Question

rationalise the denominater of

Answer 1

$\textbf{\underline{Rationalization\: of \:the \:denominator}}$

$\bf{\frac{1}{7 + {\sqrt{2}}}}$

firstly, change the sign of the denominator , i.e 7 + √2 will become 7-√2 , so multiply this term both with numerator and denominator.

= $\bf{\frac{1(7 - {\sqrt{2}})}{(7 + {\sqrt{2}})(7 - {\sqrt{2}})}}$

Multiply the terms , and we know that (a+b)(a-b) = a² - b² , use this identity for denominator!

= $\bf{\frac{7 - {\sqrt{2}}}{[{7}^{2} - {({\sqrt{2}})^{2}]}}}$

solve it more , square root and square will cancel out!

= $\bf{\frac{7 - {\sqrt{2}}}{(49 - 2)}}$

= $\bf{\frac{7 - {\sqrt{2}}}{(47)}}$

thus , after rationalising we will get :-

$\boxed{\frac{7 - {\sqrt{2}}}{47}}$

Answer 2

Here's the answer

$\sf{ \underline{Rationalizing \: the \: Denominator }}$

$\sf {\frac{1}{7 + \sqrt{2} }}$

Multiply both the denominator and numerator by the conjugate of denominator to rationalize the denominator.

$\sf{ \therefore \frac{1}{7 + \sqrt{2}} \times \frac{7 - \sqrt{2} }{7 - \sqrt{2} }}$

Remember that $\sf {({a + b)} {(a - b)} = {a^2} - {b^2})}$

Simplifying it , we get =>

$\sf{ = \frac{7 - \sqrt{2} }{(7 - \sqrt{2}) \times( 7- \sqrt{2}) } }$

$\sf {= \frac{7 - \sqrt{2} }{7 {}^{2} - (\sqrt{2} ) {}^{2} }}$

$\sf{= \frac{7 - \sqrt{2} }{49 - 2}}$

$\sf{ = \frac{7 - \sqrt{2} }{47}}$

Thus,

Final answer =>

After rationalizing denominator of 1 / 7 - √2,
We get =>

$\sf{ = \frac{7 - \sqrt{2} }{47}}$

Thanks!!