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

Question

If x+y+z=5 and xy+yz+zx=3 , then least and largest value of x are (A) (10/3),5 (B) -1,(13/3) (C) (17/3),7 (D) None of these

Tardigrade Admin · Accepted Answer

xy+yz+zx=3 &z=5-x-y so xy+y(.5-x-y.)+(.5-x-y.)x=0 ⇒ y2+y(.x-5.)+(x2 - 5 x + 3)=0 since y is real, therefore D≥ 0 (x-5-4.1 ⋅(x2-5 x+3) ≥ 0. ⇒ (.x+1.)(.3x-13.)≤ 0 ∴ -1≤ x≤ (13/3)

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

$\frac{10}{3}, 5$

$- 1, \frac{13}{3}$

$\frac{17}{3}, 7$

None of these

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

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

310​,5

−1,313​

317​,7