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If 2a+3b+6c=0, then prove that at least ...

If `2a+3b+6c=0,` then prove that at least one root of the equation `ax^(2)+bx+c=0` lies in the interval (0,1).

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Verified by Experts

`"Let "f(x) =(ax^(2))/(3)+(bx^(2))/(2)+cx`
`f(0) =0 " " and" "f(1) =(a)/(3)+(b)/(2) +c =2a +3b +6c =0`
`"if "" " f(0) =f(1) " then " f(x) =0 " for some value of " x in (0,1)`
`rArr " "ax^(2) +bx +c =0` for at least one `x in (0,1)`
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