Find the roots of quadratic equations \(\cfrac{4}{x}-3\) = \(\cfrac

1.	Find the roots of quadratic equations \(\cfrac{4}{x}-3\) = \(\cfrac{5}{2x+3}\),x ≠ 0, - \(\cfrac{3}{2}\)
Answer» \(\cfrac{4}{x}-3\) = \(\cfrac{5}{2x+3}\),x ≠ 0, - \(\cfrac{3}{2}\) ⇒ \(\cfrac{4}{x}-\)\(\cfrac{5}{2x+3}\) = 3 ⇒ \(\cfrac{8x + 12-5x}{x(2x+3)}\) = 3 ⇒ \(\cfrac{3x+12}{2x^2+3x}\)= 3 ⇒ \(\cfrac{x+4}{2x^2+3x}\) = 1 ⇒ 2x² + 3x = x + 4 (cross multiplication) ⇒ 2x² + 2x - 4 = 0 ⇒ x² + x - 2 = 0 ⇒ x² + 2x - x - 2 = 0 ⇒ x(x + 2) - 1(x + 2) = 0 ⇒ (x + 2) (x - 1) = 0 ⇒ x + 2 = 0 or x - 1 = 0 ⇒ x = - 2 or x = 1 Hence, - 2 and 1 are the roots of the given equation.

Find the roots of quadratic equations \(\cfrac{4}{x}-3\) = \(\cfrac{5}{2x+3}\),x ≠ 0, - \(\cfrac{3}{2}\)

Answer»

\(\cfrac{4}{x}-3\) = \(\cfrac{5}{2x+3}\),x ≠ 0, - \(\cfrac{3}{2}\)

⇒ \(\cfrac{4}{x}-\)\(\cfrac{5}{2x+3}\) = 3

⇒ \(\cfrac{8x + 12-5x}{x(2x+3)}\) = 3

⇒ \(\cfrac{3x+12}{2x^2+3x}\)= 3

⇒ \(\cfrac{x+4}{2x^2+3x}\) = 1

⇒ 2x² + 3x = x + 4 (cross multiplication)

⇒ 2x² + 2x - 4 = 0

⇒ x² + x - 2 = 0

⇒ x² + 2x - x - 2 = 0

⇒ x(x + 2) - 1(x + 2) = 0

⇒ (x + 2) (x - 1) = 0

⇒ x + 2 = 0 or x - 1 = 0

⇒ x = - 2 or x = 1

Hence, - 2 and 1 are the roots of the given equation.