Question

1. If $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\mathrm{and}b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, Find value of a² + b² – 5ab.
2. If $a+\frac{1}{a}=4$, than the value of $a^2-\frac{1}{a^2}$ is

Answer 1

Dear student
$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{a}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{{\left(\sqrt{3}-\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2-2\sqrt{6}}{3-2} \\ =5-2\sqrt{6} \\ \mathrm{b}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2+2\sqrt{6}}{3-2} \\ =5+2\sqrt{6} \\ \mathrm{So},{\mathrm{a}}^2+{\mathrm{b}}^2-5\mathrm{ab} \\ ={\left(5-2\sqrt{6}\right)}^2+{\left(5+2\sqrt{6}\right)}^2-5\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right) \\ =25+24-\cancel{20\sqrt{6}}+25+24+\cancel{20\sqrt{6}}-5\left(25-24\right) \\ =93 \\ \mathrm{Note}: \\ 1){\left(\mathrm{x}\pm \mathrm{y}\right)}^2={\mathrm{x}}^2+{\mathrm{y}}^2\pm 2\mathrm{xy} \\ 2){\mathrm{x}}^2-{\mathrm{y}}^2=(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}) \end{gathered}$$
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