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If $2 a+3 b+6 c=0(a, b, c \in R)$ then the quadratic equation $a x^2+b x+c=0$ has
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at least one root in [0, 1]
at least one root in [0, 1]
Let $f(x)=\frac{a x^3}{3}+\frac{b x^2}{2}+c x \Rightarrow f(0)=0$ and $f(1)=\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6}=0$
Also $f(x)$ is continuous and differentiable in $[0,1]$ and $\left[0,1]\right.$. So by Rolle's theorem, $f^{\prime}(x)=0$. i.e. $a x^2+b x+c=0$ has at least one root in $[0,1]$
Also $f(x)$ is continuous and differentiable in $[0,1]$ and $\left[0,1]\right.$. So by Rolle's theorem, $f^{\prime}(x)=0$. i.e. $a x^2+b x+c=0$ has at least one root in $[0,1]$
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