If $x + y = 17$ and $x^{2} + y^{2} = 167$ , what is the value of $x y$ ?

17 + 4 \sqrt{114}

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61

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48 + \sqrt{167}

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122

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Solution

The correct option is B $61$
Given, $x + y = 17$ and $x^{2} + y^{2} = 167$ .

On squaring the equation $x + y = 17$ both the sides, we get,
$\Rightarrow (x + y)^{2} = 17^{2}$ .

We know, $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .

Then,
$\Rightarrow x^{2} + y^{2} + 2 x y = 289$
$\Rightarrow 167 + 2 x y = 289$
$\Rightarrow 2 x y = 289 - 167$
$\Rightarrow 2 x y = 122$
$\Rightarrow x y = 61$ .

Therefore, the value of $x y$ is $61$ .
Hence, option $B$ is correct.

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