Question

If a =
$\frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} - \sqrt{5} }$

And b =
$\frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} }$

Then show that
$\sqrt{a} - \sqrt{b} - 2 \sqrt{ab} = 0$


​

Answer 1

Answer:

$a=\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}\\\\b=\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}}\\\\\implies a=\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}\times \dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}\\\\\implies a=\dfrac{(\sqrt{10}+\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}\\\\\implies \sqrt{a}=\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{5}}\\\\\text{Similarly we will have :}\\\\\implies \sqrt{b}=\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{5}}$

$\sqrt{a}-\sqrt{b}=\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{5}}-\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{5}}\\\\\implies \sqrt{a}-\sqrt{b}=\dfrac{2\sqrt{5}}{\sqrt{5}}\implies 2\\\\-2\sqrt{ab}=-2(\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{5}}\times \dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{5}})\\\\\implies -2\sqrt{ab}=-2(\dfrac{(\sqrt{10})^2-(\sqrt{5})^2}{5})\\\\\implies -2\sqrt{ab}=-2(\dfrac{10-5}{5})\\\\\implies -2\sqrt{ab}=-2\times \dfrac{5}{5}\implies -2\\\\\text{Adding both we get :}\\\\\implies 2-2=0\\\\\textbf{QED}$

Step-by-step explanation:

The basic thing involved here is to rationalize the denominators .

Use the formula :

( a + b ) ( a - b ) = a² - b²

Multiplying a surd with the conjugate surd results in a rational number .

Answer 2

ANSWER:----------------

a= 10 − 5

10 + 5

b= 10 + 5

10 −5

⟹a= 10 − 5

10+5× 10 + 5

10 +5

a= ( 10 ) 2 −( 5 ) 2( 10 +5 }

$a−b=10+55−10−55⟹a−b=255⟹2−2ab=−2(10+55×10−55)⟹−2ab=−2((10)2−(5)25)⟹−2ab=−2(10−55)⟹−2ab=−2×55⟹−2$

Add both we willget

$⟹2−2=0</p><p></p><p>[tex]\begin{lgathered}\sort{a}$-

$\sqrt{b}=\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{5}}-\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{5}}\\\\\implies \sqrt{a}$-

$\sqrt{b}=\dfrac{2\sqrt{5}}{\sqrt{5}}\implies2\\\\-2\sqrt{ab}=-2(\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{5}}\times \dfrac{\sqrt{10}$-

$\sqrt{5}}{\sqrt{5}})\\\\\implies -2\sqrt{ab}=-2(\dfrac{(\sqrt{10})^2-(\sqrt{5})^2}{5})\\\\\implies$

[TeX]-2\sqrt{ab}=-2(\dfrac{10-5}{5})\\\\\implies -2\sqrt{ab}=-2\times \dfrac{5}{5}\implies[/tex]

[TeX] -2\\\\\text{Adding both we get :}\\\\\implies 2-2=0\\\\\textbf{QED}\end{lgathered}[/tex]

=−2×

5

5

⟹−2 both we get :--