Solve the following equations by using quadratic formula and give your answer correct to 2 decimal places :

(i) 4x² - 5x - 3 = 0

(ii) $2x -\dfrac{1}{x} = 7$

(i) The given equation is 4x² - 5x - 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 4 , b = -5 , c = -3

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

we obtain:

$\Rightarrow x = \dfrac{-(-5) ± \sqrt{(-5)^2 - 4\times 4 \times -3}}{2 \times 4} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{25 + 48}}{8} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{73}}{8} \\[1em] \Rightarrow x = \dfrac{5 + \sqrt{73}}{8} \text{ or } \dfrac{5 - \sqrt{73}}{8} \\[1em] \text{ Also } \sqrt{73} = 8.54 \text{ (From tables)} \\[1em] \Rightarrow x = \dfrac{5 + 8.54}{8} \text { or } \dfrac{5 - 8.54}{8} \\[1em] \Rightarrow x = \dfrac{13.54}{8} \text{ or } \dfrac{-3.54}{8} \\[1em] \Rightarrow x = 1.69 \text{ or } -0.44$

Hence roots of the given equations are 1.69, -0.44.

(ii) Given,

$\Rightarrow 2x - \dfrac{1}{x} = 7 \\[1em] \Rightarrow \dfrac{2x^2 - 1}{x} = 7 \\[1em] \Rightarrow 2x^2 - 1 = 7x \\[1em] \Rightarrow 2x^2 - 7x - 1 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -7 , c = -1

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

we obtain:

$\Rightarrow x = \dfrac{-(-7) ± \sqrt{(-7)^2 - 4\times 2 \times -1}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{7 ± \sqrt{49 + 8}}{4} \\[1em] \Rightarrow x = \dfrac{7 ± \sqrt{57}}{4} \\[1em] \Rightarrow x = \dfrac{7 + \sqrt{57}}{4} \text{ or } \dfrac{7 - \sqrt{57}}{4} \\[1em] \text{ Also } \sqrt{57} = 7.549 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{7 + 7.549}{4} \text { or } \dfrac{7 - 7.549}{4} \\[1em] \Rightarrow x = \dfrac{14.549}{4} \text{ or } \dfrac{-0.549}{4} \\[1em] \Rightarrow x = 3.642 \text{ or } -0.142 \\[1em] x = 3.64 \text{ or } -0.14 \text{ (Correct to two decimal places.) }$

Hence roots of the given equations are 3.64, -0.14.