Question

If x= 3 + √8 and y = 3-√8 then
1+/x sq. + 1/y sq.​

Answer 1

Answer:

$\frac{1}{x {}^{2} } + \frac{1}{y {}^{2} } \\ ( \frac{1}{x} + \frac{1}{y} ) {}^{2} - 2 \times \frac{1}{x} \times \frac{1}{y} \\ ( \frac{x + y}{xy} ) {}^{2} - \frac{2}{xy} \\ \\ putting \: the \: values \\ = 6 {}^{2} - \frac{2}{1} = 34$

now ....X=3+√8

1/X=3-√8

and ...y=3-√8

therefore...1/y=3+√8

now

$\frac{1}{x} + \frac{1}{y} = 3 - \sqrt{8} + 3 + \sqrt{8} = 6 \\ \\ xy = (3 + \sqrt{8} )(3 - \sqrt{8} ) = 1$