If x, y and z are real numbers such that $x + y + z = 5$ and $x y + y z + z x = 3$ , what is the largest value of 'x' can have?

\frac{5}{3}

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\sqrt{19}

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\frac{13}{3}

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17

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Solution

The correct option is B $\sqrt{19}$
$(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)$
$\Rightarrow 25 = x^{2} + y^{2} + z^{2} + 6$
$\Rightarrow x^{2} = 19 - (y^{2} + z^{2})$
As $(y^{2} + z^{2}) \geq 0$
$\Rightarrow x_{m a x} = 19$
$\Rightarrow x_{m a x} = \sqrt{19}$

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Similar questions

x, y

z

x + y + z = 5

x y + y z + z x = 3

x

Find the value of $A + B + C$ if :

$A = (7 x y + 5 y z - 3 z x)$
$B = (4 y z + 9 z x - 4 y)$
$C = (- 3 x z + 5 x - 2 x y)$

When $7 x y + 5 y z - 3 z x$ , $4 y z + 9 z x - 4 y$ and $- 3 x z + 5 x - 2 x y$ are added, we get

x + y + z = 0

x y + y z + z x = 3

x^{3} y + y^{3} z + z^{3} x

(x, y, z)

x^{y} = z,

y^{z} = x

z^{x} = y

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