Question

Using propeties of proportion solve
1-px/1+px×√1+qx/1-qx=1​

Answer 1

∴ x = 0 or x = ± $\dfrac{1}{p}$$\sqrt{\frac{2p - q}{q}}$

Step-by-step explanation:

We have,

$\dfrac{1 - px}{1 + px}$ × $\dfrac{\sqrt{1 + qx} }{\sqrt{1 - qx}}$ = 1

⇒ $\dfrac{1 - px}{1 + px}$  = $\dfrac{\sqrt{1 - qx} }{\sqrt{1 + qx}}$

Squaring both sides, we get

$\dfrac{1 + p^{2}x^{2} -2px }{1 + p^{2}x^{2} + 2px}$ = $\dfrac{1 - qx}{1 + qx}$

Applying componendo and dividendo, we get

$\dfrac{2(1 + p^{2} x^{2}) }{2( -2px)}$ = $\dfrac{2}{-2qx}$

⇒ x($p^{2} qx$- 2px + q) = 0

⇒ x = 0 or $p^{2} qx$ - 2px + q = 0

⇒ $p^{2} qx$ - 2px + q = 0

∴ $x^{2}$ = $\dfrac{2p -q}{p^{2} q}$

x = ± $\dfrac{1}{p}$$\sqrt{\frac{2p - q}{q}}$

∴ x = 0 or x = ± $\dfrac{1}{p}$$\sqrt{\frac{2p - q}{q}}$