Question

rationalize the denominator of 1/7+3√2

Answer 1

$\frac{1}{7 + 3 \sqrt{2} } \\ = \frac{1 \times (7 - 3 \sqrt{2} )}{(7 + 3 \sqrt{2})(7 - 3 \sqrt{2} ) } \\ = \frac{7 - 3 \sqrt{2} }{49 - 12} \\ = \frac{7 - 3 \sqrt{2} }{37}$
Here, is the answer for your question.

Answer 2

Answer-

After rationalizing the denominator it becomes,

$\dfrac{7-3\sqrt2}{31}$

Solution-

The given expression,

$=\dfrac{1}{7+3\sqrt2}$

multiplying the conjugate of the denominator,

$=\dfrac{1(7-3\sqrt2)}{(7+3\sqrt2)(7-3\sqrt2)}$

$=\dfrac{7-3\sqrt2}{(7+3\sqrt2)(7-3\sqrt2)}$

Using (a+b)(a-b) = a²-b²

$=\dfrac{7-3\sqrt2}{(7)^2-(3\sqrt2)^2}$

$=\dfrac{7-3\sqrt2}{49-18}$

$=\dfrac{7-3\sqrt2}{31}$

Therefore, this is the rationalized the denominator form of the given exression.