Question

find the value of (x+y+z). if (x-3)²+(x-4)²+(z-5)²=0 and xyz are real numbers​

Answer 1

EXPLANATION.

Equations : (x - 3)² + (y - 4)² + (z - 5)² = 0.

xyz are real numbers.

We can write equations as,

⇒ (x - 3)² = 0.

⇒ x - 3 = 0.

⇒ x = 3.

⇒ (y - 4)² = 0.

⇒ y - 4 = 0.

⇒ y = 4.

⇒ (z - 5)² = 0.

⇒ z - 5 = 0.

⇒ z = 5.

Value of : x + y + z.

⇒ x + y + z = 3 + 4 + 5.

⇒ x + y + z = 12.

Answer 2

To Find :

The value of (x+y+z)

Given :

(x-3)²+(y-4)²+(z-5)² = 0

Process :

As we can clearly see that the numbers given are real number. Hence, the square of the number must be positive. As we know, sum of positive number is not 0, so the value of All the term is 0 . Now, with thus equation we will solve it and find out the value of x,y and z.

Solution:

As we have already said, value of all the terms is 0, hence :

(x-3)² = 0

Or, (x-3) = √0

Or, x = 0+3

Or, x = 3

(y-4)² = 0

Or, (y-4) = √0

Or, y = 0+4

Or, y = 4

(z-5)² = 0

Or, z-5 = √0

Or, z = 0+5

Or, z = 5

Hence, x+y+z

= 3+4+5

= 7+5

= 12

Hence, x+y+z = 12