Question

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

Answer 1

If (x-1) ^2 +(y-3) ^2 +(z-5) ^2 +(t-7) ^2=0 then find xyzt +16?

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xyzt + 16 = 121

Solution below:

(x-1) ^2 + (y-3) ^2 + (z-5) ^2 + (t-7) ^2 = 0

assuming:

x, y, z, t ∈ R

this can be re-written in the form:

a^2 + b^2 + c^2 + d^2 = 0

where,

a = (x-1)

b = (y-3)

c = (z-5)

d = (t-7)

for real numbers,

a^2 >= 0

b^2 >= 0

c^2 >= 0

d^2 >= 0

In other words,

each of a^2, b^2, c^2 and d^2 can never achieve a negative value for it to satisfy the overall equation

i.e. LHS ∈ [0, ∞)

For equation to achieve a state of LHS = 0,

each a^2, b^2, c^2 and d^2 have to be 0

Therefore,

a = 0 ⇒ x - 1 = 0 ⇒ x = 1

b = 0 ⇒ y - 3 = 0 ⇒ y = 3

c = 0 ⇒ z - 5 = 0 ⇒ z = 5

d = 0 ⇒ t - 7 = 0 ⇒ t = 7

Hence,

xyzt + 16 = 121

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