Question:
The required solution of $(x^2-y^2-z^2)p+2xyq = 2xz $, will be
The required solution of $(x^2-y^2-z^2)p+2xyq = 2xz $, will be
Updated On: Mar 21, 2024
- $x^2-y^2-z^2=zf(y/z)$
- $x^2 + y^2+z^2=zf(yz)$
- $x^2 + y^2+z^2=zf(y/z)$
- $- x^2 y^2-z^2=zf(z/y)$
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The Correct Option is C
Solution and Explanation
The Correct answer is option (C) : $x^2 + y^2+z^2=zf(y/z)$
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