If $x= \frac{\sqrt{p^{2}+q^{2}}+\sqrt{p^{2}-q^{2}}}{{\sqrt{p^{2}+q^{2}}-\sqrt{p^{2}-q^{2}}}}$ then $q^{2}x^{2}-2p^{2}x+q^{2}$ equals to :

- $3$
- $-1$
- $-2$
- $0$

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1
votes

If $x= \frac{\sqrt{p^{2}+q^{2}}+\sqrt{p^{2}-q^{2}}}{{\sqrt{p^{2}+q^{2}}-\sqrt{p^{2}-q^{2}}}}$ then $q^{2}x^{2}-2p^{2}x+q^{2}$ equals to :

- $3$
- $-1$
- $-2$
- $0$

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1
votes

$ \begin{align} &\frac{x}{1}= \frac{\sqrt{p^{2}+q^{2}}+\sqrt{p^{2}-q^{2}}}{{\sqrt{p^{2}+q^{2}}-\sqrt{p^{2}-q^{2}}}} \\ \implies &\frac{x+1}{x-1} = \frac{\sqrt{p^2 + q^2} }{\sqrt{p^2-q^2}} \qquad \qquad \rightarrow \text{ apply Componendo and Dividendo} \\ \implies & \frac{(x+1)^2}{(x-1)^2} = \frac{p^2+q^2}{p^2-q^2} \qquad \qquad \rightarrow \text{take square on both side} \\ \implies & \frac{(x+1)^2 + (x-1)^2}{(x+1)^2 - (x-1)^2} = \frac{p^2}{q^2} \qquad \rightarrow \text{ apply Componendo and Dividendo} \\ \implies & \frac{x^2 + 1}{2x} = \frac{p^2}{q^2} \\ \implies& q^2(x^2+1) = 2xp^2 \\ \implies & q^2x^2 -2p^2x +q^2 = 0 \end{align}$

**Option D.**

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