Question

If a, b, c and d are in proportion prove that `(13a + 17b)/(13c + 17d) = sqrt((2ma^2 - 3nb^2)/(2mc^2 - 3nd^2)`

Answer 1

a, b, c and d are in proportion

`a/b = c/d = k(say)`

Then a = bk and c = dk

`L.H.S =  (13a + 17b)/(13c + 17d) = (13(bk) + 17b)/(13(dk) + 17d) = (b(13k + 17))/(d(13k + 17)) = b/d`

R.H.S = `sqrt((2ma^2 - 3nb^2)/(2mc^2 - 3nd^2)) = sqrt((2m(bk)^2 - 3nb^2)/(2m(dk)^2 - 3nd^2)) = sqrt((b^2(2mk^2 - 3n))/(d^2(2mk^2 - 3n))) = b/d`

Hence L.H.S = R.H.S