If one root of a quadratic equation is $\dfrac { 1 }{ 1+\sqrt { 3 } } $, then the quadratic equation is
- $2{ x }^{ 2 }+x-1=0$
- $2{ x }^{ 2 }-2x-1=0$
- $2{ x }^{ 2 }+2x+1=0$
- $2{ x }^{ 2 }+x+1=0$
- $2{ x }^{ 2 }+2x-1=0$
The root is 1/(1+sqrt(3)). Rationalizing the denominator gives (sqrt(3)-1)/(3-1) = (sqrt(3)-1)/2. If x = (sqrt(3)-1)/2, then 2x + 1 = sqrt(3). Squaring both sides: (2x+1)^2 = 3, so 4x^2 + 4x + 1 = 3, which simplifies to 4x^2 + 4x - 2 = 0, or 2x^2 + 2x - 1 = 0.
If a quadratic equation with rational coefficients has one root of 1 / (1 + sqrt(3)), its other root must be the conjugate 1 / (1 - sqrt(3)). The sum of the roots is 1 / (1 + sqrt(3)) + 1 / (1 - sqrt(3)), which equals (1 - sqrt(3) + 1 + sqrt(3)) / ((1 + sqrt(3))(1 - sqrt(3))). The denominator simplifies to 1 - 3 = -2, and the numerator simplifies to 2, making the sum -1. The product of the roots is (1 / (1 + sqrt(3))) * (1 / (1 - sqrt(3))) = 1 / (1 - 3) = -1/2. Using the standard form x^2 - (sum)x + product = 0, we get x^2 - (-1)x + (-1/2) = 0, which simplifies to x^2 + x - 1/2 = 0. Multiplying the entire equation by 2 results in 2x^2 + 2x - 1 = 0.