Solve the following quadratic equations and give your answer correct to two significant figures :

(i) x² - 4x - 8 = 0

(ii) $x -\dfrac{18}{x} = 6$

(i) The given equation is x² - 4x - 8 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -4 , c = -8

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-4) ± \sqrt{(-4)^2 - 4\times 1 \times -8}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{4 ± \sqrt{16 + 32}}{2} \\[1em] \Rightarrow x = \dfrac{4 ± \sqrt{48}}{2} \\[1em] \Rightarrow x = \dfrac{4 + \sqrt{48}}{2} \text{ or } \dfrac{4 - \sqrt{48}}{2} \\[1em] \text{ Also } \sqrt{48} = 6.928 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{4 + 6.928}{2} \text { or } \dfrac{4 - 6.928}{2} \\[1em] \Rightarrow x = \dfrac{10.928}{2} \text{ or } \dfrac{-2.928}{2} \\[1em] \Rightarrow x = 5.464 \text{ or } -1.464 \\[1em] x = 5.5 \text{ or } -1.5 \text{ (correct to two significant figures) }$

Hence roots of the given equations are 5.5 , -1.5.

(ii) Given,

$\Rightarrow x - \dfrac{18}{x} = 6 \\[1em] \Rightarrow \dfrac{x^2 - 18}{x} = 6 \\[1em] \Rightarrow x^2 - 18 = 6x \\[1em] \Rightarrow x^2 - 6x - 18 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -6 , c = -18

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times -18}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 + 72}}{2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{108}}{2} \\[1em] \Rightarrow x = \dfrac{6 + \sqrt{108}}{2} \text{ or } \dfrac{6 - \sqrt{108}}{2} \\[1em] \text{ Also } \sqrt{108} = 10.392 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{6 + 10.392}{2} \text { or } \dfrac{6 - 10.392}{2} \\[1em] \Rightarrow x = \dfrac{16.392}{2} \text{ or } \dfrac{-4.392}{2} \\[1em] \Rightarrow x = 8.196 \text{ or } -2.195 \\[1em] \Rightarrow x = 8.2 \text{ or } -2.2 \text{ (correct to two significant figures) }$

Hence roots of the given equations are 8.2 , -2.2.