Question
Rationales the denominator and simplify:
$\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}$
$\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}$
✓
Answer
$\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt{48}-\sqrt{18}$
$=\frac{\big(4\sqrt3+5\sqrt2\big)\big(\sqrt{48}-\sqrt{18}\big)}{\big(\sqrt{48}+\sqrt{18}\big)\big(\sqrt{48}-\sqrt{18}\big)}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(4\sqrt3+5\sqrt2\big)\big(\sqrt{48}-\sqrt{18}\big)}{48-18}$
$=\frac{48-12\sqrt6+20\sqrt6-30}{30}$
$=\frac{18+8\sqrt6}{30}$
$=\frac{9+4\sqrt6}{15}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt{48}-\sqrt{18}$
$=\frac{\big(4\sqrt3+5\sqrt2\big)\big(\sqrt{48}-\sqrt{18}\big)}{\big(\sqrt{48}+\sqrt{18}\big)\big(\sqrt{48}-\sqrt{18}\big)}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(4\sqrt3+5\sqrt2\big)\big(\sqrt{48}-\sqrt{18}\big)}{48-18}$
$=\frac{48-12\sqrt6+20\sqrt6-30}{30}$
$=\frac{18+8\sqrt6}{30}$
$=\frac{9+4\sqrt6}{15}$
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