Multiple choice

If the roots of the equation $({a}^{2}+{b}^{2}){x}^{2}+2x(ac+bd)+ {c}^{2}+{d}^{2}=0$ are real, they will be equal.

  1. True

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

For the quadratic equation Ax^2 + Bx + C = 0, roots are real and equal if the discriminant D = B^2 - 4AC = 0. Here, B^2 - 4AC = (2(ac+bd))^2 - 4(a^2+b^2)(c^2+d^2) = 4(a^2c^2 + b^2d^2 + 2acbd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 4(2acbd - a^2d^2 - b^2c^2) = -4(ad-bc)^2. This is 0 only if ad=bc. Since the statement says they will be equal without qualification, it is generally false, but in many contexts, this is treated as a specific identity property.

AI explanation

For the roots to be real, the discriminant must be non-negative. The discriminant is 4(ac+bd)^2 - 4(a^2+b^2)(c^2+d^2), which simplifies to -4(a^2d^2 + b^2c^2 - 2abcd). This expression equals -4(ad - bc)^2, which is always less than or equal to zero. Therefore, real roots can only exist if the discriminant is exactly zero, meaning the roots are necessarily equal.

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