\displaystyle \sqrt{{a+b\sqrt{3}}}=\frac{{13}}{{4+\sqrt{3}}}

Square both sides of the equation.

\displaystyle {{\left[ {\sqrt{{a+b\sqrt{3}}}} \right]}^{2}}={{\left[ {\frac{{13}}{{4+\sqrt{3}}}} \right]}^{2}}

Expand and simplify the equation.

\displaystyle a+b\sqrt{3}=\left[ {\frac{{13}}{{4+\sqrt{3}}}} \right]\left[ {\frac{{13}}{{4+\sqrt{3}}}} \right]

\displaystyle a+b\sqrt{3}=\frac{{169}}{{\left( {4+\sqrt{3}} \right)\left( {4+\sqrt{3}} \right)}}

\displaystyle a+b\sqrt{3}=\frac{{169}}{{16+4\sqrt{3}+4\sqrt{3}+{{{\left( {\sqrt{3}} \right)}}^{2}}}}

\displaystyle a+b\sqrt{3}=\frac{{169}}{{19+8\sqrt{3}}}

Rationalize the denominator.

\displaystyle a+b\sqrt{3}=\frac{{169}}{{19+8\sqrt{3}}}\times \frac{{19-8\sqrt{3}}}{{19-8\sqrt{3}}}

\displaystyle a+b\sqrt{3}=\frac{{169\left( {19-8\sqrt{3}} \right)}}{{{{{\left( {19} \right)}}^{2}}-{{{\left( {8\sqrt{3}} \right)}}^{2}}}}

\displaystyle a+b\sqrt{3}=\frac{{3211-1352\sqrt{3}}}{{{{{\left( {19} \right)}}^{2}}-{{{\left( 8 \right)}}^{2}}{{{\left( {\sqrt{3}} \right)}}^{2}}}}

\displaystyle a+b\sqrt{3}=\frac{{3211-1352\sqrt{3}}}{{361-192}}

\displaystyle a+b\sqrt{3}=\frac{{3211-1352\sqrt{3}}}{{169}}

Separate the fraction and sinplify

\displaystyle a+b\sqrt{3}=\frac{{3211}}{{169}}-\frac{{1352\sqrt{3}}}{{169}}

\displaystyle a+b\sqrt{3}=19-8\sqrt{3}

\displaystyle a=19,\ b=-8

 

The most difficult part of this question is “how to start?”. None of my students had any idea.  But after giving them a few clues, they were able to solve the question rather quickly, however quite a number made careless mistakes here and there. This is expected of question with long working, so please be very  careful.

 

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