The roots of the following quadratic equation are Not real 2$(a^2 + b^2)x^2$ + 2(a + b) x + 1 = 0

Using the discriminant formula D = b^2 - 4ac for 2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0, we substitute the coefficients. The discriminant evaluates to [2(a + b)]^2 - 4(2(a^2 + b^2))(1) = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2 = -4(a^2 - 2ab + b^2) = -4(a - b)^2. Since the square of any real number is non-negative, multiplying by -4 makes the discriminant less than or equal to zero, meaning the roots are never real.

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