Question

find the value of a and b in the following ​

Answer 1

Solutions!!

(1)

$\sf \dfrac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\dfrac{19}{11}$

$\sf \dfrac{3-\sqrt{5}}{3+2\sqrt{5}}\times \dfrac{3-2\sqrt{5}}{3-2\sqrt{5}}=a\sqrt{5}-\dfrac{19}{11}$

$\sf \dfrac{(3-\sqrt{5})(3-2\sqrt{5})}{(3+2\sqrt{5})(3-2\sqrt{5})}=a\sqrt{5}-\dfrac{19}{11}$

$\sf \dfrac{(3-\sqrt{5})(3-2\sqrt{5})}{9-20}=a\sqrt{5}-\dfrac{19}{11}$

$\sf \dfrac{(3-\sqrt{5})(3-2\sqrt{5})}{-11}=a\sqrt{5}-\dfrac{19}{11}$

$\sf \dfrac{9-6\sqrt{5}-3\sqrt{5}+10}{-11}=a\sqrt{5}-\dfrac{19}{11}$

$\sf \dfrac{19-9\sqrt{5}}{-11}=a\sqrt{5}-\dfrac{19}{11}$

$\sf -a\sqrt{5}=-\dfrac{19}{11}+\dfrac{19-9\sqrt{5}}{11}$

$\sf -a\sqrt{5}=\dfrac{-19+19-9\sqrt{5}}{11}$

$\sf -a\sqrt{5}=\dfrac{-9\sqrt{5}}{11}$

$\sf a=\dfrac{-9\sqrt{5}}{11\times (-\sqrt{5})}$

$\sf a=\dfrac{9}{11}$

(2)

$\sf \dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}=a+\dfrac{7}{11}\sqrt{5}b$

Taking LHS,

$\sf =\dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}$

$\sf =\dfrac{(7+\sqrt{5})(7+\sqrt{5})}{44}-\dfrac{(7-\sqrt{5})(7-\sqrt{5})}{44}$

$\sf =\dfrac{(7+\sqrt{5})^{2}}{44}-\dfrac{(7-\sqrt{5})^{2}}{44}$

$\sf =\dfrac{49+5+14\sqrt{5}}{44}-\dfrac{49+5-14\sqrt{5}}{44}$

$\sf =\dfrac{54+14\sqrt{5}}{44}-\dfrac{54-14\sqrt{5}}{44}$

$\sf =\dfrac{54+14\sqrt{5}-54+14\sqrt{5}}{44}$

$\sf =\dfrac{28\sqrt{5}}{44}$

$\sf =\dfrac{7\sqrt{5}}{11}$

Comparing with RHS,

$\sf \dfrac{7\sqrt{5}}{11}=a + \dfrac{7}{11}\sqrt{5}b$

$\sf 0+\dfrac{7\sqrt{5}(1)}{11}=a+\dfrac{7}{11}\sqrt{5}b$

Divide $\sf \dfrac{7\sqrt{5}}{11}$ from both sides of the equation.

$\sf 0 + 1 = a + b$

Hence,

a = 0

b = 1