Question

find three numbers in the ratio 3:4:6 the sum of whose squares is 976

Answer 1

Hi...☺️

Here is your answer...✌️

Given that,

Three numbers are in the ratio
3:4:6

Let the common multiple be x

Then the numbers are
3x , 4x , 6x

Now
According to the question,

(3x)² + (4x)² + (6x)² = 976

9x² + 16x² + 36x² = 976

61x² = 976

x² = 976/61

x² = 16

=> x = ±4

If x = 4

3x = 3×4 = 12
4x = 4×4 = 16
6x = 6×4 = 24

If x = -4

3x = 3×(-4) = -12
4x = 4×(-4) = -16
6x = 6×(-4) = -24

HENCE,

The numbers are
12 , 16 , 24

OR

-12, -16 , -24

Answer 2

Let the common multiple be x

thus

the numbers are

3x
4x
6x

From the given condition

${(3x)}^{2} + {(4x)}^{2} + {(6x)}^{2} = 976$
${9x}^{2} + {16x}^{2} + {36x}^{2} = 976$
$61 {x}^{2} = 976$
${x}^{2} = \frac{976}{61}$
${x}^{2} = 16$
$x = 4$
thus the required numbers are

3x = 3(4)=12

4x = 4(4)=16

6x = 6(4)=24