We are given that ,
[tex]{:\implies \quad \sf \dfrac{12+\sqrt{128}}{1-\sqrt{2}}=a+b\sqrt{2}}[/tex]
Consider LHS;
[tex]{:\implies \quad \sf \dfrac{12+\sqrt{128}}{1-\sqrt{2}}}[/tex]
Can be further written as ;
[tex]{:\implies \quad \sf \dfrac{12+8\sqrt{2}}{1-\sqrt{2}}}[/tex]
Rationalizing the denominator we have ;
[tex]{:\implies \quad \sf \dfrac{12+8\sqrt{2}}{1-\sqrt{2}}\times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}}[/tex]
[tex]{:\implies \quad \sf \dfrac{(12+8\sqrt{2})(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}}[/tex]
[tex]{:\implies \quad \sf \dfrac{12(1+\sqrt{2})+8\sqrt{2}(1+\sqrt{2})}{1-2}}[/tex]
[tex]{:\implies \quad \sf \dfrac{12+12\sqrt{2}+8\sqrt{2}+16}{-1}}[/tex]
[tex]{:\implies \quad \sf \dfrac{28+20\sqrt{2}}{-1}}[/tex]
[tex]{:\implies \quad \sf -28-20\sqrt{2}}[/tex]
Hence, we can conclude that ;
- [tex]{\bf a=-28}[/tex]
- [tex]{\bf b=-20}[/tex]
