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Mathematics

If (x1)12+(y2)12+(z3)12=0(x - 1)^{\dfrac{1}{2}} + (y - 2)^{\dfrac{1}{2}} + (z - 3)^{\dfrac{1}{2}} = 0, then find the values of x, y, z.

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Answer

Given,

(x1)12+(y2)12+(z3)12=0\Rightarrow (x - 1)^{\dfrac{1}{2}} + (y - 2)^{\dfrac{1}{2}} + (z - 3)^{\dfrac{1}{2}} = 0

We can write,

(x1)12(x - 1)^{\dfrac{1}{2}}= 0 ….(1)

(y2)12(y - 2)^{\dfrac{1}{2}} = 0 ….(2)

(z3)12(z - 3)^{\dfrac{1}{2}} = 0 ….(3)

Solving equation 1,

(x1)12(x - 1)^{\dfrac{1}{2}}= 0

Squaring both sides,

⇒ (x - 1) = 0

⇒ x = 1

Solving equation 2,

(y2)12(y - 2)^{\dfrac{1}{2}} = 0

Squaring both sides,

⇒ (y - 2) = 0

⇒ y = 2

Solving equation 3,

(z3)12(z - 3)^{\dfrac{1}{2}} = 0

Squaring both sides,

⇒ (z - 3) = 0

⇒ z = 3

Hence, the values of x, y, z are 1, 2, 3.

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