Question

Using quadratic formula, solve the following equation for x:

abx²+ (b²- ac) x - bc = 0


Answer fast I need it For exam.

Answer 1

$\boxed{\bold{\large{Answer:}}}$

$\textsf{We\:have, \: abx^2+ (b^2 - ac) x - bc = 0}$

$\textsf{Here,\: A\:=\:ab,\:B\:= b^2\:-\:ac, \:C\:=\:-ac}$

$\textsf{Now,}$

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

Answer 2

$\fbox{ \fbox{\bold{ \large{ \: \: Given \: \: \: }}}}$

$\to$ abx² + (b² - ac)x - bc = 0

$\fbox{ \fbox{\bold{ \large{ \: \: To \: Find \: \: }}}}$

$\to$ x = ?

$\fbox{ \fbox{\large{ \bold{ \: \: Solution \: \: \: }}}}$

$\bold{ \underline{We \: know \: that \: , \: \: }}$

$\bold{ x = \frac{ - b \: ± \: \sqrt{ {b}^{2} \: - \: 4ac \: \: } }{2a} }$

$\bold{ \underline{According \: to \: the \: question \: , \: \: }}$

$\bold{x = \frac{ - ( {b}^{2} - ac) \: ± \: \sqrt{ {( {b}^{2} - ac)}^{2} - 4a {b}^{2}c \: \: } }{2ab} }\\ \\ \bold{x = \frac{ - ( {b}^{2} \: - \: ac) \: ± \: \sqrt{ {b}^{4} \: + \: {a}^{2} {c}^{2} \: - \: 2a {b}^{2}c \: + \: 4(a {b}^{2}c ) \: \: } }{2ab}} \\ \\ \bold{x = \frac{ - ( {b}^{2} \: - \: ac) \: ± \: \sqrt{ {b}^{4} + {a}^{2} {c}^{2} + 2a {b}^{2} c \: \: } }{2ab}} \\ \\ \bold{x = \frac{- ( {b}^{2} - ac)± \sqrt{ {( {b}^{2} + ac)}^{2} } }{2ab} } \\ \\ \bold{x = \frac{ - {b}^{2} + ac + {b}^{2} + ac}{2ab} } \: \: \: \bold{Or }\: \: \: \bold{x = \frac{ - {b}^{2} + ac - {b}^{2} + ac}{2ab}} \\ \\ \bold{x = \frac{2ac}{2ab} } \: \: \: \bold{ Or }\: \: \: \bold{x = \frac{ - 2 {b}^{2} }{2ab}}$