Question

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?

Answer 1

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` ]

Answer 2

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.

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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)`