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Let $p, q, r \in R$ and $r>p>0$. If the quadratic equation $p x^2+q x+r=0$ has two complex roots $\alpha$ and $\beta$, then $|\alpha|+|\beta|$ is
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Verified Answer
The correct answer is:
greater than 2
greater than 2
Given quadratic equation is
$$
\begin{aligned}
& p x^2+q x+r=0 \\
& D=q^2-4 p r
\end{aligned}
$$
Since $\alpha$ and $\beta$ are two complex root
$$
\begin{array}{ll}
\therefore \beta=\bar{\alpha} \Rightarrow|\beta|=|\bar{\alpha}| \Rightarrow|\beta|=|\alpha| \\
(\because|\bar{\alpha}|=|\alpha|) \\
\text { Consider } \\
|\alpha|+|\beta|=|\alpha|+|\alpha| & (\because|\beta|=|\alpha|) \\
=2|\alpha|>2.1=2 & (\because|\alpha|>1)
\end{array}
$$
Hence, $|\alpha|+|\beta|$ is greater than 2 .
$$
\begin{aligned}
& p x^2+q x+r=0 \\
& D=q^2-4 p r
\end{aligned}
$$
Since $\alpha$ and $\beta$ are two complex root
$$
\begin{array}{ll}
\therefore \beta=\bar{\alpha} \Rightarrow|\beta|=|\bar{\alpha}| \Rightarrow|\beta|=|\alpha| \\
(\because|\bar{\alpha}|=|\alpha|) \\
\text { Consider } \\
|\alpha|+|\beta|=|\alpha|+|\alpha| & (\because|\beta|=|\alpha|) \\
=2|\alpha|>2.1=2 & (\because|\alpha|>1)
\end{array}
$$
Hence, $|\alpha|+|\beta|$ is greater than 2 .
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