Question 1184300




{{{ax^2-2bx+c=0 }}}

Since {{{a}}}, {{{b}}} and {{{c}}} are in {{{AP}}}

=> {{{2b=a+c}}}

Given, quadratic equation,

{{{ax^2-2bx+c=0 }}}

{{{ax^2-(a+c)x+c=0}}} .................substitute {{{2b=a+c }}}

{{{ ax^2-ax-cx+c=0}}}

 {{{(ax^2-ax)-(cx-c)=0}}}

{{{ ax(x-1)-c(x-1)=0}}}

{{{(x-1)(ax-c)=0}}}

roots:

{{{x=1}}}, or {{{x=c/a}}}
 
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