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.The discr

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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.The discriminants are given by ${D_1} = b_1^2 - 4{c_1}$ and ${D_2} = b_2^2 - 4{c_2}$.Adding these,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,we have ${D_1} + {D_2} \ge 0$.If the sum of two real numbers is non-negative,then at least one of them must be non-negative. Therefore,${D_1} \ge 0$ or ${D_2} \ge 0$.$A$ quadratic equation has real roots if its discriminant is greater than or equal to zero. Thus,at least one of the equations must have real roots.

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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