Question

please answer this question​

Answer 1

Answer:

Given: abx

2

+(b

2

−ac)x−bc=0

Factorizing the given equation,

⟹abx

2

+b

2

x−acx−bc=0

⟹bx(ax+b)−c(ax+b)=0

⟹(bx−c)(ax+b)=0

Solving the above equation for x, we get

⟹(bx−c)=0 or (ax+b)=0

⟹x=

b

c

or x=

a

−b

HOPE IT HELPS YOU ✌✌✌. PLZZ MARK AS BRAINLIEST.

Answer 2

$\huge\purple {SOLUTION}$

》》Given Equation :-

${abx}^{2} + ( {b}^{2} - ac)x - bc = 0$

》》Finding the value of ${x}$ :-

$d = {b}^{2} - 4ac$

${ {(b}^{2 } - ac)}^{2} - 4(ab)( - bc)$

${b}^{4} + {a}^{2} {c}^{2} - {2b}^{2} ac + {4b}^{2} ac$

${b}^{4} + {a}^{2} {c}^{2} + {2b}^{2} ac$

${ {(b}^{2} + ac)}^{2}$

$x = { - b}^{2} + \sqrt{ {b}^{2} - 4ac} \: \: \: \: \: \: ...........(quadratic \: formula)$

$x = { - b}^{2} + ac + \sqrt{ {(b}^{2} + {ac}^{2)} }$

$x = \frac{ { - b}^{2} + ac - {b}^{2} - ac}{2ab}$

$x = \frac{ { - 2b}^{2} }{2ab}$

$x = \frac{ - b}{a}$

and

$x = \frac{ - b + ac + {b}^{2} + ac }{2ab}$

$x = \frac{2ac}{2ab}$

$x = \frac{c}{b}$

⏩Therefore,

$x \frac{ - b}{a} \: \: \: \: and \: \: \: \: x = \frac{c}{b}$

$\huge\mathcal {It\:Might\:Help\:You}$