Question

Find the value of a and b if 7+3under root 5/ 3+under root 5 - 7-3under root5/3- under root 5 =a+b under root 5​

Answer 1

Given:

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

To Find:

Value of 'a' & 'b'.

Solution:

$\sf \Longrightarrow \dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}$

$\sf \Longrightarrow \dfrac{\Big(7 + 3\sqrt{5}\Big) \Big(3 - \sqrt{5}\Big) - \Bigg[ \Big(7 - 3\sqrt{5}\Big) \Big(3 + \sqrt{5} \Big) \Bigg] }{\Big(3 + \sqrt{5}\Big) \Big(3 - \sqrt{5} \Big)}$

$\sf \Longrightarrow \dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 3(5) - \Big( 21 + 7\sqrt{5} - 9\sqrt{5} - 3(5) \Big)}{\Big(3\Big)^2 - \Big(\sqrt{5}\Big)^2}$

$\sf \Longrightarrow \dfrac{21 + 2\sqrt{5} - 15 - \Big( 21 - 2\sqrt{5} - 15 \Big)}{9 - 5}$

$\sf \Longrightarrow \dfrac{6 + 2\sqrt{5} - \Big(6 - 2\sqrt{5} \Big)}{4}$

$\sf \Longrightarrow \dfrac{6 + 2\sqrt{5} - 6 + 2\sqrt{5}}{4}$

$\sf \Longrightarrow \dfrac{4\sqrt{5} }{4}$

$\sf \Longrightarrow \sqrt{5}$

ATQ;

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

$\sf \therefore \ \sqrt{5} = a + b\sqrt{5}$

Final answer:

⇒ a = 0

⇒ b = 1

Answer 2

Step-by-step explanation:

⭐AnswEr⭐

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

$\implies \frac{7 + 3 \sqrt{5} (3 - \sqrt{5} ) - (7 - 3 \sqrt{5}(3 + \sqrt{5})) }{(3 + \sqrt{5})(3 - \sqrt{5} }$

$\implies \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - (21 + 7 \sqrt{5} - 9 \sqrt{5} - 15)}{9 - 5}$

$\implies \frac{6 + 2 \sqrt{5} - (6 - 2 \sqrt{5} )}{4}$

$\implies \frac{6 + 2 \sqrt{5} - 6 + 2 \sqrt{5} }{4}$

$\implies \frac{4 \sqrt{5} }{4}$

$\implies \sqrt{5}$

Therefore,

√5=a+b√5.

and

♣️ Value of a and b is 0 and 1 respectively ♣️