If sqrt 50 - K sqrt 8 frac 2 sqrt 2 find K - WAEC Mathematics 2013 Question
If \(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\), find K
A
-2
B
-1
C
1
D
2
correct option: d
\(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\)
\(\sqrt{50} - \frac{2}{\sqrt{2}}\) = K\(\sqrt{8}\)
= \(\sqrt{2} \times 25 - \frac{2}{\sqrt{2}}\)
= K \(\sqrt{4 \times 2}\)
\(\frac{5\sqrt{2}}{1} - \frac{2}{\sqrt{2}}\) = 2K\(\sqrt{2}\)
\(\frac{5\sqrt{4} - 2}{\sqrt{2}} = 2K\sqrt{2}\)
\(\frac{10 - 2}{\sqrt{2}} = 2K \sqrt{2}\)
\(\frac{8}{\sqrt{2}} = \frac{2K\sqrt{2}}{1}\)
= 2k\(\sqrt{2} \times \sqrt{2}\) = 8
2k \(\sqrt{4}\) = 8
2k x 2 = 8
4k = 8
k = \(\frac{8}{4}\)
k = 2
\(\sqrt{50} - \frac{2}{\sqrt{2}}\) = K\(\sqrt{8}\)
= \(\sqrt{2} \times 25 - \frac{2}{\sqrt{2}}\)
= K \(\sqrt{4 \times 2}\)
\(\frac{5\sqrt{2}}{1} - \frac{2}{\sqrt{2}}\) = 2K\(\sqrt{2}\)
\(\frac{5\sqrt{4} - 2}{\sqrt{2}} = 2K\sqrt{2}\)
\(\frac{10 - 2}{\sqrt{2}} = 2K \sqrt{2}\)
\(\frac{8}{\sqrt{2}} = \frac{2K\sqrt{2}}{1}\)
= 2k\(\sqrt{2} \times \sqrt{2}\) = 8
2k \(\sqrt{4}\) = 8
2k x 2 = 8
4k = 8
k = \(\frac{8}{4}\)
k = 2
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