Question

If |x-1| + |y-2|+ (z-3)^2 ≤ 0 then find the value of x + y + z (where x, y, z €
R)
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Answer 1

Answer: x + y + z = 6

Concept: Value of mod function is always positive

Value of square of a term is always positive.

Given: |x-1| + |y-2|+ (z-3)^2 ≤ 0

To find: x + y + z

Step-by-step explanation:

Value of mod function is always positive.

Value of square of any number is always positive

|x-1| + |y-2|+ (z-3)^2 ≤ 0

In the following inequality only mod and squares are present

So the above expression can never become negative

hence,

|x-1| + |y-2|+ (z-3)^2 = 0

Now value of mod function and square can never be negative,

So for the expression to become zero individual terms needs to be zero,

Hence,

|x-1| = 0

|y-2| = 0

(z-3)^2 = 0

Hence,

x-1 = 0

x = 1

y - 2 =0

y = 2

z-3 = 0

z = 3

Now we have to find x + y + z

x + y + z = 1 + 2 + 3

x + y + z = 6

Answer: x + y + z = 6

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Answer 2

Answer:

x+y+z=6

Step-by-step explanation:

Given that |x-1| + |y-2|+ (z-3)^2 ≤ 0.

Now, anything under mod is considered as positive. Also squares are positive. Hence, keeping these points in mind, we get,

x-1=0

=> x=1

Again,

y-2=0

=> y=2

Again,

z-3=0

=> z=3

Therefore, x+y+z = 1+2+3 = 6