Question

If x = ${\displaystyle \frac{1}{(3-2\sqrt{2})}}$ and y = ${\displaystyle \frac{1}{(3+2\sqrt{2})}}$, find the values of

x² + y²

Answer 1

x² + y²

(x² + y²) = (x + y)² - 2xy    ----(1)

Now, x + y = ${\displaystyle \frac{1}{(3-2\sqrt{2})}+\frac{1}{(3+2\sqrt{2})}}$

= ${\displaystyle \frac{(3+2\sqrt{2})+(3-2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}}$

= ${\displaystyle \frac{6}{9-8}}$

= 6

and xy = ${\displaystyle \frac{1}{(3-2\sqrt{2})}\times \frac{1}{(3+2\sqrt{2})}}$

= ${\displaystyle \frac{1}{9-8}}$

= 1

substituting the valuesin (1), we get
(x² + y²)
= (x + y)² - 2xy
= 36 - 2
= 34
(x² + y²)
= 34