Question

x^2+3x+3/x+1/x^2= 26 and x can be written as a+/sqrt(b) where a and b are positive integers, then find a+b

Answer 1

\[x^2+3x+\frac{ 3 }{ x }+\frac{ 1 }{ x^2 }=26\]
\[\left( x^2+\frac{ 1 }{x^2 } \right)+3\left( x+\frac{ 1 }{x } \right)=26\]
\[put~x+\frac{ 1 }{x }=t,squaring,\]
\[x^2+\frac{ 1 }{ x^2 }+2*x*\frac{ 1 }{x }=t^2,x^2+\frac{ 1 }{x^2 }=t^2-2\]
\[t^2-2+3t=26\]
\[t^2+3t-28=0\]
\[t=\frac{ -3\pm \sqrt{9-4*1*-28} }{2*1 }\]
\[t=\frac{ -3\pm \sqrt{9+112} }{2 },t=\frac{ -3\pm11 }{ 2 }=4,-7\]
\[x+\frac{ 1 }{x}=4,x^2-1-4x=0,solve~ for~ x.\]
similarly solve \[x+\frac{ 1 }{ x }=-7\]

Answer 2

For four I got \[2+\sqrt{3}, 2-\sqrt{3}\]
and -7 I got \[\frac{ -7\pm3\sqrt{5} }{ 2 }\]

Answer 3

only \[2+\sqrt{3}~satisfies.~hence~a=2,b=3,a+b=2+3=5\]