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Mathematics

Rationalize the denominator of :

13−2+1\dfrac{1}{\sqrt{3} - \sqrt{2} + 1}

Rational Irrational Nos

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Answer

Rationalizing,

⇒13−2+1×3−2−13−2−1⇒3−2−1(3−2)2−12⇒3−2−1(3)2+(2)2−2×3×2−1⇒3−2−13+2−26−1⇒3−2−14−26⇒3−2−12(2−6)\Rightarrow \dfrac{1}{\sqrt{3} - \sqrt{2} + 1} \times \dfrac{\sqrt{3} - \sqrt{2} - 1}{\sqrt{3} - \sqrt{2} - 1} \\[1em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2} - 1}{(\sqrt{3} - \sqrt{2})^2 - 1^2} \\[1em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2} - 1}{(\sqrt{3})^2 + (\sqrt{2})^2 - 2 \times \sqrt{3} \times \sqrt{2} - 1} \\[1em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2} - 1}{3 + 2 - 2\sqrt{6} - 1} \\[1em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2} - 1}{4 - 2\sqrt{6}} \\[1em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2} - 1}{2(2 - \sqrt{6})}

Rationalizing again,

⇒3−2−12(2−6)×2+62+6⇒23−22−2+18−12−62(22−62)⇒23−22−2+32−23−62(4−6)⇒23−2(2−3)−2−23−62(4−6)⇒2−2−62×−2⇒−14(2−2−6)⇒14(2+6−2).\Rightarrow \dfrac{\sqrt{3} - \sqrt{2} - 1}{2(2 - \sqrt{6})} \times \dfrac{2 + \sqrt{6}}{2 + \sqrt{6}} \\[1em] \Rightarrow \dfrac{2\sqrt{3} - 2\sqrt{2} - 2 + \sqrt{18} - \sqrt{12} - \sqrt{6}}{2(2^2 - \sqrt{6}^2)} \\[1em] \Rightarrow \dfrac{2\sqrt{3} - 2\sqrt{2} - 2 + 3\sqrt{2} - 2\sqrt{3} - \sqrt{6}}{2(4 - 6)} \\[1em] \Rightarrow \dfrac{\cancel{2\sqrt{3}} - \sqrt{2}(2 - 3) - 2 - \cancel{2\sqrt{3}} - \sqrt{6}}{2(4 - 6)} \\[1em] \Rightarrow \dfrac{\sqrt{2} - 2 - \sqrt{6}}{2 \times -2} \\[1em] \Rightarrow -\dfrac{1}{4}(\sqrt{2} - 2 - \sqrt{6}) \\[1em] \Rightarrow \dfrac{1}{4}(2 + \sqrt{6} - \sqrt{2}).

Hence, solution = 14(2+6−2)\dfrac{1}{4}(2 + \sqrt{6} - \sqrt{2}).

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