# 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

The three numbers are

Hence the largest number is either

#### Explanation:

If the smallest is

#(2n)^2+(3n)^2+(4n)^2 = (2^2+3^2+4^2)n^2 = 29n^2#

Since we are told that this is

#n^2 = 116/29 = 4#

Hence

So the three numbers are

So the largest number is either

[ Note that the greatest number in the two cases is

I got

If you label your unknown variable in lieu of the ratios as

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

To check our work:

#2x = 2(pm2) = pm4#

#3x = 3(pm2) = pm6#

#4x = 4(pm2) = pm8#

Indeed,

#(pm4)^2 + (pm6)^2 + (pm8)^2 = 16 + 36 + 64#

#= 52 + 64 = color(green)(116)#