Question

find zeroes (x^2/a)+(b/ac) *(-x/b) - (1/c)​

Answer 1

$\textbf{Given:}$

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

$\textbf{To find:}$

$\text{Zeros of f(x)}$

$\textbf{Solution:}$

$\text{Consider,}$

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

$f(x)=\frac{x^2c+bx}{ac}-(\frac{cx+b}{bc})$

$f(x)=x(\frac{xc+b}{ac})-(\frac{cx+b}{bc})$

$f(x)=\frac{x}{ac}(xc+b)-\frac{1}{bc}(cx+b)$

$f(x)=(\frac{x}{ac}-\frac{1}{bc})(cx+b)$

$\text{Now, f(x)=0 }$

$\implies\,(\frac{x}{ac}-\frac{1}{bc})(cx+b)=0$

$\implies\,\frac{x}{ac}-\frac{1}{bc}=0\;\text{(or)}\;cx+b=0$

$\implies\,\frac{x}{ac}=\frac{1}{bc}\;\text{(or)}\;cx=-b$

$\implies\,x=\frac{ac}{bc}\;\text{(or)}\;x=\frac{-b}{c}$

$\implies\bf\,x=\frac{a}{b}\;\text{(or)}\;x=\frac{-b}{c}$

$\therefore\textbf{The zeros of f(x) are \bf\frac{a}{b},\frac{-b}{c}\; }$

Find more:

Obtain other zeroes of the polynomial

f(x) = 2x4 + 3x3 - 5x2 - 9x - 3

if two of its zeroes are √3 and - √3.​

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