ABC26GN1212 · Simplification
Subject: General Aptitude · Chapter: Simplification · Exam: · Marks: · Difficulty:
If $\frac{x^{2}+y^{2}+z^{2}-64}{x y-y z-z x}=-2$ and $x+y=3 z$, then the value of $z$ is
(a)2
(b)3
(c)4
(d)None of these
Answer
Explanation
$$\begin{aligned} & \text { Given: } x^{2}+y^{2}+z^{2}-64=-2(x y-y z-z x) \\ & \text { Now, }[x+y+(-z)]^{2}=x^{2}+y^{2}+z^{2}+2(x y-y \\ & \Rightarrow \quad(3 z-z)^{2}=x^{2}+y^{2}+z^{2}+2(x y-y z-z x) \\ & \Rightarrow \quad-2(x y-y z-z x)=\left(x^{2}+y^{2}+z^{2}\right)-(2 z)^{2} \end{aligned}$$ From (i) and (ii), we get: $(2 z)^{2}=64$ $$\begin{aligned} & \Leftrightarrow z^{2}=16 \\ & \Leftrightarrow z=4 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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