Question

Please answer it as fast as possible and please define it.

Answer 1

Answer:

a = $\sf\frac{47}{2}$ and b = $\sf\frac{21}{2}$

Step-by-step explanation:

Given

$\sf\frac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} = a + b\sqrt{5}$

Let's do some rationalisation.

For rationalisation, we will multiply rationalising factor with both numerator and denominator. Rationalising factor of 7 - 3√5 is 7 + 3√5. So we get

$\sf\frac{7+3\sqrt{5}}{7-3\sqrt{5}}=a + b\sqrt{5}\\\\= \frac{(7+3\sqrt{5})(7+3\sqrt{5})}{(7-3\sqrt{5})(7+3\sqrt{5})}=a + b\sqrt{5}\\\\=\frac{(7+3\sqrt{5})^2}{7^2 - (3\sqrt{5})^2}=a + b\sqrt{5}\\\\= \frac{49+45+42\sqrt{5}}{49-45}= a + b\sqrt{5}\\\\\frac{94+42\sqrt{5}}{4} = a + b\sqrt{5}\\\\= \frac{94}{4}+\frac{42\sqrt{5}}{4}=a + b\sqrt{5}\\\\= \frac{47}{2}+ \frac{21}{2}\sqrt{5}= a + b\sqrt{5}$

Now, on comparing both sides, we can say that, a = 47/2 and b = 21/2

Answer 2

Answer:-

Given,

$\mathbf{ \frac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} } = a + b \sqrt{5} }$

We have to rationalize the denominator.

Rationalizing factor of 7-3√5 is 7+3√5.

Multiply 7+3√5 with both numerator and denominator.

$\mathbf{ = \frac{(7 + 3 \sqrt{5})(7 + 3 \sqrt{5} }{(7 - 3 \sqrt{5}(7 + 3 \sqrt{5} } = a + b \sqrt{5} }$

$\mathbf{ \frac{ {(7 + 3 \sqrt{5}) }^{2} }{ {(7)}^{2} - ( {3 \sqrt{5} }^{2}) } = a + b \sqrt{5} }$

$\mathbf{[ {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}]}$

$\mathbf{[(a + b)(a - b) = {a}^{2} - {b}^{2}]}$

$\mathbf{ \frac{49 + 42 \sqrt{5} + 45 }{49 - 45} = a + b \sqrt{5} }$

$\mathbf{ \frac{94 + 42 \sqrt{5} }{49 - 45} = a + b \sqrt{5} }$

$\mathbf{ \frac{94 + 42 \sqrt{5} }{4} = a + b \sqrt{5} }$

$\mathbf{ \frac{94}{4} + \frac{42 \sqrt{5} }{4} = a + b \sqrt{5} }$

$\mathbf{ \frac{47}{2} + \frac{21 }{2} \sqrt{5} = a + b \sqrt{5} }$

On comparing both the sides ,

$\mathbf{a = \frac{47}{2} ,b \sqrt{5} = \frac{21}{2} \sqrt{5} }$

[√5 gets cancelled]

$\boxed{Therefore \: a = \frac{47}{2} ,b = \frac{21}{2} }$