If 2a+3b+6c=0, [( a,b,c R )] then the quadratic equation ax^2 + bx + c = 0 has

Question

If $2a+3b+6c=0,$ $$\left( {a,b,c \in R} \right)$$ then the quadratic equation $a{x^2} + bx + c = 0$ has

Accepted Answer

Let $$f\left( x \right) = {{a{x^3}} \over 3} + {{b{x^2}} \over 2} + cx \Rightarrow f\left( 0 \right) = 0$$ and $f(1)$

$$ = {a \over 3} + {b \over 2} + c = {{2a + 3b + 6c} \over 6} = 0$$

Also $f(x)$ is continuous and differentiable in $$\left[ {0,1} \right]$$ and

$$\left[ {0,1\left[ {.\,\,} \right.} \right.$$ So by Rolle's theorem $$f'\left( x \right) = 0.$$

i.e $$\,\,a{x^2} + bx + c = 0$$ has at least one root in $$\left[ {0,1} \right].$$

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