Question

State which of the following are irrational numbers.
$(3–\sqrt{6}{)}^2$

Answer 1

$(3–\sqrt{6}{)}^2$

Let us simplify,

By using $(a–b{)}^2=a^2+b^2–2ab$

$(3–\sqrt{6}{)}^2=3^2+(\sqrt{6}{)}^2–\mathrm{2.3.}\sqrt{6}$

$=9+6–6\sqrt{6}$

$=15–6\sqrt{6}$

Hence it is an irrational number.