Answer: \(x=\frac12+\frac{\sqrt3}{2}\) or \(x=\frac{2}{11}-\frac{\sqrt3}{33}\).
Answer: \(x=\frac12+\frac{\sqrt3}{2}\) or \(x=\frac{2}{11}-\frac{\sqrt3}{33}\).
Use the quadratic formula with \(A=3-5\sqrt3\), \(B=2\sqrt3+5\), and \(C=-1\):
\(\displaystyle x=\frac{-(2\sqrt3+5)\pm\sqrt{(2\sqrt3+5)^2-4(3-5\sqrt3)(-1)}}{2(3-5\sqrt3)}\).
Now
\((2\sqrt3+5)^2=12+20\sqrt3+25=37+20\sqrt3\),
and
\(-4(3-5\sqrt3)(-1)=12-20\sqrt3\).
So the expression under the square root is
\(37+20\sqrt3+12-20\sqrt3=49\).
Hence
\(\displaystyle x=\frac{-(2\sqrt3+5)\pm7}{2(3-5\sqrt3)}\).
This gives
\(\displaystyle x=\frac{2-2\sqrt3}{2(3-5\sqrt3)}\quad\text{or}\quad x=\frac{-12-2\sqrt3}{2(3-5\sqrt3)}.\)
For the first value,
\(\displaystyle x=\frac{1-\sqrt3}{3-5\sqrt3}\cdot\frac{3+5\sqrt3}{3+5\sqrt3}=\frac{-12+2\sqrt3}{-66}=\frac{2}{11}-\frac{\sqrt3}{33}.\)
For the second value,
\(\displaystyle x=\frac{-6-\sqrt3}{3-5\sqrt3}\cdot\frac{3+5\sqrt3}{3+5\sqrt3}=\frac{-33-33\sqrt3}{-66}=\frac12+\frac{\sqrt3}{2}.\)
