Multiple choice

If the quadratic equation $ax^{2}+(b-c)x+a-b-c=0$ has unequal real roots for every $c\ \epsilon R$ (the set of all real numbers), then $ba>0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the quadratic ax^2 + (b-c)x + (a-b-c) = 0 to have real roots, the discriminant D must be > 0. D = (b-c)^2 - 4a(a-b-c) = b^2 - 2bc + c^2 - 4a^2 + 4ab + 4ac. This expression must be positive for all c, which implies the coefficient of c^2 is positive and the discriminant of this quadratic in c is negative. The provided statement is a known property for this specific structure.

AI explanation

For the quadratic equation ax^2 + (b-c)x + a-b-c = 0 to have unequal real roots for every real number c, its discriminant must be strictly positive for all values of c. The discriminant evaluates to (b-c)^2 - 4a(a-b-c), which expands and simplifies to the quadratic expression c^2 + (4a-2b)c + (b^2 + 4bc - 4a^2). To properly analyze when this expression is strictly positive for all c, we consider the condition for the leading coefficient and its own discriminant. Formulating the standard parameter boundaries reveals that this strict positivity is guaranteed only when the relationship b < a < 0 or b > a > 0 is satisfied. Therefore, the given statement is true.

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