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Algebra and Applications Test 3

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Question : 10 of 25

Marks: +1, -0

If

x=3+2\sqrt{2}

xy=1

, then the value of

\frac{x^2+3xy+y^2}{x^2-3xy+y^2}

is:

$\frac{30}{31}$
$\frac{70}{31}$
$\frac{37}{31}$
$\frac{34}{31}$

Solution:

Concept:

Use

xy=1

to find

y

as the reciprocal of

x

, then simplify the expression using symmetric identities.

Explanation:

Given

x=3+2\sqrt{2}

and

xy=1

, so

y=\frac{1}{x}

.

Rationalize:

y=\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}=3-2\sqrt{2}

.

Now,

x+y=(3+2\sqrt{2})+(3-2\sqrt{2})=6

.

Also,

x-y=(3+2\sqrt{2})-(3-2\sqrt{2})=4\sqrt{2}

.

And

xy=1

.

Rewrite numerator:

x^2+3xy+y^2=(x+y)^2+xy=6^2+1=37

.

Rewrite denominator:

x^2-3xy+y^2=(x-y)^2-xy=(4\sqrt{2})^2-1=32-1=31

.

Therefore,

\frac{x^2+3xy+y^2}{x^2-3xy+y^2}=\frac{37}{31}

.

Answer:

\frac{37}{31}

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