Question

(6^1/2-3/6^1/2+3) + (6^1/2+3/6^1/2-3) = a+b5^1/2​
find values of a and b

Answer 1

Dear Student,
Here is the solution of your asked query:
$$\begin{gathered} \left(\frac{6^{\frac{1}{2}}-3}{6^{\frac{1}{2}}+3}\right)+\left(\frac{6^{\frac{1}{2}}+3}{6^{\frac{1}{2}}-3}\right)=\mathrm{a}+\mathrm{b}{5}^{\frac{1}{2}} \\ \Rightarrow \left(\frac{\sqrt{6}-3}{\sqrt{6}+3}\right)+\left(\frac{\sqrt{6}+3}{\sqrt{6}-3}\right)=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \Rightarrow \left(\frac{\sqrt{6}-3}{\sqrt{6}+3}\times \frac{\sqrt{6}-3}{\sqrt{6}-3}\right)+\left(\frac{\sqrt{6}+3}{\sqrt{6}-3}\times \frac{\sqrt{6}+3}{\sqrt{6}+3}\right)=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \Rightarrow \left[\frac{{\left(\sqrt{6}-3\right)}^2}{{\left(\sqrt{6}\right)}^2-3^2}\right]+\left[\frac{{\left(\sqrt{6}+3\right)}^2}{{\left(\sqrt{6}\right)}^2-3^2}\right]=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \Rightarrow \left[\frac{{\left(\sqrt{6}\right)}^2+3^2-2\times 3\times \sqrt{6}}{6-3}\right]+\left[\frac{{\left(\sqrt{6}\right)}^2+3^2+2\times 3\times \sqrt{6}}{6-3}\right]=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \Rightarrow \left[\frac{6+9-6\sqrt{6}}{3}\right]+\left[\frac{6+9+6\sqrt{6}}{3}\right]=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \Rightarrow \frac{15-6\sqrt{6}+15+6\sqrt{6}}{3}=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \Rightarrow 10=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \Rightarrow 10+0\sqrt{5}=\mathrm{a}+\mathrm{b}\sqrt{5} \\ \mathrm{Comparing}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}; \\ \mathrm{a}=10\mathrm{and}\mathrm{b}=0 \end{gathered}$$
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