4 Marks Questions

Question 14 Marks

Find the values of $a$ and $b$ in the following: $\frac{7+\sqrt{5}}{7-\sqrt{5}}=\frac{7-\sqrt{5}}{7+\sqrt{5}}=\text{a}+\frac{7}{11}\sqrt{5}\text{b}$

Answer

$\frac{7+\sqrt{5}}{7-\sqrt{5}}\times\frac{7+\sqrt{5}}{7+\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}\times\frac{7-\sqrt{5}}{7-\sqrt{5}}=\text{a}+\frac{7}{11}\sqrt{5}\text{b}$
$\frac{(7+\sqrt{5})^2}{(7)^2-(\sqrt{5})^2}-\frac{(7-\sqrt{5})^2}{(7)^2-(\sqrt{5})^2}=\text{a}+\frac{7}{11}\sqrt{5}\text{b}$
$=\frac{49+5+14\sqrt{5}}{49-5}-\frac{49+5-14\sqrt{5}}{49-5}=\text{a}+\frac{7}{11}\sqrt{5}\text{b}$
$=\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}=\text{a}+\frac{7}{11}\sqrt{5}\text{b}=\frac{28\sqrt{5}}{44}$
$\Rightarrow\ \frac{7\sqrt{5}}{11}=\text{a}+\frac{7}{11}\sqrt{5}\text{b}$
$\Rightarrow\ 0+\frac{7\sqrt{5}}{11}=\text{a}+\frac{7}{11}\sqrt{5}\text{b}$ Thus, $a = 0$ and $b = 1.$

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Question 24 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

$\text{x}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{3+2+2\sqrt{3\times2}}{3-2}$
$\text{(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\Rightarrow\ \text{x}=\frac{5+2\sqrt{6}}{1}=5+2\sqrt{6}$
$\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$\text{y}=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{3+2-2\sqrt{2\times3}}{3-2}$
$\text{(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\text{y}=5-2\sqrt{6}$ 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, $\text{(x+y})^2=\text{x}^2+\text{y}^2+2\text{xy}$
$\text{x}^2+\text{y}^2=\text{(x+y})^2-2\text{xy}$
$\therefore\ \text{x}^2+\text{y}^2=(10)^2-(1)^2=100-2=98$

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