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 that $x$ 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 C $\sqrt{19}$
$(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + x z)$
$\Rightarrow 25 = x^{2} + y^{2} + z^{2} + 6$
$\Rightarrow x^{2} + y^{2} + z^{2} = 19$
Here the maximum value which $x^{2}$ can take is $19$ i.e when $y^{2} + z^{2} = 0$
$∴ x^{2} = 19 \Rightarrow x = \sqrt{19}$

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Similar questions

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

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

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

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 = 6

x y + y z + z x = 12

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

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