Question

$find \: the \: zeroes \: of \: the \: polynomial \: f(x) = abx {}^{2} + (b {}^{2} - ac)x - bc$
​

Answer 1

Answer:

$\alpha = \frac{c}{b} \\ \beta = \frac{ - b}{a}$

Step-by-step explanation:

$ab {x}^{2} + ( {b}^{2} - ac)x - bc \\ = ab {x}^{2} + {b}^{2} x - acx - bc \\ = bx(ax + b) - c(ax + b) \\ =( bx - c)(ax + b)$

Either,

$bx - c = 0 \\ = > bx = c \\ = > x = \frac{c}{b}$

OR,

$ax + b = 0 \\ = > ax = - b \\ = > x = \frac{ - b}{a}$

Answer 2

${\Huge{\mathfrak{\red{Answer:}}}}$

$\begin{lgathered}x = \frac{ -B \pm \sqrt{ {B}^{2} - 4 AC} }{2 A} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) \pm \sqrt{( {b}^{2} - ac {)}^{2} } - 4(ab) ( - bc) }{2ab} \\ \\ \implies \: x \: = \frac{ - ( {b}^{2} - ac) \pm \sqrt{( {b}^{2} - ac {) }^{2} +4a {b}^{2}c } }{2ab} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) \pm \sqrt{ {b}^{4} - 2a {b}^{2}c + {a}^{2} {c}^{2} + 4a {b}^{2}c } }{2ab} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) \pm \sqrt{( {b}^{2} + ac {)}^{2} } }{2ab} \: \: \: \implies \: x = \frac{ - ( {b}^{2} - ac) \pm( {b}^{2} + ac) }{2ab} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) + ( {b}^{2} + ac) }{2ab} \: \: \: \: or \: \: \: \: x = \frac{ - ( {b}^{2} -ac) - ( {b}^{2} + ac) }{2ab} \\ \\ \implies \: x = \frac{2ac}{2ab} \: \: \: \: \: \: or \: \: \: \: x = \frac{ - 2 {b}^{2} }{2ab} \: \: \: \: \: \implies \: x = \frac{c}{b} \: \: \: \: \: \: or \frac{ - b}{a}\end{lgathered} </p><p>$