Question
Examine whether the following number are rational or irrational:
$\big(\sqrt{3}+2\big)^2$
$\big(\sqrt{3}+2\big)^2$
✓
Answer
$\big(\sqrt{3}+2\big)^2$
$=\big(\sqrt{3}\big)^2+2\times2\times\sqrt{3}+(2)^2$
$=3+4\sqrt{3}+4$
$=7+4\sqrt{3}$
Clearly, the given number is irrational.
$=\big(\sqrt{3}\big)^2+2\times2\times\sqrt{3}+(2)^2$
$=3+4\sqrt{3}+4$
$=7+4\sqrt{3}$
Clearly, the given number is irrational.
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