ABC26GN0998 · Simplification
Subject: General Aptitude · Chapter: Simplification · Exam: · Marks: · Difficulty:
If $(x+y)^{2}-z^{2}=4,(y+z)^{2}-x^{2}=9,(z+x)^{2}-y^{2}$ $=36$, what is/are the value(s) of $x+y+z$ ?
Answer
Explanation
$\left[(x+y)^{2}-z^{2}\right]+\left[(y+z)^{2}-x^{2}\right]+\left[(z+x)^{2}-y^{2}\right]$ $$=4+9+36 \begin{aligned} \Rightarrow \quad & (x+y+z)(x+y-z)+(x+y+z) \\ & (y+z-x)+(x+y+z)(z+x-y)=49 \end{aligned}$$ $\Rightarrow \quad(x+y+z)[(x+y-z)+(y+z-x)+(z+x-y)]=49$ $$\Rightarrow \quad(x+y+z)(x+y+z)=49 \Rightarrow(x+y+z)^{2}=49$$ $\Rightarrow \quad(x+y+z)= \pm 7$.
Explanation as extracted from the printed page; notation may be imperfect.
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