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

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.

Small Group Tuition Additional Math (A Math), Combine Science Physics Chemistry  Elementary Math (E Math) Singapore Syllabus at Admiralty, Yew Tee, Choa Chu Kang Woodlands and johor Bahru