l^2 - 2nl) - 4(ml - m^2 - nl + mn) = 0$$n^2 + l^2 - 2nl - 4ml + 4m^2 + 4nl - 4mn = 0$Rearranging the terms:$l^2 + n^2 + (2m)^2 + 2nl - 4mn - 4ml = 0$This expression is the expansion of $(l + n - 2m)^2 = 0$.Therefore,$l + n - 2m = 0$,which implies $2m = n + l$.This indicates that $l, m, n$ are in Arithmetic Progression $(A.P.)$.

If the equation $(m - n)x^2 + (n - l)x + l - m = 0$ has equal roots,then $l, m$ and $n$ satisfy:

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If the equation $(m - n)x^2 + (n - l)x + l - m = 0$ has equal roots,then $l, m$ and $n$ satisfy:

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Solve the given two equations and select the correct answer from the given options.$I.$ $(17)^{2} + 144 \div 18 = x$$II.$ $(26)^{2} - 18 \times 21 = y$

Solve the given equations and choose the correct option.$I.$ $x = \sqrt{(36)^{1/2} \times (1296)^{1/4}}$$II.$ $2y + 3z = 33$$III.$ $6y + 5z = 71$

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