Question

rationalize the denominator of 1/6+3√2 .
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Answer 1

Hope it helps you...........

Answer 2

To rationalize:

$\Longrightarrow \sf \dfrac{1}{6 + 3\sqrt{2}}$

What we mean when we try to rationalize something, is to eliminate the root [square or cube root] from the denominator to the numerator.

What we'll do to rationalize the denominator is to multiply both the denominator and the numerator by the conjugate of the denominator.

Here, the conjugate of the denominator [6 + 3√2] is 6 - 3√2.

$\Longrightarrow \sf \dfrac{1}{6 + 3\sqrt{2}}$

$\Longrightarrow \sf \dfrac{1}{6 + 3\sqrt{2}} \times \dfrac{6 - 3\sqrt{2}}{6 - 3\sqrt{2}}$

$\Longrightarrow \sf \dfrac{6 - 3\sqrt{2}}{(6 + 3\sqrt{2})(6 - 3\sqrt{2})}$

Using the algebraic identity (a + b)(a - b) = a² - b² we get:

$\Longrightarrow \sf \dfrac{6 - 3\sqrt{2}}{6^{2} - (3\sqrt{2})^2}$

$\Longrightarrow \sf \dfrac{6 - 3\sqrt{2}}{36 - (9 \times 2)}$

$\Longrightarrow \sf \dfrac{6 - 3\sqrt{2}}{36 - 18}$

$\Longrightarrow \sf \dfrac{6 - 3\sqrt{2}}{ 18}$

$\Longrightarrow \sf \dfrac{3(2 - \sqrt{2})}{ 18}$

$\Longrightarrow \sf \dfrac{2 - \sqrt{2}}{6}$

Hence rationalized.