Selina Solutions Class 9 Mathematics Solutions for Exercise 1(C) in Chapter 1 - Chapter 1- Rational and Irrational Numbers

Question 6 Exercise 1(C)

If x= =\frac{\sqrt{5}-2}{\sqrt{5}+2} \text { and } y=\frac{\sqrt{5}+2}{\sqrt{5}-2} Find:

(i) x²

(ii) y²

(iii) xy

(iv) x² +y² =xy


(i) \begin{array}{l} x^{2}=\left(\frac{\sqrt{5}-2}{\sqrt{5}+2}\right)^{2}=\frac{5+4-4 \sqrt{5}}{5+4+4 \sqrt{5}}=\frac{9-4 \sqrt{5}}{9+4 \sqrt{5}} \\ =\frac{9-4 \sqrt{5}}{9+4 \sqrt{5}} \times\left(\frac{9-4 \sqrt{5}}{9-4 \sqrt{5}}\right)=\frac{(9-4 \sqrt{5})^{2}}{(9)^{2}-(4 \sqrt{5})^{2}} \\ =\frac{81+80-72 \sqrt{5}}{81-80}=161-72 \sqrt{5} \end{array}

\begin{aligned} &\text { (ii) }\\ &\begin{array}{l} y^{2}=\left(\frac{\sqrt{5}+2}{\sqrt{5}-2}\right)^{2}=\frac{5+4+4 \sqrt{5}}{5+4-4 \sqrt{5}}=\frac{9+4 \sqrt{5}}{9-4 \sqrt{5}} \\ =\frac{9+4 \sqrt{5}}{9-4 \sqrt{5}} \times \frac{9+4 \sqrt{5}}{9+4 \sqrt{5}}=\frac{(9+4 \sqrt{5})^{2}}{(9)^{2}-(4 \sqrt{5})^{2}}=\frac{81+80+72 \sqrt{5}}{81-80} \\ =161+72 \sqrt{5} \end{array} \end{aligned}

\begin{aligned} &\text { (iii) }\\ &\frac{(\sqrt{5}-2)}{(\sqrt{5}+2)} \frac{(\sqrt{5}+2)}{(\sqrt{5}-2)}=1 \end{aligned}

\begin{aligned} &\text { (iv) }\\ &x^{2}+y^{2}+x y=161-72 \sqrt{5}+161+72 \sqrt{5}+1\\ &=322+1=323 \end{aligned}

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