Answer:
[tex]a=11, b=2[/tex]
Step-by-step explanation:
Keep in mind that [tex]4-\sqrt{5}[/tex] and [tex]4+\sqrt{5}[/tex] are conjugates of each other. While it's typically known for rationalization of the denominator, getting an integer answer helps to simplify the fraction as shown in steps 5-7:
[tex]\frac{(4-\sqrt{5})(4+\sqrt{5})}{2\sqrt{11}}[/tex]
[tex]\frac{4(4)+4(\sqrt{5})-4(\sqrt{5})-\sqrt{5}(\sqrt{5})}{2\sqrt{11}}[/tex]
[tex]\frac{16+4\sqrt{5}-4\sqrt{5}-5}{2\sqrt{11}}[/tex]
[tex]\frac{16-5}{2\sqrt{11}}[/tex]
[tex]\frac{11}{2\sqrt{11}}[/tex]
[tex]\frac{\sqrt{11}\sqrt{11}}{2\sqrt{11}}[/tex]
[tex]\frac{\sqrt{11}}{2}[/tex]
Therefore, [tex]a=11[/tex] and [tex]b=2[/tex]
