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Substitute $\frac{P}{Q}\; = \; \frac{4 +\sqrt2}{2+\sqrt2}$
Then rationalize the denominator by multiplying
numerator by conjugate of
2 + $\sqrt2 \;which\;is\;2\;-\sqrt2$
therefore $\frac{4\;+\sqrt2}{2 + \sqrt2}$ x $\frac{2 - \sqrt2}{2\;-\sqrt2}$
= $\frac{8 - 4\sqrt2\; + 2\sqrt2 \; -2}{4\; -2\sqrt2 \; + 2\sqrt2 - 2}$
= $\frac{6\; -2\sqrt2}{2}$ = 3 - $\sqrt2$
= a + b$\sqrt c$
equating corresponding elements
johnmulu answered the question on March 8, 2017 at 06:45
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