Question

If a=√3-√2 \ √3+√2 b=1\a find the value of aa+bb-5ab​

Answer 1

Answer:

$a=\frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2} } \\b=\frac{\sqrt{3} + \sqrt{2} }{\sqrt{3} - \sqrt{2} } \\\\a^2=(\frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2} })^2=\frac{3+2-2\sqrt{6} }{3+2+2\sqrt{6}} =\frac{5-2\sqrt{6}}{5+2\sqrt{6}} \\\\b^2=(\frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2} })^2=\frac{3+2+2\sqrt{6} }{3+2-2\sqrt{6}} =\frac{5+2\sqrt{6}}{5-2\sqrt{6}} \\ab=\frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2} } *\frac{\sqrt{3} + \sqrt{2} }{\sqrt{3} - \sqrt{2} } = 1$

$a^2+b^2-5ab=\frac{5-2\sqrt{6}}{5+2\sqrt{6}} + \frac{5+2\sqrt{6}}{5-2\sqrt{6}} -5 \\\\= \frac{(5-2\sqrt{6)}^2+(5+2\sqrt{6})^2-5(5-2\sqrt{6})(5+2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})} \\= \frac{25+24-20\sqrt{6}+25+24+20\sqrt{6}-5(25-24) }{25-24}\\= \frac{49-20\sqrt{6}+49+20\sqrt{6}-5(1) }{1}\\ \\=49+49+5\\a^2+b^2-5ab=103$

used formulas

(a+b)(a-b)=a^2-b^2

(a+b)^2=a^2+b^2+2ab

(a-b)^2=a^2+b^2-2ab