Question

( 3 + under root 7) + (3 - under root 7 )/ (3 - under root 7 )+( 3 + under root 7) = a + b under root 7

Answer 1

Given

$\dfrac{(3+\sqrt{7}) +(3-\sqrt{7})}{(3-\sqrt{7})+(3+\sqrt{7}) } =a+b\sqrt{7}$

Solving the LHS side, we get,

$\implies \dfrac{3+\sqrt{7} +3-\sqrt{7}}{3-\sqrt{7}+3+\sqrt{7} }$

$\implies \dfrac{6}{6}$

$\implies 1$

Now,

$1=a+b\sqrt{7}$

$\implies 1+0\sqrt{7}=a+b\sqrt{7}$

$\implies a = 1$

$\implies b = 0$

$\boxed{\boxed{\bold{Therefore, \ the \ values \ of \ a \ and \ b \ are \ 1 \ and \ 0 \ respectively}}}}}}}}$

                                                                         

Answer 2

Correct question :-

$\sf{If\ \dfrac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7},\ find\ a\ and\ b. }$

Solution :-

$\displaystyle{\sf{\longmapsto\frac{(3+\sqrt{7})(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})} =a+b\sqrt{7} }}\\\\\\\sf{(By\ rationalisation)}\\\\\displaystyle{\sf{\longmapsto \frac{(3+\sqrt{7})^2}{(3)^2-(\sqrt{7})^2}=a+b\sqrt{7} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{(3)^2+(\sqrt{7})^2+2\times3\times\sqrt{7}}{9-7}=a+b\sqrt{7} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{9+7+6\sqrt{7}}{2}=a+b\sqrt{7} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{16+6\sqrt{7}}{2}=a+b\sqrt{7} }}$

$\displaystyle{\sf{\longmapsto \frac{2(8+3\sqrt{7})}{2}=a+b\sqrt{7} }}\\\\\\\displaystyle{\sf{\longmapsto 8+3\sqrt{7}=a+b\sqrt{7} }}$

Comparing both the sides, we get :-

a = 8

b = 3