If the equation x^2 - m(2x - 8) - 15 = 0 has | vedclass

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(C) Given the quadratic equation $x^2 - m(2x - 8) - 15 = 0$. Expanding the equation: $x^2 - 2mx + 8m - 15 = 0$. Comparing this with the standard form

Vedclass · Accepted Answer

$3, 5$

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(C) Given the quadratic equation $x^2 - m(2x - 8) - 15 = 0$. Expanding the equation: $x^2 - 2mx + 8m - 15 = 0$. Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 1$,$b = -2m$,and $c = 8m - 15$. For the equation to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$. Substituting the values: $(-2m)^2 - 4(1)(8m - 15) = 0$. $4m^2 - 32m + 60 = 0$. Dividing by $4$: $m^2 - 8m + 15 = 0$. Factoring the quadratic: $(m - 3)(m - 5) = 0$. Therefore,$m = 3$ or $m = 5$.

If the equation $x^2 - m(2x - 8) - 15 = 0$ has equal roots,then $m = ......$

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