Multiple choice

If ${ b }{ 1 }{ b }{ 2 }=2\left( { c }{ 1 }+{ c }{ 2 } \right) $ and ${ b }{ 1, }{ b }{ 2 },{ c }{ 1 },{ c }{ 2 }\quad $ are all real numbers, then atleast one of the equations ${ x }^{ 2 }+{ b }{ 1 }x+c_1=0$ and ${ x }^{ 2 }+{ b }{ 2 }x+c_2=0$ has

  1. real roots

  2. purely imaginary roots
  3. roots of the form $a+ib(a,b\in R,ab\ne 0$)

  4. rational roots
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A Correct answer
Explanation

The sum of the discriminants of the two equations is D1 + D2 = (b1^2 - 4c1) + (b2^2 - 4c2) = b1^2 + b2^2 - 4(c1 + c2). Given b1*b2 = 2(c1 + c2), we substitute 2(c1 + c2) = b1*b2 into the expression: D1 + D2 = b1^2 + b2^2 - 2(b1*b2) = (b1 - b2)^2. Since (b1 - b2)^2 >= 0, at least one discriminant must be non-negative, ensuring at least one equation has real roots.

AI explanation

Assume for the sake of contradiction that neither equation has real roots, which requires their discriminants to be negative: b1^2 - 4c1 < 0 and b2^2 - 4c2 < 0. Multiplying the inequalities by positive numbers and adding them leads to the condition b1^2 * b2^2 < 16(c1 * c2). Substituting the given relation b1 * b2 = 2(c1 + c2) transforms this into (c1 + c2)^2 < 4c1c2, which simplifies to the impossible inequality (c1 - c2)^2 < 0. Because this leads to a logical contradiction, our initial assumption is false and at least one of the equations must have real roots.

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