Question

x
$\frac{x}{3} + \frac{3}{x} = 4 \frac{1}{4}$
​

Answer 1

SOLUTION :

$\qquad\quad{:}\longrightarrow\sf \dfrac {x}{3}+{\dfrac {3}{x}}=4 {\dfrac {1}{4}}$

$\qquad\quad{:}\longrightarrow\sf \dfrac {x^2+9}{3x}={\dfrac {17}{4}}$

• By cross multiplication :

$\qquad\quad{:}\longrightarrow\sf 4 (x^2+9)=17 (3x)$

$\qquad\quad{:}\longrightarrow\sf 4x^2+36=51x$

$\qquad\quad{:}\longrightarrow\sf 4x^2-51x+36=0$

Let the roots are $\bf{\alpha} \:and \bf{\beta}$

Finding $\bf{\alpha}$.

$\qquad\quad{:}\longrightarrow\bf \alpha={\dfrac {-b+\sqrt {b^2-4ac}}{2a}}$

$\qquad\quad{:}\longrightarrow\sf \alpha = \dfrac { - ( - 51)+\sqrt{(-51)^2 - 4 \times 4 \times 36}}{2 \times 4}$

$\qquad\quad{:}\longrightarrow\sf \alpha = \dfrac {51+\sqrt {2601 - 576}}{8}$

$\qquad\quad{:}\longrightarrow\sf \alpha = \dfrac {51+\sqrt {2025}}{8}$

$\qquad\quad{:}\longrightarrow\sf \alpha = \dfrac {51+45}{8}$

$\qquad\quad{:}\longrightarrow\sf \alpha = \cancel{\dfrac {96}{8}}$$\qquad\quad{:}\longrightarrow\bf \alpha = 12$

Finding $\bf{\beta}$.

$\qquad\quad{:}\longrightarrow\bf \beta={\dfrac {-b-\sqrt {b^2-4ac}}{2a}}$

$\qquad\quad{:}\longrightarrow\sf \beta = \dfrac { - (- 51)-\sqrt{(-51)^2 - 4 \times 4 \times 36}}{2 \times 4}$

$\qquad\quad{:}\longrightarrow\sf \beta = \dfrac {51-\sqrt {2601 -576 }}{8}$

$\qquad\quad{:}\longrightarrow\sf \beta = \dfrac {51-\sqrt {2025}}{8}$

$\qquad\quad{:}\longrightarrow\sf \beta = \dfrac {51 - 45}{8}$

$\qquad\quad{:}\longrightarrow\sf \beta = \cancel{\dfrac {6}{8}}$

$\qquad\quad{:}\longrightarrow\bf \beta = \dfrac {3}{4}$