Question
Rationales the denominator and simplify:$\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
✓
Answer
$\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt3-\sqrt2$
$=\frac{\big(\sqrt3-\sqrt2\big)\big(\sqrt3-\sqrt2\big)}{\big(\sqrt3+\sqrt2\big)\big(\sqrt3-\sqrt2\big)}$
As we know, $(\text{a}-\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}$
As we know, $(\text{a}-\text{b})^2=(\text{a}^2-2\times\text{a}\times\text{b}+\text{b}^2)$
$=\frac{3-2\sqrt3\sqrt2+2}{1}=5-2\sqrt6$
$=\frac{\big(\sqrt3-\sqrt2\big)\big(\sqrt3-\sqrt2\big)}{\big(\sqrt3+\sqrt2\big)\big(\sqrt3-\sqrt2\big)}$
As we know, $(\text{a}-\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}$
As we know, $(\text{a}-\text{b})^2=(\text{a}^2-2\times\text{a}\times\text{b}+\text{b}^2)$
$=\frac{3-2\sqrt3\sqrt2+2}{1}=5-2\sqrt6$
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