The quadratic equation whose one root is $\cfrac{3+\sqrt{5}}{2-\sqrt{5}}$ is
- ${x}^{2}+22x+4=0$
- ${x}^{2}+22x-4=0$
- ${x}^{2}+11x+8=0$
- ${x}^{2}+11x-8=0$
Rationalize the root: (3 + sqrt(5)) / (2 - sqrt(5)) * (2 + sqrt(5)) / (2 + sqrt(5)) = (6 + 3*sqrt(5) + 2*sqrt(5) + 5) / (4 - 5) = (11 + 5*sqrt(5)) / -1 = -11 - 5*sqrt(5). For a quadratic with rational coefficients, the conjugate -11 + 5*sqrt(5) must also be a root. The sum of roots is -22 and the product is (-11)^2 - (5*sqrt(5))^2 = 121 - 125 = -4. The equation is x^2 - (sum)x + (product) = 0, which is x^2 + 22x - 4 = 0.
To rationalize the given root, multiply the numerator and denominator by the conjugate 2+sqrt(5), which gives ((3+sqrt(5))(2+sqrt(5)))/(4-5) = -11-5sqrt(5). Because rational coefficients mean irrational roots occur in conjugate pairs, the second root must be -11+5sqrt(5). Using the sum and product of roots to form the quadratic equation, we find the sum is -22 and the product is (-11-5sqrt(5))(-11+5sqrt(5)) = 121-25 = -4. Thus, the required quadratic equation is x^2 + 22x - 4 = 0.