If sqrt 72 sqrt 32 - 3 sqrt 18 x sqrt 8 Find th... - WAEC Mathematics 2011 Question
If \(\sqrt{72} + \sqrt{32} - 3 \sqrt{18} = x \sqrt{8}\), Find the value of x
A
1
B
\(\frac{3}{4}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{4}\)
correct option: c
\(\sqrt{2} + \sqrt{32} - 3\sqrt{18} = x\sqrt{8}\)
= \(\sqrt{36 \times 2} + \sqrt{16 \times 2} - 3\sqrt{2 \times 9}\)
= x\(\sqrt{2 \times 4}\)
= 6\(\sqrt{2} + 4\sqrt{2} - 9\sqrt{2} = 2 \times \sqrt{2}\)
\(\sqrt{2} (6 + 4 - 9) = 2x\sqrt{2}\)
\(\sqrt{2} = 2x \sqrt{2}\) divide both sides by \(\sqrt{2}\)
\(\frac{\sqrt{2}}{\sqrt{2}} = \frac{2 \times \sqrt{2}}{\sqrt{2}}\)
1 = 2x
2x = 1
x = \(\frac{1}{2}\)
= \(\sqrt{36 \times 2} + \sqrt{16 \times 2} - 3\sqrt{2 \times 9}\)
= x\(\sqrt{2 \times 4}\)
= 6\(\sqrt{2} + 4\sqrt{2} - 9\sqrt{2} = 2 \times \sqrt{2}\)
\(\sqrt{2} (6 + 4 - 9) = 2x\sqrt{2}\)
\(\sqrt{2} = 2x \sqrt{2}\) divide both sides by \(\sqrt{2}\)
\(\frac{\sqrt{2}}{\sqrt{2}} = \frac{2 \times \sqrt{2}}{\sqrt{2}}\)
1 = 2x
2x = 1
x = \(\frac{1}{2}\)
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