Question 14 Marks
If $\text{x}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ then find the value of $x^2 + y^2.$
Answer
View full question & answer→$\text{x}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{3+2+2\sqrt{3\times2}}{3-2}$ $\text{(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\Rightarrow\ \text{x}=\frac{5+2\sqrt{6}}{1}=5+2\sqrt{6}$
$\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$\text{y}=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{3+2-2\sqrt{2\times3}}{3-2}$ $\text{(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\text{y}=5-2\sqrt{6}$
Now, $\text{x+y}=5+2\sqrt{6}+5-2\sqrt{6}=10$
And, $\text{xy}=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}\times\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})}=1$
We know that,
$\text{(x+y})^2=\text{x}^2+\text{y}^2+2\text{xy}$
$\text{x}^2+\text{y}^2=\text{(x+y})^2-2\text{xy}$
$\therefore\ \text{x}^2+\text{y}^2=(10)^2-(1)^2=100-2=98$
$=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{3+2+2\sqrt{3\times2}}{3-2}$ $\text{(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\Rightarrow\ \text{x}=\frac{5+2\sqrt{6}}{1}=5+2\sqrt{6}$
$\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$\text{y}=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{3+2-2\sqrt{2\times3}}{3-2}$ $\text{(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\text{y}=5-2\sqrt{6}$
Now, $\text{x+y}=5+2\sqrt{6}+5-2\sqrt{6}=10$
And, $\text{xy}=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}\times\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})}=1$
We know that,
$\text{(x+y})^2=\text{x}^2+\text{y}^2+2\text{xy}$
$\text{x}^2+\text{y}^2=\text{(x+y})^2-2\text{xy}$
$\therefore\ \text{x}^2+\text{y}^2=(10)^2-(1)^2=100-2=98$