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

(i) 4x² - 5x - 3 = 0

(ii) x² - 7x + 3 = 0

(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) For a quadratic equation in the form :

ax² + bx + c = 0

The solutions are :

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

Comparing equation x² - 7x + 3 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -7 and c = 3

$\Rightarrow x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] = \dfrac{7 \pm \sqrt{49 - 12}}{2} \\[1em] = \dfrac{7 \pm \sqrt{37}}{2} \\[1em] = \dfrac{7 \pm 6.08}{2} \\[1em] = \dfrac{13.08}{2} \text{ or } \dfrac{0.92}{2}\\[1em] = 6.54 \text{ or } 0.46$

Hence, the value of x = 6.54 or 0.46