Question

If √3x = √2x+1, then x is equal to ​

Answer 1

Answer:

x = $\sqrt{3} +\sqrt{2}$

Step-by-step explanation:

Given:

$\sqrt{3} x=\sqrt{2} x+1$

$\sqrt{3} x-\sqrt{2} x=1$

$x(\sqrt{3}-\sqrt{2})=1$

$x=\frac{1}{\sqrt{3}-\sqrt{2} }$

Rationalizing the denominator and numerator:

x = $\frac{1}{\sqrt{3}-\sqrt{2} }$  ×  $\frac{\sqrt{3}+\sqrt{2} }{\sqrt{3}+\sqrt{2} }$

x = $\frac{\sqrt{3} +\sqrt{2} }{1}$

Answer 2

$\green{\underline{\sf Given:}}$

$\sf\leadsto\sqrt{3} x = \sqrt{2} x + 1$

$\green{\underline{\sf Find:}}$

$\sf \leadsto value \: of \: x$

$\green{\underline{\sf Solution:}}$

$\sf :\implies\sqrt{3} x = \sqrt{2} x + 1 \\ \\ \sf :\implies\sqrt{3} x - \sqrt{2} x = 1 \\ \\ \sf :\implies x(\sqrt{3} - \sqrt{2})+ 1 \\ \\ \sf :\implies x = \frac{1}{ \sqrt{3} - \sqrt{2} } \\ \\ \sf \quad \star Rationalise \: the \: denominator\star \\ \\ \sf :\implies x = \frac{1}{ \sqrt{3} - \sqrt{2} } \\ \\ \sf :\implies x = \frac{1}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \sf :\implies x = \frac{(\sqrt{3} + \sqrt{2})}{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ \sf \quad \star by \: using \: {(a)}^{2} - {(b)}^{2} = (a + b)(a - b) \star \\ \\ \sf :\implies x = \frac{ \sqrt{3} + \sqrt{2} }{3 - 2} \\ \\ \sf :\implies x = \frac{ \sqrt{3} + \sqrt{2} }{1} = \sqrt{3} + \sqrt{2} \\ \\ \boxed{\sf \to x = \sqrt{3} + \sqrt{2}}$