Expressions with rationalization (review)

Text

Evaluate, rationalizing where necessary:

• (a) $\dfrac{\left(\dfrac{1+\sqrt{3}}{\sqrt{3}-1}\right)^2-\dfrac{\left(\sqrt{3}-1\right)^2}{\left(1+\sqrt{3}\right)^2}}{8\sqrt{3}}$;
• (b) $\dfrac{2}{\left(\sqrt{2}-1\right)^2}+\dfrac{\sqrt{3}+\sqrt{6}}{\sqrt{6}-\sqrt{3}}-\dfrac{\left(\sqrt{3}+\sqrt{6}\right)^2}{\left(\sqrt{2}+1\right)^2}$;
• (c) $\dfrac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}+\dfrac{\left(\sqrt{2}-\sqrt{5}\right)^2\left(\sqrt{2}+\sqrt{5}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2\left(\sqrt{2}+\sqrt{3}\right)^2}$;
• (d) $\left(2+\sqrt{2}\right)^2+\left(\sqrt{2}+1\right)^2\left(1+\dfrac{\left(2-\sqrt{2}\right)\left(2\sqrt{6}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{3}}\right)}{\sqrt{2}+\sqrt{3}}\right)$.

Solution

(a) Rationalize the inner fraction by the conjugate $\sqrt3+1$: $\frac{1+\sqrt3}{\sqrt3-1}=\frac{(1+\sqrt3)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\frac{(1+\sqrt3)^2}{2}=2+\sqrt3,$ so the first term of the numerator is $(2+\sqrt3)^2=7+4\sqrt3$. The second is its reciprocal: $\frac{(\sqrt3-1)^2}{(1+\sqrt3)^2}=\frac{1}{(2+\sqrt3)^2}=\frac{1}{7+4\sqrt3}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3.$ The numerator is $(7+4\sqrt3)-(7-4\sqrt3)=8\sqrt3$, and dividing by $8\sqrt3$ gives $1$.

(b) $\dfrac{2}{(\sqrt2-1)^2}=\dfrac{2}{3-2\sqrt2}=\dfrac{2(3+2\sqrt2)}{9-8}=6+4\sqrt2$; $\dfrac{\sqrt3+\sqrt6}{\sqrt6-\sqrt3}=\dfrac{\sqrt3(1+\sqrt2)}{\sqrt3(\sqrt2-1)}=\dfrac{1+\sqrt2}{\sqrt2-1}=\dfrac{(1+\sqrt2)^2}{2-1}=3+2\sqrt2$; $\dfrac{(\sqrt3+\sqrt6)^2}{(\sqrt2+1)^2}=\dfrac{9+6\sqrt2}{3+2\sqrt2}=\dfrac{3(3+2\sqrt2)}{3+2\sqrt2}=3$. Adding: $(6+4\sqrt2)+(3+2\sqrt2)-3=6+6\sqrt2$.

(c) $\dfrac{(\sqrt2+\sqrt3)^2}{(\sqrt2-\sqrt3)^2}=\left(\dfrac{(\sqrt2+\sqrt3)^2}{2-3}\right)^2=\big(-(5+2\sqrt6)\big)^2=49+20\sqrt6$; the second fraction has numerator $\big[(\sqrt2-\sqrt5)(\sqrt2+\sqrt5)\big]^2=(2-5)^2=9$ and denominator $\big[(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)\big]^2=(2-3)^2=1$, so it equals $9$. Total: $49+20\sqrt6+9=58+20\sqrt6$.

(d) Rationalize $\dfrac{\sqrt3-\sqrt2}{\sqrt2+\sqrt3}=\dfrac{(\sqrt3-\sqrt2)^2}{3-2}=5-2\sqrt6$, hence $2\sqrt6+\big(5-2\sqrt6\big)=5$. Then $(2-\sqrt2)\cdot5=10-5\sqrt2$ and, dividing by $\sqrt2+\sqrt3$ (multiply by $\sqrt3-\sqrt2$, denominator $=1$): $\frac{10-5\sqrt2}{\sqrt2+\sqrt3}=(10-5\sqrt2)(\sqrt3-\sqrt2)=10+10\sqrt3-10\sqrt2-5\sqrt6.$ Add $1$: $11+10\sqrt3-10\sqrt2-5\sqrt6$; multiply by $(\sqrt2+1)^2=3+2\sqrt2$ to get $-7+10\sqrt3-8\sqrt2+5\sqrt6$; finally add $(2+\sqrt2)^2=6+4\sqrt2$: $-1+10\sqrt3-4\sqrt2+5\sqrt6.$

$\boxed{(a)\ 1\qquad (b)\ 6+6\sqrt2\qquad (c)\ 58+20\sqrt6\qquad (d)\ 10\sqrt3+5\sqrt6-4\sqrt2-1}$