Question

Using quadratic formula solve abx^2+(b^2-ac)x-bc=0

Answer 1

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Answer 2

This one is very good problem lets try to figure out this problem :

abx2+(b2−ac)x+bc=0abx2+(b2−ac)x⏟+bc=0

Evaluation the upper bold term ,we get

abx2+b2x−acx−bc=0abx2+b2x⏟−acx−bc⏟=0

Now taking common in paired terms:

⟹bx(ax+b)−c(ax+b)=0⟹bx(ax+b)−c(ax+b)=0

Now just taking (ax+b)(ax+b) as common ,so have

⟹(ax+b)(bx−c)=0⟹(ax+b)(bx−c)=0

So,the value of x=−ba,x=cbx=−ba,x=cb⏟.

Hence,the roots of the equations are −ba,cb