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?

2 Answers
Aug 22, 2016

Answer:

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

Hence the largest number is either #8# or #-8#

Explanation:

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# ]

Aug 22, 2016

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:#.

So, what you have 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 check 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)#