Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x + \frac{1}{x} = 3, x \neq 0$

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Solution

The given equation is

$x + \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} + 1}{x} = 3 \Rightarrow x^{2} + 1 = 3 x \Rightarrow x^{2} - 3 x + 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times 1 = 9 - 4 = 5 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{5}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{5}}{2 \times 1} = \frac{3 + \sqrt{5}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{5}}{2 \times 1} = \frac{3 - \sqrt{5}}{2}$

Hence, $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$ are the roots of the given equation.

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