Question

The quadratic equation whose one root is $\frac{3+\sqrt{5}}{2-\sqrt{5}}$ is:

B. $x^{2}+22x-4=0$

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Answer

To find the quadratic equation whose one root is $\frac{3 + \sqrt{5}}{2 - \sqrt{5}}$, we can follow these steps:

  1. Simplify the Root: Given root is: $$ \alpha = \frac{3 + \sqrt{5}}{2 - \sqrt{5}} $$ To simplify, rationalize the denominator: $$ \alpha = \frac{3 + \sqrt{5}}{2 - \sqrt{5}} \cdot \frac{2 + \sqrt{5}}{2 + \sqrt{5}} = \frac{(3+\sqrt{5})(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})} $$ Simplify the numerator: $$ (3+\sqrt{5})(2+\sqrt{5}) = 3 \cdot 2 + 3 \sqrt{5} + \sqrt{5} \cdot 2 + \sqrt{5} \cdot \sqrt{5} = 6 + 3\sqrt{5} + 2\sqrt{5} + 5 = 11 + 5\sqrt{5} $$ Simplify the denominator using $(a - b)(a + b) = a^2 - b^2$: $$ (2 - \sqrt{5})(2 + \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1 $$ Thus: $$ \alpha = \frac{11 + 5\sqrt{5}}{-1} = -11 - 5\sqrt{5} $$

  2. Conjugate Root: For a quadratic equation, if one root is $\alpha = -11 - 5\sqrt{5}$, the other root $\beta$ (conjugate) will be: $$ \beta = -11 + 5\sqrt{5} $$

  3. Form the Quadratic Equation: The sum of the roots ($ \alpha + \beta $) and the product of the roots ($ \alpha \beta $) are: $$ \alpha + \beta = (-11 - 5\sqrt{5}) + (-11 + 5\sqrt{5}) = -22 $$ $$ \alpha \beta = (-11 - 5\sqrt{5})(-11 + 5\sqrt{5}) = (-11)^2 - (5\sqrt{5})^2 = 121 - 25 \cdot 5 = 121 - 125 = -4 $$ Therefore, the quadratic equation is: $$ x^2 - (\alpha + \beta)x + \alpha \beta = 0 $$ Substituting the values, we get: $$ x^2 - (-22)x + (-4) = 0 $$ Simplifying: $$ x^2 + 22x - 4 = 0 $$

Thus, the quadratic equation is: $$ \mathbf{x^2 + 22x - 4 = 0} $$

Answer:

The correct option is (B) $x^2 + 22x - 4 = 0$.


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