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If $b_{1} b_{2}=2\left(c_{1}+c_{2}\right)$ and $b_{1}, b_{2}, c_{1}, c_{2}$ are all real numbers, then at least one of the equations $\bar{x}^{2}+h_{1} x+c_{1}=0$ and $x^{2}+b_{2} x+c_{2}=0$ has
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Verified Answer
The correct answer is:
real roots
We have equations
$$
\begin{aligned}
x^{2}+b_{1} x+c_{1} &=0 \\
D_{1} &=b_{1}^{2}-4 c_{1} \\
x^{2}+b_{2} x+c_{2} &=0
\end{aligned}
$$
$$
\begin{array}{c}
D_{2}=b_{2}^{2}-4 c_{2} \\
\text { Now, } D_{1}+D_{2}=b_{1}^{2}+b_{2}^{2}-4\left(c_{1}+c_{2}\right) \\
=b_{1}^{2}+b_{2}^{2}-2 b_{1} b_{2} \quad\left[\because b_{1} b_{2}=2\left(c_{1}+c_{2}\right)\right] \\
=\left(b_{1}-b_{2}\right)^{2} \geq 0
\end{array}
$$
$\Rightarrow$ At least one of $D_{1}$ and $D_{2}$ are non-negative
real roots.
$$
\begin{aligned}
x^{2}+b_{1} x+c_{1} &=0 \\
D_{1} &=b_{1}^{2}-4 c_{1} \\
x^{2}+b_{2} x+c_{2} &=0
\end{aligned}
$$
$$
\begin{array}{c}
D_{2}=b_{2}^{2}-4 c_{2} \\
\text { Now, } D_{1}+D_{2}=b_{1}^{2}+b_{2}^{2}-4\left(c_{1}+c_{2}\right) \\
=b_{1}^{2}+b_{2}^{2}-2 b_{1} b_{2} \quad\left[\because b_{1} b_{2}=2\left(c_{1}+c_{2}\right)\right] \\
=\left(b_{1}-b_{2}\right)^{2} \geq 0
\end{array}
$$
$\Rightarrow$ At least one of $D_{1}$ and $D_{2}$ are non-negative
real roots.
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