Solve the following equation by using quadratic formula :

$x + \dfrac{1}{x} = 3$, x ≠ 0, giving answer correct to two significant figures.

Given,

Equation :

$\Rightarrow x + \dfrac{1}{x} = 3 \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = 3 \\[1em] \Rightarrow x^2 + 1 = 3x \\[1em] \Rightarrow x^2 - 3x + 1 = 0.$

Comparing above equation, with ax² + bx + c = 0, we get :

a = 1, b = -3 and c = 1.

By formula,

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 1}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{3 \pm \sqrt{9 - 4}}{2} \\[1em] \Rightarrow x = \dfrac{3 \pm \sqrt{5}}{2} \\[1em] \Rightarrow x = \dfrac{3 \pm 2.24}{2} \\[1em] \Rightarrow x = \dfrac{3 + 2.24}{2}, \dfrac{3 - 2.24}{2} \\[1em] \Rightarrow x = \dfrac{5.24}{2}, \dfrac{0.76}{2} \\[1em] \Rightarrow x = 2.62, 0.38$

Correcting to two significant figures.

Hence, x = 2.6, 0.38