Question

If x= 7+4√3 and xy=1, then find 1/x² + 1/y²​

Answer 1

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$\huge\bold\purple{Answer:-}$

⏩ Given:-

* x= 7+4√3

* xy=1 or y= 1/x

Now, taking y=1/x and substituting the value of x, we get,

y= 1/7+4√3

=> y= 1× (7-4√3) / 7+4√3 × (7-4√3)

=> y= 7-4√3 / (7)² - (4√3)²

=> y= 7- 4√3/ 49 -48

=> y= 7 - 4√3

Then, x+y will be,

7+4√3 + (7-4√3)

= 14

Now, finding the value of 1/x² + 1/y²,

(Write all the numerators above the least common denominator x²y²)

= y² + x²/ x²y²

= (y+x)² -2(xy )/(xy)²

Now, substitute the given values.

= (14)² - 2(1)/ (1)²

= 196 - 2

= 194 → The value of 1/x² + 1/y²

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Answer 2

$\huge\underline\mathfrak\purple{Solution}$