Question

am^2+(b^2-ac)x-bc=0, x=?​

Answer 1

Answer:

Step-by-step explanation:

Given that, ax² + ( b²- ac ) x - bc = 0

Comparing the given equation with ax² + bx + c = 0,

We have a = a , b = b²- ac and c = -bc.

Discriminant = D = b² - 4ac = ( b² - ac )² - 4 ( a ) ( - bc )

= ( b² - ac )² + 4abc

∴ x = - b ± √ b² - 4ac /2a

= - ( b² - ac ) ± √ ( b² - ac )² + 4abc / 2a

= - b² - ac ± ( b² - ac ) + 2√abc / 2a

= - b² - ac + b² - ac + 2√abc / 2a (or) -b² - ac - b² + ac - 2√abc / 2a

= - 2ac + 2√abc / 2a ( or ) - 2b² - 2√abc / 2a

= 2 ( - ac + √abc ) / 2a ( or ) 2 ( - b² - √abc ) /2a

= - ac + √abc / a ( or ) - b² - √abc / a

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