Question

x= 7+4√3

xy=1
then
$\frac{1}{ {x}^{2} } + \: \frac{1}{ {y}^{2} } \: = \:$

Options-
1. 194
2. 19
3. 192
4. 14​

Answer 1

If x = 7 + 4√3 and xy = 1 then 1/x² + 1/y².

➜ 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

Answer 2

Answer:

option 1) 194

i hope it will help u mate❤❤