Rationalise the denominaor in $\dfrac{7}{\sqrt{3}+2}$
$-7\sqrt{3}+14$
$7\sqrt{3}+14$
$-7\sqrt{3}-14$
$7\sqrt{3}-14$
Explanation:
$\dfrac{7}{\sqrt{3}+2}=\dfrac{7}{\sqrt{3}+2} \times \dfrac{\sqrt{3}-2}{\sqrt{3}-2}$ [multiply the numerator and denominator by $\sqrt{3}-2$]
=$\dfrac{7\sqrt{3}-14}{3-4}$
=$\dfrac{7\sqrt{3}-14}{-1}$
=$-7\sqrt{3}+14$
$\dfrac{7}{\sqrt{3}+2}=\dfrac{7}{\sqrt{3}+2} \times \dfrac{\sqrt{3}-2}{\sqrt{3}-2}$ [multiply the numerator and denominator by $\sqrt{3}-2$]
=$\dfrac{7\sqrt{3}-14}{3-4}$
=$\dfrac{7\sqrt{3}-14}{-1}$
=$-7\sqrt{3}+14$