The sum of the squares of three numbers is #116#, and the ratio of these numbers is #2:3:4#. What is the largest number?

Question

Abigail Allen · Accepted Answer

The three numbers are #4#, #6# and #8# or #-4#, #-6# and #-8#.

Hence the largest number is either #8# or #-8# If the smallest is #2n# then the sum of the squares is: #(2n)^2+(3n)^2+(4n)^2 = (2^2+3^2+4^2)n^2 = 29n^2# Since we are told that this is #116# we find: #n^2 = 116/29 = 4# Hence #n = +-2# So the three numbers are #4#, #6# and #8# or #-4#, #-6# and #-8#. So the largest number is either #8# or #-8#. [ Note that the greatest number in the two cases is #8# or #-4# ]

Ethan Adkins · Answer

I got #8#, assuming positivity. If you wish, you could say that since #-2:-3:-4 = 2:3:4#, your highest number is #pm8#, in magnitude. If you label your unknown variable in lieu of the ratios as #x#, then you can have the ratio represented as #2x:3x:4x:#. Thus, what you possess is: #(2x)^2 + (3x)^2 + (4x)^2 = 116# #4x^2 + 9x^2 + 16x^2 = 116# #29x^2 = 116# #x^2 = 116/29 = 4# #color(blue)(x = pm2)# Therefore, the largest number is #4x = 4(pm2) = bb(pm8)#. To verify our work: #2x = 2(pm2) = pm4#
#3x = 3(pm2) = pm6#
#4x = 4(pm2) = pm8# Indeed, #pm4:pm6:pm8 = pm2:pm3:pm4#, and: #(pm4)^2 + (pm6)^2 + (pm8)^2 = 16 + 36 + 64# #= 52 + 64 = color(green)(116)#

Abigail Allen · Answer

undefined The largest number is $ 4 \times \sqrt{\frac{116}{29}} $, which is approximately $ 4.$