# 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?

Aug 22, 2016

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 $2 n$ then the sum of the squares is:

${\left(2 n\right)}^{2} + {\left(3 n\right)}^{2} + {\left(4 n\right)}^{2} = \left({2}^{2} + {3}^{2} + {4}^{2}\right) {n}^{2} = 29 {n}^{2}$

Since we are told that this is $116$ we find:

${n}^{2} = \frac{116}{29} = 4$

Hence $n = \pm 2$

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 $\pm 8$, in magnitude.

If you label your unknown variable in lieu of the ratios as $x$, then you can have the ratio represented as $2 x : 3 x : 4 x :$.

So, what you have is:

${\left(2 x\right)}^{2} + {\left(3 x\right)}^{2} + {\left(4 x\right)}^{2} = 116$

$4 {x}^{2} + 9 {x}^{2} + 16 {x}^{2} = 116$

$29 {x}^{2} = 116$

${x}^{2} = \frac{116}{29} = 4$

$\textcolor{b l u e}{x = \pm 2}$

Therefore, the largest number is $4 x = 4 \left(\pm 2\right) = \boldsymbol{\pm 8}$.

To check our work:

$2 x = 2 \left(\pm 2\right) = \pm 4$
$3 x = 3 \left(\pm 2\right) = \pm 6$
$4 x = 4 \left(\pm 2\right) = \pm 8$

Indeed, $\pm 4 : \pm 6 : \pm 8 = \pm 2 : \pm 3 : \pm 4$, and:

${\left(\pm 4\right)}^{2} + {\left(\pm 6\right)}^{2} + {\left(\pm 8\right)}^{2} = 16 + 36 + 64$

$= 52 + 64 = \textcolor{g r e e n}{116}$