JAMB - Mathematics (1999 - No. 3)
Find the value of X if \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\)
3√2+4
3√2-4
3-2√2
4+2√2
Explanation
Start your solution by cross-multiplying,
\(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\)
[x - √2] √2 = x + √2
where √2×√2=2
x√2 - √2 * √2 = x + √2
then collect like terms
x√2 - x = √2 + 2
and factorize accordingly to get the unknown.
x(√2 - 1) = √2 + 2
x = \(\frac{√2 + 2}{√2 - 1}\)
rationalize
x = \(\frac{√2 + 2}{√2 - 1}\) * \(\frac{√2 + 1}{√2 + 1}\)
x = \(\frac{√4 + √2 + 2√2 + 2}{√4 + √2 - √2 - 1}\)
x = \(\frac{2 + 3√2 + 2}{2 - 1}\)
x = \(\frac{3√2 + 4}{1}\)
x = 3√2 + 4
Comments (0)
