Question

Positive integers x, y, z are in the ratio 2 : 3 : 5. If
${ x}^{2} + \: {y}^{2} + {z}^{2} + 8 \: = 616$
find the values of x, y, z.​

Answer 1

Topic :-

Algebra

Given :-

Positive integers x, y and z are in ratio 2 : 3 : 5.

x² + y² + z² + 8 = 616

To Find :-

Values of x, y and z.

Solution :-

Let us assume that,

x = 2k

y = 3k

z = 5k

2k : 3k : 5k : : 2 : 3 : 5

Hence, our assumption is correct.

Now, it is given that,

x² + y² + z² + 8 = 616

Substituting values,

(2k)² + (3k)² + (5k)² + 8 = 616

4k² + 9k² + 25k² + 8 = 616

38k² + 8 = 616

38k² = 616 - 8

38k² = 608

k² = 608/38

k² = 16

k = ± 4

Since, x, y and z are positive integers, we will take only positive value of 'k'.

Hence, k = 4.

x = 2k = 2(4) = 8

y = 3k = 3(4) = 12

z = 5k = 5(4) = 20

Answer :-

Value of x, y and z are 8, 12 and 20 respectively.

Verification :-

8 : 12 : 20 : : 2 : 3 : 5 and

8² + 12² + 20² + 8

64 + 144 + 400 +8

616

Checked conditions match with given conditions.

Hence, verified!

Answer 2

$\huge\mathtt\blue{A} \mathtt\pink{N}\mathtt\red{S}\mathtt\purple{W}\mathtt\green{E}\mathtt\orange{R}$

$\pink{X=8}$

$\blue{Y=12}$

$\green{Z=20}$

Step-by-step explanation:

Let X+Y+Z=K

Now X,Y,Z are in ratio than there individual value =

X=1/5k

Y=3/10k

Z=k/2

Now put the values in the equation given

$\rightrightarrows (1/5k)²+(3/10k)²+(k/2)²+8=616$

$\rightrightarrows k²/25+9k²/100+k²/4=608$

$\rightrightarrows k²=1600$

$\rightrightarrows k=±40$

But k should be positive

[tex]\rightrightarrows k=40 [/tex]

Now values of X,Y,Z⏬

X=k/5 =8

Y=3k/10= 12

Z=k/2=20