Question

If one root of a quadratic equation is \( \frac{1}{1+\sqrt{3}} \), then the quadratic equation is

• A. \( 2x^2 + x - 1 = 0 \)
• B. \( 2x^2 - 2x - 1 = 0 \)
• C. \( 2x^2 + 2x + 1 = 0 \)
• D. \( 2x^2 + x + 1 = 0 \)
• E. \( 2x^2 + 2x - 1 = 0 \)

Hint:

Always rationalize irrational roots and use conjugates to form quadratic equations.

Answer 1

The Correct Option is A

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: \[ 2x^2 + x - 1 = 0 \] \[ \boxed{2x^2 + x - 1 = 0} \]