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If $a, b, c \in R$ are such that $4 a+2 b+c>0$ and $a x^2+b x+c=0$ has no real roots, then the value of $(c+a)(c+b)$ is
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Verified Answer
The correct answer is:
greater than ab
Since, the equation $a x^2+b x+c=0$ have no real roots and $4 a+2 b+c>0$
$$
\begin{aligned}
\therefore \quad a+b+c & >0 \\
& \left\{\because a x^2+b x+c>0, \forall x \in R\right\}
\end{aligned}
$$
So,
$$
\begin{aligned}
\text { So, } & & c+a>-b \text { and } c+b & >-a \\
\Rightarrow & & & (c+a)(c+b)>a b
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad a+b+c & >0 \\
& \left\{\because a x^2+b x+c>0, \forall x \in R\right\}
\end{aligned}
$$
So,
$$
\begin{aligned}
\text { So, } & & c+a>-b \text { and } c+b & >-a \\
\Rightarrow & & & (c+a)(c+b)>a b
\end{aligned}
$$
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