Quadratic Equations — Analytical & Application Based Questions

Solve for x, if $\dfrac{5}{x} + 4\sqrt3 = \dfrac{2\sqrt3}{x^2},$ x ≠ 0

Answer

Given,

$\dfrac{5}{x} + 4\sqrt3 = \dfrac{2\sqrt3}{x^2}$

Substituting $\dfrac{1}{x} = t$ in the given equation, we get :

$\Rightarrow 5t + 4\sqrt3 = 2\sqrt3t^2 \\[1em] \Rightarrow 5t + 4\sqrt3 - 2\sqrt3t^2 = 0 \\[1em] \Rightarrow 2\sqrt3t^2 - 5t - 4\sqrt3 = 0 \\[1em] \Rightarrow 2\sqrt3t^2 - 8t + 3t - 4\sqrt3 = 0 \\[1em] \Rightarrow 2t(\sqrt3t - 4) + \sqrt3(\sqrt3t - 4) = 0 \\[1em] \Rightarrow (2t + \sqrt3) = 0 \text{ or } (\sqrt3t - 4) = 0 \\[1em] \Rightarrow 2t = -\sqrt3 \text{ or } \sqrt3t = 4 \\[1em] \Rightarrow t = \dfrac{-\sqrt3}{2} \text{ or } t = \dfrac{4}{\sqrt3} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-\sqrt3}{2} \text{ or } \dfrac{1}{x} = \dfrac{4}{\sqrt3} \\[1em] \Rightarrow x = \dfrac{2}{-\sqrt3} \text{ or } x = \dfrac{\sqrt3}{4}$

Hence, $x = \dfrac{2}{-\sqrt3} \text{ or } x = \dfrac{\sqrt3}{4}$.

Determine whether the following quadratic equation has real roots.

5x² - 9x + 4 = 0.

(a) Give reason for your answer

(b) If the equation has real roots, identify them.

Answer

(a) Given, equation : 5x² − 9x + 4 = 0

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

a = 5, b = -9 and c = 4.

Discriminant (D) = b² - 4ac = (-9)² - 4 × 5 × 4 = 81 - 80 = 1.

Since, D > 0 and a perfect square.

Hence, equation 5x² − 9x + 4 = 0 has real roots.

(b) Solving,

⇒ 5x² − 9x + 4 = 0

⇒ 5x² - 5x - 4x + 4 = 0

⇒ 5x(x - 1) - 4(x - 1) = 0

⇒ (5x - 4)(x - 1) = 0

⇒ 5x - 4 = 0 or x - 1 = 0

⇒ 5x = 4 or x = 1

⇒ x = $\dfrac{4}{5}$ or x = 1.

Hence, roots of the equation are 1 and $\dfrac{4}{5}$.