If $b_1b_2=2(c_1+c_2)$, then at least one of the equations $x^2+b_1x+c_1=0$ and $x^2+b_2x+c_2=0$ has?
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Real roots
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Purely imaginary roots
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Imaginary roots
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None of these
The discriminant of the first equation is D1 = b1^2 - 4c1 and the second is D2 = b2^2 - 4c2. Summing them: D1 + D2 = b1^2 + b2^2 - 4(c1 + c2). Given b1*b2 = 2(c1 + c2), we have D1 + D2 = b1^2 + b2^2 - 2(b1*b2) = (b1 - b2)^2. Since the sum of discriminants is a square, it is >= 0, implying at least one discriminant must be >= 0, meaning at least one equation has real roots.
To find the nature of the roots, we analyze the discriminants of both equations by assuming neither has real roots, meaning their discriminants are both negative. This gives b1^2 - 4c1 < 0 and b2^2 - 4c2 < 0, which implies b1^2 < 4c1 and b2^2 < 4c2. Multiplying these inequalities yields b1^2 * b2^2 < 16(c1)(c2). Since we are given b1b2 = 2(c1 + c2), substituting it into the squared form gives 4(c1 + c2)^2 < 16c1c2, which simplifies to (c1 - c2)^2 < 0. Because the square of any real number cannot be strictly negative, our initial assumption is false, proving at least one equation must have real roots. The correct result is Real roots.