If b_{1} b_{2} = 2(c_{1} + c_{2}) and b_{1}, | vedclass

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(A) Consider the two quadratic equations:$x^{2} + b_{1}x + c_{1} = 0$ and $x^{2} + b_{2}x + c_{2} = 0$.Let $D_{1}$ and $D_{2}$ be the discriminants of

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(A) Consider the two quadratic equations:$x^{2} + b_{1}x + c_{1} = 0$ and $x^{2} + b_{2}x + c_{2} = 0$.Let $D_{1}$ and $D_{2}$ be the discriminants of these equations respectively.$D_{1} = b_{1}^{2} - 4c_{1}$$D_{2} = b_{2}^{2} - 4c_{2}$Adding the two discriminants, we get:$D_{1} + D_{2} = b_{1}^{2} + b_{2}^{2} - 4(c_{1} + c_{2})$Given that $b_{1}b_{2} = 2(c_{1} + c_{2})$, we substitute $4(c_{1} + c_{2}) = 2b_{1}b_{2}$ into the equation:$D_{1} + D_{2} = b_{1}^{2} + b_{2}^{2} - 2b_{1}b_{2}$$D_{1} + D_{2} = (b_{1} - b_{2})^{2}$Since $(b_{1} - b_{2})^{2} \geq 0$ for all real $b_{1}, b_{2}$, it follows that $D_{1} + D_{2} \geq 0$.If the sum of two real numbers is non-negative, then at least one of them must be non-negative.Therefore, at least one of $D_{1}$ or $D_{2}$ is $\geq 0$, which implies that at least one of the equations has real roots.

If $b_{1} b_{2} = 2(c_{1} + c_{2})$ and $b_{1}, b_{2}, c_{1}, c_{2}$ are all real numbers, then at least one of the equations $x^{2} + b_{1} x + c_{1} = 0$ and $x^{2} + b_{2} x + c_{2} = 0$ has

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