If 2a+3b+6c = 0, then show that the equation `a x^2 + bx + c = 0` has atleast one real root between 0 to 1.

Explanation

Given `2a+3b+6c=0`
`impliesa/3+b/2+c=0`…..i
Let `f'(x)=ax^(2)+bx+c`
Then `f(x)=(ax^(3))/3+(bx^(2))/2+cx+d`
Now `f(0)=d` and `f(1)=a/3+b/2+c+d`
`=0+d` [from Eq(i) ]
Since `f(x)` is a polynomial of three degree, then `f(x)` is continuous and differentiable everywhere and `f(0)=f(1)`, then by Rolle's theorem `f'(x)=0` i.e. `ax^(2)+bx+c=0` has atleast one real root between 0 and 1.