Question
Examine whether the following number are rational or irrational:
$\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
$\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
✓
Answer
$\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
$=\sqrt{4\times2}+4\sqrt{16\times2}-6\sqrt{2}$
$=2\sqrt{2}+16\sqrt{2}-6\sqrt{2}$
$=12\sqrt{2}$
Thus, the given number is irrational.
$=\sqrt{4\times2}+4\sqrt{16\times2}-6\sqrt{2}$
$=2\sqrt{2}+16\sqrt{2}-6\sqrt{2}$
$=12\sqrt{2}$
Thus, the given number is irrational.
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