Question

Rationalize the denominator 1/7+3 root over 2

Answer 1

$\frac{1}{7 + 3 \sqrt{2} } \\ \frac{1}{7 + 3 \sqrt{2} } \times \frac{7 - 3 \sqrt{2} }{7 - 3 \sqrt{2} } \\ \frac{7 - 3 \sqrt{2} }{ {(7)}^{2} - {(3 \sqrt{2)} }^{2} } \\ \frac{7 - 3 \sqrt{2} }{49 - 9(2)} \\ \frac{7 - 3 \sqrt{2} }{49 - 18} \\ \frac{7 - 3 \sqrt{2} }{31}$

Answer 2

Answer:

The rationalized answer of the given expression is found to be: $\frac{7-3\sqrt2}{31}$

Step-by-step explanation:

The given expression is:  $\frac{1}{7+3\sqrt2}$

To rationalize the denominator of any given expression, we know that we need to multiply and divide the expression with the conjugate of the denominator,

The conjugate of the denominator in the given expression is: $7-3\sqrt2$,

So, multiplying and dividing the given expression by $7-3\sqrt2$, we get:

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

We know the identity that says: $(a+b)(a-b)=a^2-b^2$,

Using this identity, we get:

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

Simplifying further, we get:

$=\frac{7-3\sqrt2}{49-9(2)}$

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

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

which is the correct answer.