Question

Let $\left(x_1, x_2\right),\left(x_2, x_3\right)$ and $\left(x_3, x_1\right)$ are respectively the roots of $x^2-2 a x+2=0, x^2-2 b x+$ $3=0$ and $x^2-2 c x+6=0$, where $x_1, x_2, x_3>0$. Find the value of $(a+b+c)$.

Answer 1

Clearly, $x _1 x _2=2, x _2 x _3=3$ and $x _3 x _1=6$
$\Rightarrow\left( x _1 x _2\right)\left( x _2 x _3\right)\left( x _3 x _1\right)=36$ or $x _1 x _2 x _3=6$ (As, $x _1, x _2, x _3$ are all positive.)
So, $x_1=\frac{x_1 x_2 x_3}{x_2 x_3}=\frac{6}{3}=2$
Similarly, $x_2=\frac{x_1 x_2 x_3}{x_1 x_3}=\frac{6}{6}=1$
and $x_3=\frac{x_1 x_2 x_3}{x_1 x_2}=\frac{6}{2}=3$
Also, we have $2 a = x _1+ x _2=3 ; 2 b = x _2+ x _3=4$ and $2 c = x _3+ x _1=5$
Hence, $2(a+b+c)=12$.
$\Rightarrow( a + b + c )=6$.