Question

If $x+y+z=5$ and $xy+yz+zx=3$ , then least and largest value of $x$ are

• A. $\frac{10}{3},5$
• B. $-1,\frac{13}{3}$
• C. $\frac{17}{3},7$
• D. None of these

Answer 1

Answer: (B)

$xy+yz+zx=3\&z=5-x-y$
so $xy+y\left(\right.5-x-y\left.\right)+\left(\right.5-x-y\left.\right)x=0$
$\Rightarrow y^{2}+y\left(\right.x-5\left.\right)+\left(x^{2} - 5 x + 3\right)=0$
since $y$ is real, therefore $D\geq 0$
$\left(x-5-4.1 \cdot\left(x^2-5 x+3\right) \geq 0\right.$
$\Rightarrow \left(\right.x+1\left.\right)\left(\right.3x-13\left.\right)\leq 0$
$\therefore -1\leq x\leq \frac{13}{3}$