# If X = 7 + 4 √ 3 and Xy =1, Then 1 X 2 + 1 Y 2 = - Mathematics

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MCQ

If $x = 7 + 4\sqrt{3}$ and xy =1, then $\frac{1}{x^2} + \frac{1}{y^2} =$

#### Options

• 64

• 134

• 194

• 1/49

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#### Solution

Given that x=7+4sqrt3, xy = 1

Hence  y is given as

y=1/x

1/x = 1/(7+4sqrt3)

We need to find  1/x^2+ 1/y^2

We know that rationalization factor for   7+4sqrt3 is 7-4sqrt3. We will multiply numerator and denominator of the given expression 1/(7+4sqrt3)by,7-4sqrt3 to get

1/x = 1/(7+4sqrt3) xx (7-4sqrt3)/(7-4sqrt3)

= (7-4sqrt3)/((7)^2 (4sqrt3)^2)

 = (7-4sqrt3)/(49 - 48)

 = 7-4sqrt3

Since xy=1so we have

x=1/y

Therefore,

1/x^2 + 1/y^2 = ( 7 - sqrt3)^2 + (7+4sqrt3)^2

 = 7^2 + (4sqrt3)^2 - 2 xx 7 xx 4sqrt3 + 7 ^2 +(4 sqrt3)^2 + 2 xx 7 xx 4sqrt3

= 49 + 48 - 14 sqrt3 + 49 +48 +14sqrt3

= 194

Concept: Laws of Exponents for Real Numbers
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#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 3 Rationalisation
Exercise 3.4 | Q 11 | Page 17

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