Question

Find the zeros of f(x) =x^2/a+b/ac*x/b-1/c​

Answer 1

Given:

$\bold{f(x) =\frac{\frac{x^2}{a+b}}{ac}\times \frac{x}{b}-\frac{1}{c}}$

To find:

Zeros= ?

Solution:

$\Rightarrow f(x)=\frac{\frac{x^2}{a+b}}{\frac{ac}{1}}\times \frac{x}{b}-\frac{1}{c}\\\\\Rightarrow f(x)=\frac{x^2}{ac(a+b)}\times \frac{x}{b}-\frac{1}{c}\\\\\Rightarrow f(x)=\frac{x^3}{abc(a+b)}-\frac{1}{c}\\\\\Rightarrow f(x)=\frac{x^3-ab(a+b)}{abc(a+b)}\\\\\Rightarrow f(x)=\frac{x^3-a^2b-ab^2}{a^2bc+ab^2c}$

The answer is:

$\boxed{ f(x)=\frac{x^3-a^2b-ab^2}{a^2bc+ab^2c}}$