Question

Find x $\dfrac{x+\sqrt{2}}{x-\sqrt{2}}=\dfrac{5}{4}$​

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★$\dfrac{x+\sqrt{2}}{x-\sqrt{2}}=\dfrac{5}{4}$

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★The value of x.

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We would be using the concept of Componendo and Dividendo.

$\huge\tt{\red{\underline{Answer:}}}$

We have,

$\implies \dfrac{x+\sqrt{2}}{x-\sqrt{2}}=\dfrac{5}{4}$

Using componendo and dividendo,

$\implies \dfrac{(x+\sqrt{2})+(x-\sqrt{2})}{(x+\sqrt{2})-(x-\sqrt{2})}=\dfrac{5+4}{5-4}$

$\implies \dfrac{x+\cancel{\sqrt{2}}+x-\cancel{\sqrt{2}}}{\cancel{x}+\sqrt{2}-\cancel{x}+\sqrt{2}}=\dfrac{9}{1}$

$\implies \dfrac{\cancel{2}x}{\cancel{2}\sqrt{2}}=9$

$\implies\dfrac{x}{\sqrt{2}}=9$

.°.${\underline{\boxed{x =9\sqrt{2}}}}$

Therefore the value of x is 9√2.

Answer 2

Step-by-step explanation:

$4(x + \sqrt{2} ) = 5(x - \sqrt{2} )$

$4x + 4 \sqrt{2} = 5x - 5 \sqrt{2}$

$4x - 5x = - 5 \sqrt{2} - 4 \sqrt{2}$

$- x = - 9 \sqrt{2}$

$x = 9 \sqrt{2}$