find the roots of the quadratic equation.16/x - 1 = 15/x + 1 ,x ≠ 0, - 1 ​

Question

Find the roots of the quadratic equation.\(\frac{16}{x}-1\) = \(\frac{15}{x+1}\),x ≠ 0, - 1

Answer 1

\(\cfrac{16}{x}-1\) = \(\cfrac{15}{x+1}\),x ≠ 0, - 1

⇒ \(\cfrac{16}{x}-\)\(\cfrac{15}{x+1}\) = 1

⇒ \(\cfrac{16x + 16 - 15x}{x(x+1)}\) = 1

⇒  \(\cfrac{x+16}{x^2+x}\) = 1

⇒  x² + x = x + 16   (Cross multiplication)

⇒  x² - 16 = 0

⇒ (x + 4) (x - 4) = 0

⇒  x + 4 = 0 or x - 4 = 0

⇒ x = - 4 or x = 4

Hence, - 4 and 4 are the roots of the given equation.