Question

abx²+(b²-ac)x-bc=0 (cbse 2005)

Answer 1

hello there !!!!!

ATQ

abx² + ( b² - ac ) x - bc = 0
abx² + b²x - acx - bc = 0
bx ( ax + b) - c ( ax + b ) = 0
( ax+b ) ( bx- c ) = 0
ax+b =0 ; bx- c =0
x = - b/a ; x = c/ b

Hope it helps you

Answer 2

Answer:

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