If alpha, beta are the roots of the equation | vedclass

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Question

(A) Given the quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$,we can write $ax^2+bx+c = a(x-\alpha)(x-\beta)$.Let $f(x) = ax^2+bx+c$.

Vedclass · Accepted Answer

$\frac{a^2(\alpha-\beta)^2}{2}$

Vedclass · Answer

(A) Given the quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$,we can write $ax^2+bx+c = a(x-\alpha)(x-\beta)$.Let $f(x) = ax^2+bx+c$. As $x ightarrow \alpha$,$f(x) ightarrow 0$.Using the limit formula $\lim_{	heta ightarrow 0} \frac{1-\cos 	heta}{	heta^2} = \frac{1}{2}$,we have:$\lim_{x ightarrow \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2} = \lim_{x ightarrow \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(a(x-\alpha)(x-\beta))^2} 	imes \frac{a^2(x-\alpha)^2(x-\beta)^2}{(x-\alpha)^2}$$= \frac{1}{2} 	imes \lim_{x ightarrow \alpha} a^2(x-\beta)^2$$= \frac{1}{2} a^2(\alpha-\beta)^2$.

If $\alpha, \beta$ are the roots of the equation $ax^2+bx+c=0$,then $\lim_{x \rightarrow \alpha} \frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2} =$

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