Question

Rationalize the denominator of :
7+3√5/3+√5 + 7-3√5/3-√5
Plzz also explain the question fully (step by step).....full explanation
I want it fast (exam time)...plzz​

Answer 1

$\tt{ \dfrac{7 + 3\sqrt{5}}{3+\sqrt{5}} + \dfrac{7 - 3\sqrt{5}}{3-\sqrt{5}}}$

If we carefully observe the given problem ; the denominators are in the form of the identity : (a+b)(a-b).

So we will use the same to solve the given problem.

$\tt{ a \implies 3}$

$\tt{ b \implies \sqrt{5}}$

In the form of identity : (a+b)(a-b)

$\tt{\implies (3 + \sqrt{5})(3-\sqrt{5})}$

We know that the expansion of the identity :

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

So we will solve it further along with the numerator.

$\longrightarrow{\tt{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} + \dfrac{7 + 3\sqrt{5}}{3 - \sqrt{5}}}}$

Adding up the numerator simply gives us :

$\longrightarrow{\tt{\dfrac{14 + 6\sqrt{10}}{3\sqrt{5}- 3\sqrt{5}}}}$

Now taking the identity as in form of the expansion we get :

$\longrightarrow{\tt{\dfrac{14+6\sqrt{10}}{{(3)}^{2} - {(\sqrt{5})}^{2}}}}$

The power 2 of and '√' get cancelled and we simply take the square of 3.

$\longrightarrow{\tt{\dfrac{14+6\sqrt{10}}{9 - 5}}}$

Further ; subtracting 5 from 9 gives :

$\huge{\boxed{\tt{\dfrac{14+6\sqrt{10}}{4}}}}$

Thus we have rationalised the denominator.