How can you write the expression with a rationalized denominator? sqrt3 - sqrt6 / sqrt 3 + sqrt 6?

Solution:

Given, the expression is sqrt3 - sqrt6 / sqrt 3 + sqrt 6

We have to write the expression with a rationalized denominator.

The expression can be written as \(\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}}\)

Multiplying both numerator and denominator with the conjugate of the denominator,

Conjugate of the denominator = \(\sqrt{3}-\sqrt{6}\)

The expression becomes \(\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}}\times \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\)

By using algebraic identity,

(a - b)(a - b) = (a - b)²

(a - b)² = a² - 2ab + b²

Numerator = \((\sqrt{3}-\sqrt{6})\times (\sqrt{3}-\sqrt{6})\)

= \((\sqrt{3}-\sqrt{6})^{2}\)

= \((\sqrt{3})^{2}-2(\sqrt{3}\sqrt{6})+(\sqrt{6})^{2}\)

= \(3-2\sqrt{18}+6\)

= 9 - 2√18

= 9 - 2(3√2)

= 9 - 6√2

Taking out common term,

= 3(3 - 2√2)

By using algebraic identity,

(a - b)(a + b) = (a² - b²)

Denominator = \((\sqrt{3}-\sqrt{6})(\sqrt{3}+\sqrt{6})\)

= \((\sqrt{3})^{2}-(\sqrt{6})^{2}\)

= 3 - 6

= -3

Now, \(\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}}\) = \(\frac{3(3-2\sqrt{2})}{-3}\\=-(3-2\sqrt{2})\\=2\sqrt{2}-3\)

Therefore, the expression with a rationalized denominator is \(2\sqrt{2}-3\).

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How can you write the expression with rationalized denominator? sqrt3 - sqrt6 / sqrt 3 + sqrt 6?

Summary:

The expression sqrt3 - sqrt6 / sqrt 3 + sqrt 6 with rationalized denominator is \(2\sqrt{2}-3\).