Question

rationalize the denominator of 1/ √3-√2

Answer 1

Rationalising the denominator:

1/✓3-✓2
Denominator is ✓3-✓2

Rationalising factor is ✓3+✓2

1/✓3-✓2 * ✓3+✓2/✓3+✓2
= ✓3+✓2/(✓3-✓2)(✓3+✓2). [Using identity. (a+b)(a-b) = a^2-b^2 ]
= ✓3+✓2/(✓3)^2-(✓2)^2
= ✓3+✓2= 3-2
✓3+✓2/1
✓3+✓2 is the answer

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Answer 2

$\bf \red{\frac{1}{\sqrt{3} - \sqrt{2} } =\sqrt{3} + \sqrt{2} }$

Given:

• $\frac{1}{ \sqrt{3} - \sqrt{2} }$

To find:

• Rationalize the denominator.

Solution:

Concept to be used:

• Rationalization is a process to free the denominator from any radical sign.
• For that multiply both numerator and denominator by rationalization factor.
• Rationalization factor is conjugate of denominator.
• If denominator is $(a + \sqrt{b} )$ then RF is $(a - \sqrt{b} )$.

Step 1:

Find RF of denominator.

As denominator is $\sqrt{3} - \sqrt{2}$

Thus,

RF is $\bf \sqrt{3} +\sqrt{2}$.

Step 2:

Multiply by RF.

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

apply Identity $\bf (x - y)(x + y) = {x}^{2} - {y}^{2}$

or

$= \frac{ \sqrt{3}+ \sqrt{2} }{( {\sqrt{3})}^{2} - ( { \sqrt{2}) }^{2} }$

or

$= \frac{ \sqrt{3} +\sqrt{2}}{3-2}$

or

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

Thus,

After rationalization, it becomes

$\bf \red{\frac{1}{\sqrt{3} - \sqrt{2} } =\sqrt{3} + \sqrt{2} }$

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