Question

Find a and b (4 + 3 root 5 /4 - 3 root 5) + (4- 3 root 5 / 4 + 3 root 5) = a + b root 5​

Answer 1

Answer:

$\sf{The \ value \ of \ a \ is \ -\dfrac{122}{29} \ and}$

$\sf{b \ is \ 0.}$

Given:

$\sf{\dfrac{4+3\sqrt5}{4-3\sqrt5}+\dfrac{4-3\sqrt5}{4+3\sqrt5}=a+b\sqrt5}$

To find:

• Value of a and b.

Solution:

$\sf{a+b\sqrt5=\dfrac{4+3\sqrt5}{4-3\sqrt5}+\dfrac{4-3\sqrt5}{4+3\sqrt5}}$

$\sf{=\dfrac{(4+3\sqrt5)^{2}+(4-3\sqrt5)^{2}}{(4+3\sqrt5)(4-3\sqrt5)}}$

$\sf{=\dfrac{2(4)^{2}+2(3\sqrt5)^{2}}{4^{2}-(3\sqrt5)^{2}}}$

$\sf{=\dfrac{2(16)+2(45)}{16-45}}$

$\sf{=\dfrac{32+90}{-29}}$

$\sf{\therefore{a+b\sqrt5=-\dfrac{122}{29}}}$

$\sf{On \ comparing \ we \ get}$

$\sf{a=-\dfrac{122}{29} \ and \ b\sqrt5=0}$

$\sf{\therefore{a=-\dfrac{122}{29} \ and \ b=0}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ a \ is \ -\dfrac{122}{29} \ and}}}$

$\sf\purple{\tt{b \ is \ 0.}}$

____________________________

$\sf\blue{Extra \ identities:}$

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

$\sf{(a+b)^{2}-4ab=(a-b)^{2}}$

$\sf{(a-b)^{2}+4ab=(a+b)^{2}}$