If frac 1 sqrt 2 1 - sqrt 2 is expressed in the... - JAMB Mathematics 2008 Question
If \(\frac{1 + \sqrt{2}}{1 - \sqrt{2}}\) is expressed in the form x + y\(\sqrt{2}\). Find the values of x and y
A
(-3, -2)
B
(-2, 3)
C
(3, 2)
D
(2, -3)
correct option: a
\(\frac{1 + \sqrt{2}}{1 - \sqrt{2}}\) x \(\frac{1 + \sqrt{2}}{1 + \sqrt{2}}\)
= \(\frac{(1 + \sqrt{2}) (1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}\)
= \(\frac{1 + \sqrt{2} + \sqrt{2} + (\sqrt{2})^2}{1 + \sqrt{2} - \sqrt{2} - (\sqrt{2})^2}\)
= \(\frac{1 + 2\sqrt{2} + \sqrt{2}}{1^2 - 2}\)
= \(\frac{3 + 2\sqrt{2}}{-1}\)
-3 - 2\(\sqrt{2}\) = x + y\(\sqrt{2}\)
x = -3, y = -2
= (-3, -2)
= \(\frac{(1 + \sqrt{2}) (1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}\)
= \(\frac{1 + \sqrt{2} + \sqrt{2} + (\sqrt{2})^2}{1 + \sqrt{2} - \sqrt{2} - (\sqrt{2})^2}\)
= \(\frac{1 + 2\sqrt{2} + \sqrt{2}}{1^2 - 2}\)
= \(\frac{3 + 2\sqrt{2}}{-1}\)
-3 - 2\(\sqrt{2}\) = x + y\(\sqrt{2}\)
x = -3, y = -2
= (-3, -2)
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