If {b_1}{b_2} = 2({c_1} + {c_2}),then at leas | vedclass

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(A) Let ${D_1}$ and ${D_2}$ be the discriminants of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ respectively.Then,${D_

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(A) Let ${D_1}$ and ${D_2}$ be the discriminants of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ respectively.Then,${D_1} = b_1^2 - 4{c_1}$ and ${D_2} = b_2^2 - 4{c_2}$.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 can substitute $4({c_1} + {c_2}) = 2{b_1}{b_2}$ into the equation:${D_1} + {D_2} = b_1^2 + b_2^2 - 2{b_1}{b_2} = {(b_1 - b_2)^2}$.Since the square of any real number is non-negative,${(b_1 - b_2)^2} \ge 0$,which implies ${D_1} + {D_2} \ge 0$.If the sum of two numbers is non-negative,at least one of them must be non-negative.Therefore,${D_1} \ge 0$ or ${D_2} \ge 0$,which means at least one of the equations has real roots.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

If ${b_1}{b_2} = 2({c_1} + {c_2})$,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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