If \frac{x^2+y^2+z^2-64}{xy-yz-zx}=-2 and x + y = 32, then find the value of z.

Simplification Test 5

Question : 26 of 40

Marks: +1, -0

{x^2+y^2+z^2-64}/{xy-yz-zx}=-2

and

x + y = 32

, then find the value of z.

Solution:

{x^2+y^2+z^2-64}/{xy-yz-zx}=-2

x^2 + y^2 + z^2 - 64

= - 2 (xy - yz - zx)

...(i)
Now,

(x + y- z) ^2

= x^2 + y^2 + z^2 + 2 (xy - yz -zx)

⇒ (3z -z)^2

= x^2 + y^2 + z^2 + 2 (xy - yz - zx)

(∵ x + y = 3z)

∴ x^2 + y^2 + z^2 - 4z^2

= -2 (xy - yz - zx)

...(ii)
From (i) and (ii), we get,

4z^2 = 64 ⇒ z^2 = 16

∴ z = 4

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