If 2a+3b+6c=0, then at least one root of the equation ax^2 + bx + c = 0 lies in the interv…

Question

If $2a+3b+6c=0$,  then at least one root of the equation

$a{x^2} + bx + c = 0$ lies in the interval

Accepted Answer

Let us define a function

$$f\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx$$

Being polynomial, it is continuous and differentiable, also,

$$f\left( 0 \right) = 0\,$$ and $$\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c$$

$$ \Rightarrow f\left( 1 \right) = {{2a + 3b + 6c} \over 6} = 0$$ (given)

$$\therefore$$ $$f\left( 0 \right) = f\left( 1 \right)$$

$$\therefore$$ $f(x)$ satisfies all conditions of Rolle's theorem

therefore $$f'\left( x \right) = 0$$ has a root in $$\left( {0,1} \right)$$

i.e. $a{x^2} + bx + c = 0$ has at lease one root in $(0, 1)$

Solutions · 1