Algebra and Applications Test 3

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Question : 10 of 25

Marks: +1, -0

If

x=3+2√2

and

xy=1

, then the value of

{x^2+3xy+y^2}/{x^2-3xy+y^2}

is:

$30/31$
$70/31$
$37/31$
$34/31$

Solution:

x=3+2√2

and

xy=1

⇒y=1/x=1/{3+2√2}×{3-2√2}/{3-2√2}

=3-2√2

∴ {x^2+3xy+y^2}/{x^2-3xy+y^2}

={(x+y)^2+xy}/{ (x-y)^2-xy}

=37/31

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