Question

If 4x2+9y2+z2=108 and 3xy+3/2yz+zx=54 then find value of x2+y2+z2

Answer 1

Answer:

49

Step-by-step explanation:

Hi,

Given that 4x² + 9y² + z² = 108------(1)

3xy+3/2yz+zx=54-------(2)

Multiplying the above equation(2) by 4 we get

12xy + 6yz + 4zx = 216-----(3)

Multiplying equation (1) by 2, we get

8x² + 18y² + 2z² = 216-----(4)

Subtracting equation (3) from (4), we get

8x² + 18y² + 2z² - (12xy + 6yz + 4zx) = 0

4x² - 12xy + 9y² + 4x² - 4zx + z² + 9y² - 6yz + z² = 0

(2x - 3y)² + (2x - z)² + (3y - z)² = 0

But square of a number is always ≥ 0.

Sum of the squares equal to zero, it means that each term is equal to 0

So, we get 2x - 3y = 0, 2x -z = 0

2x = 3y = z

Substituting the above values in equation (1), we get

4x² + 9y² + z² = 108

z² + z² + z² = 108

z² = 36

z = ±6,

2x = z

x = z/2 = ±3

3y = z

y = z/3 = ±2

So, the value of x² + y² + z²

= 6² + 3² + 2²

= 49

Hope, it helps !