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(D) Given equation: $x + \frac{1}{x} = 3$.Multiply the entire equation by $x$ to clear the denominator: $x^2 + 1 = 3x$.Rearrange into the standard qua

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$\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}$

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(D) Given equation: $x + \frac{1}{x} = 3$.Multiply the entire equation by $x$ to clear the denominator: $x^2 + 1 = 3x$.Rearrange into the standard quadratic form $ax^2 + bx + c = 0$: $x^2 - 3x + 1 = 0$.Here,$a = 1$,$b = -3$,and $c = 1$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Calculate the discriminant $D = b^2 - 4ac = (-3)^2 - 4(1)(1) = 9 - 4 = 5$.Since $D > 0$,real roots exist.Substituting the values into the formula: $x = \frac{-(-3) \pm \sqrt{5}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$.Thus,the solutions are $x = \frac{3+\sqrt{5}}{2}$ and $x = \frac{3-\sqrt{5}}{2}$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $x + \frac{1}{x} = 3, x \neq 0$.

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