Understanding the Concept:
For quadratic equations with irrational roots, the conjugate is also a root.
Step 1: Rationalize given root
\[
x = \frac{1}{1+\sqrt{3}}
\]
Multiply numerator and denominator:
\[
x = \frac{1-\sqrt{3}}{(1+\sqrt{3})(1-\sqrt{3})}
\]
\[
= \frac{1-\sqrt{3}}{1-3} = \frac{1-\sqrt{3}}{-2}
\]
\[
= \frac{\sqrt{3}-1}{2}
\]
Step 2: Identify second root
Conjugate:
\[
\frac{-\sqrt{3}-1}{2}
\]
Step 3: Sum of roots
\[
\frac{\sqrt{3}-1}{2} + \frac{-\sqrt{3}-1}{2} = \frac{-2}{2} = -1
\]
Step 4: Product of roots
\[
\frac{\sqrt{3}-1}{2} \cdot \frac{-\sqrt{3}-1}{2}
= \frac{-3 +1}{4} = \frac{-2}{4} = -\frac{1}{2}
\]
Step 5: Form quadratic
\[
x^2 - (\text{sum})x + (\text{product}) = 0
\]
\[
x^2 + x - \frac{1}{2} = 0
\]
Multiply by 2:
\[
2x^2 + x - 1 = 0
\]
\[
\boxed{2x^2 + x - 1 = 0}
\]