If the roots of the quadratic equation $m{ x }^{ 2 }-nx+k=0$ are $\tan { { 33 }^{ o } } $ and $\tan { { 12 }^{ o } } $, then the value of $\cfrac { 2m+n+k }{ m } $ is equal to

By Vieta's formulas, the sum of the roots is tan(33) + tan(12) = n/m and the product is tan(33) * tan(12) = k/m. Using the tangent addition formula for 45 degrees, we know tan(45) = tan(33 + 12) = (tan(33) + tan(12)) / (1 - tan(33)*tan(12)). Since tan(45) = 1, we have the equation 1 = (n/m) / (1 - k/m), which simplifies to 1 - k/m = n/m, or m - k = n. Substituting n = m - k into the given expression (2m + n + k) / m yields (2m + m - k + k) / m, which equals 3m / m = 3.

• Thanks
• Wrong explanation