Question

a=3+2root 2,find a square +1 by a square​

Answer 1

Answer:

Step-by-step explanation:

BRAINLIEST BRAINLIEST BRAINLIEST

Answer 2

Answer: 34

Step-by-step explanation:

Given,

a = 3 + 2√2  ..........i(

$\frac{1}{a}=\frac{1}{3+2\sqrt{2}}\\\;\\\frac{1}{a}=\frac{(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}\\\;\\\frac{1}{a}=\frac{3-2\sqrt{2}}{3^2-(2\sqrt{2})^2}\\\;\\\frac{1}{a}=\frac{3-2\sqrt{2}}{9-8}\\\;\\\frac{1}{a}=\frac{3-2\sqrt{2}}{1}\\\;\\\frac{1}{a}=3-2\sqrt{2}\;\;\;...........ii)$

Adding Eq. i) and ii)

$a+\frac{1}{a}=3+2\sqrt{2}+3-2\sqrt{2}\\\;\\a+\frac{1}{a}=6\\\;\\\text{Making Sqaure of Both Sides}\\\;\\(a+\frac{1}{a})^2=6^2\\\;\\a^2+\frac{1}{a^2}+(2\times a\times\frac{1}{a})=36\\\;\\a^2+\frac{1}{a^2}+2=36\\\;\\a^2+\frac{1}{a^2}=36-2\\\;\\a^2+\frac{1}{a^2}=34$