Answer:
[tex]17+\sqrt{288}[/tex]
Step-by-step explanation:
Given expression:
[tex]\dfrac{3+\sqrt{8}}{(\sqrt{2}-1)^2}[/tex]
To express the given expression in the form p + √q where p and q are integers, begin by expanding the denominator using the FOIL method:
[tex]\dfrac{3+\sqrt{8}}{(\sqrt{2}-1)(\sqrt{2}-1)}\\\\\\\\\dfrac{3+\sqrt{8}}{\sqrt{2}\cdot \sqrt{2}-1\cdot \sqrt{2}-1\cdot \sqrt{2}-1\cdot (-1)}\\\\\\\\\dfrac{3+\sqrt{8}}{2-2\sqrt{2}+1}\\\\\\\\\dfrac{3+\sqrt{8}}{3-2\sqrt{2}}[/tex]
Simplify the numerator:
[tex]\dfrac{3+\sqrt{8}}{3-2\sqrt{2}}\\\\\\\\\dfrac{3+\sqrt{4\cdot 2}}{3-2\sqrt{2}}\\\\\\\\\dfrac{3+\sqrt{4}\sqrt{2}}{3-2\sqrt{2}}\\\\\\\\\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}\\\\\\\\[/tex]
Now, rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator:
[tex]\dfrac{(3+\sqrt{8})(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}\\\\\\\\ \dfrac{9+6\sqrt{2}+3\sqrt{8}+2\sqrt{2}\sqrt{8}}{9+6\sqrt{2}-6\sqrt{2}-4\sqrt{2}\sqrt{2}}\\\\\\\\ \dfrac{9+6\sqrt{2}+3\sqrt{8}+2\sqrt{16}}{9-8}\\\\\\\\\dfrac{9+6\sqrt{2}+3\cdot 2\sqrt{2}+2(4)}{1}\\\\\\\\9+6\sqrt{2}+6\sqrt{2}+8\\\\\\\\17+12\sqrt{2}[/tex]
Finally, rewrite 12 as the square root of 144:
[tex]17+\sqrt{144}\sqrt{2}\\\\\\17+\sqrt{144\cdot 2}\\\\\\\Large\boxed{\boxed{17+\sqrt{288}}}[/tex]
Therefore, the values of p and q are:
