The equation $2(a^{2} + b^{2}) x^{2} + 2(a + b)x + 1 = 0$ has which type of roots, when $a\neq b$.
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Two distinct real roots
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No real roots
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Two equal roots
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Two real roots
Discriminant D = b^2 - 4ac. Here, b = 2(a+b), a = 2(a^2+b^2), c = 1. D = [2(a+b)]^2 - 4 * 2(a^2+b^2) * 1 = 4(a^2 + 2ab + b^2) - 8(a^2 + b^2) = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2 = -4a^2 + 8ab - 4b^2 = -4(a^2 - 2ab + b^2) = -4(a-b)^2. Since a != b, (a-b)^2 > 0, so D < 0. Thus, no real roots.
To determine the nature of the roots, we calculate the discriminant, D = b^2 - 4ac. Substituting the given values gives D = [2(a + b)]^2 - 4 * 2(a^2 + b^2) * 1. This simplifies to 4(a^2 + 2ab + b^2) - 8(a^2 + b^2), which equals -4a^2 + 8ab - 4b^2. Factoring out -4 yields D = -4(a^2 - 2ab + b^2) = -4(a - b)^2. Since the square of any real number is non-negative, -4(a - b)^2 is strictly negative for a not equal to b, meaning the equation has no real roots.