Question

Solve for x using quadratic formula : abx^2 + (b^2 - ac)x - bc = 0​

Answer 1

• (ab)x^2 + (b^2 - ac)x -bc= 0

• a = ab
• b = b² - ac
• c = -bc

Apply quadratic formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$

$\\ \sf\Rrightarrow x = \dfrac{-(b^2 - ac) \pm \sqrt{(b^2 -ac)^2-4(ab)(-bc)} }{2(ab)}$

$\\ \sf\Rrightarrow x= \dfrac{-b^2 + ac \pm \sqrt{\left(b^2-ac\right)^2+4abbc} }{2ab}$

$\\ \sf\Rrightarrow x = \dfrac{-b^2 + ac \pm \sqrt{b^4+2b^2ac+a^2c^2} }{2ab}$

$\\ \sf\Rrightarrow x = \dfrac{-b^2 + ac \pm \sqrt{\left(b^2+ac\right)^2} }{2ab}$

$\\ \sf\Rrightarrow x = \dfrac{-b^2 + ac \pm( b^2+ac )}{2ab}$

$\\ \sf\Rrightarrow x = \dfrac{-b^2 + ac +( b^2+ac )}{2ab} \quad or \quad \dfrac{-b^2 + ac -( b^2+ac )}{2ab}$

$\\ \sf\Rrightarrow x = \dfrac{2ac}{2ab} \quad or \quad \dfrac{-2b^2}{2ab}$

$\\ \sf\Rrightarrow x = \dfrac{c}{b} \quad or \quad \dfrac{-b}{a}$

Answer 2

Answer:

$\boxed{\sf x = \dfrac{c}{b} \quad or \quad \dfrac{-b}{a}}$

Explanation:

Given expression: (ab)x^2 + (b^2 - ac)x + (-bc) = 0

Here given:

• a = ab

• b = b² - ac

• c = -bc

Apply quadratic formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \quad when \ ax^2 + bx + c = 0$

Insert values:

$\sf x = \dfrac{-(b^2 - ac) \pm \sqrt{(b^2 -ac)^2-4(ab)(-bc)} }{2(ab)}$

$\sf x = \dfrac{-b^2 + ac \pm \sqrt{\left(b^2-ac\right)^2+4abbc} }{2ab}$

$\sf x = \dfrac{-b^2 + ac \pm \sqrt{b^4+2b^2ac+a^2c^2} }{2ab}$

$\sf x = \dfrac{-b^2 + ac \pm \sqrt{\left(b^2+ac\right)^2} }{2ab}$

$\sf x = \dfrac{-b^2 + ac \pm( b^2+ac )}{2ab}$

$\sf x = \dfrac{-b^2 + ac +( b^2+ac )}{2ab} \quad or \quad \dfrac{-b^2 + ac -( b^2+ac )}{2ab}$

$\sf x = \dfrac{2ac}{2ab} \quad or \quad \dfrac{-2b^2}{2ab}$

$\sf x = \dfrac{c}{b} \quad or \quad \dfrac{-b}{a}$