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 = ( -b \pm √(b^2 - 4ac))/(2a)`

`\\ \sf\Rrightarrow x = (-(b^2 - ac) \pm √((b^2 -ac)^2-4(ab)(-bc)) )/(2(ab))`

`\\ \sf\Rrightarrow x= (-b^2 + ac \pm √(\left(b^2-ac\right)^2+4abbc) )/(2ab)`

`\\ \sf\Rrightarrow x = (-b^2 + ac \pm √(b^4+2b^2ac+a^2c^2) )/(2ab)`

`\\ \sf\Rrightarrow x = (-b^2 + ac \pm √(\left(b^2+ac\right)^2) )/(2ab)`

`\\ \sf\Rrightarrow x = (-b^2 + ac \pm( b^2+ac ))/(2ab)`

`\\ \sf\Rrightarrow x = (-b^2 + ac +( b^2+ac ))/(2ab) \quad or \quad (-b^2 + ac -( b^2+ac ))/(2ab)`

`\\ \sf\Rrightarrow x = (2ac)/(2ab) \quad or \quad (-2b^2)/(2ab)`

`\\ \sf\Rrightarrow x = (c)/(b) \quad or \quad (-b)/(a)`

Answer 2

`\boxed{\sf x = (c)/(b) \quad or \quad (-b)/(a)}`

Step-by-step 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 = ( -b \pm √(b^2 - 4ac))/(2a) \quad when \ ax^2 + bx + c = 0`

Insert values:

`\sf x = (-(b^2 - ac) \pm √((b^2 -ac)^2-4(ab)(-bc)) )/(2(ab))`

`\sf x = (-b^2 + ac \pm √(\left(b^2-ac\right)^2+4abbc) )/(2ab)`

`\sf x = (-b^2 + ac \pm √(b^4+2b^2ac+a^2c^2) )/(2ab)`

`\sf x = (-b^2 + ac \pm √(\left(b^2+ac\right)^2) )/(2ab)`

`\sf x = (-b^2 + ac \pm( b^2+ac ))/(2ab)`

`\sf x = (-b^2 + ac +( b^2+ac ))/(2ab) \quad or \quad (-b^2 + ac -( b^2+ac ))/(2ab)`

`\sf x = (2ac)/(2ab) \quad or \quad (-2b^2)/(2ab)`

`\sf x = (c)/(b) \quad or \quad (-b)/(a)`