Question

what is the rationalizing factor for the denominator of the expression 1/3+√5​

Answer 1

$\underline{\mathfrak{\huge{The\:Question:}}}$

What is the rationalizing factor for the denominator of the expression :-

$\tt{\frac{1}{3+\sqrt{5}}}$

$\underline{\mathfrak{\huge{Your\:Answer:}}}$

$\tt{\frac{1}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}}$

Multiply and divide the equation by $3 - \sqrt{5}$ :-

=》 $\tt{\frac{3-\sqrt{5}}{(3+\sqrt{5})(3-\sqrt{5})}}$

Use the identity :-

$\tt{(a+b)(a-b) = a^{2} - b^{2}}$

In the formed equation and then solve it :-

=》 $\tt{\frac{3-\sqrt{5}}{9-5}}$

Subtract the denominator and then get your answer :-

=》 $\tt{\frac{3-\sqrt{5}}{4}}$

There's your answer !

So, solving it further, we get that the rationalizing factor of the given equation will be :-

$\boxed{\tt{3 - \sqrt{5}}}$

That's your final solution

Answer 2

What is the rationalizing factor for the denominator of the expression 1/3+√5 ?

Step by step explanation :

$\tt{ \frac{1}{3 + \sqrt{5}} }$

Multiplying the given fraction by the conjugate of the denominator,

i.e multiply the fraction by

$\tt{ \frac{3 - \sqrt{5} }{3 - \sqrt{5}} }$

$\tt {\therefore \frac{1}{3 + \sqrt{5} } \times \frac{3 - \sqrt{5} }{3 - \sqrt{5}} }$

Now, Using the identity,

$\tt{(a + b)(a - b) = {a}^{2} - b {}^{2} }$

$\tt{ \therefore \frac{3 - \sqrt{5} }{(3) {}^{2} - (\sqrt{5}) {}^{2} }}$

$\tt{ \therefore \frac{3 - \sqrt{5} }{9 - 5}}$

$\tt{ = \frac{3 - \sqrt{5} }{4}}$

This is the rationalised form for the denominator of the given expression.