Question on: JAMB Mathematics - 1998
Find the value of k if \(\frac{k}{\sqrt{3} + \sqrt{2}}\) = k\(\sqrt{3 - 2}\)
A
3
B
2
C
\(\sqrt{3}\)
D
\(\sqrt 2\)
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Correct Option: D
\(\frac{k}{\sqrt{3} + \sqrt{2}}\) = k\(\sqrt{3 - 2}\)
\(\frac{k}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)
= k\(\sqrt{3 - 2}\)
= k(\(\sqrt{3} - \sqrt{2}\))
= k\(\sqrt{3 - 2}\)
= k\(\sqrt{3}\) - k\(\sqrt{2}\)
= k\(\sqrt{3 - 2}\)
k2 = \(\sqrt{2}\)
k = \(\frac{2}{\sqrt{2}}\)
= \(\sqrt{2}\)
\(\frac{k}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)
= k\(\sqrt{3 - 2}\)
= k(\(\sqrt{3} - \sqrt{2}\))
= k\(\sqrt{3 - 2}\)
= k\(\sqrt{3}\) - k\(\sqrt{2}\)
= k\(\sqrt{3 - 2}\)
k2 = \(\sqrt{2}\)
k = \(\frac{2}{\sqrt{2}}\)
= \(\sqrt{2}\)
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