Given quadratic equation is

2(a² + b² )x² + 2(a + b)x + 1 = 0

Comparing with Ax² + Bx + C = 0, we get

A = 2(a² + b² ), B = 2(a + b), C = 1

Now, calculate D = B² - 4ac

=> D = {2(a + b)}² - 4 * 2(a² + b² ) * 1

=> D = 4(a² + b² + 2ab) - 8(a² + b² )

=> D = 4a² + 4b² + 8ab - 8a² - 8b²

=> D = -4a² - 4b² + 8ab

=> D = -(4a² + 4b² - 8ab)

=> D = -4(a² + b² - 2ab)

=> D = -4(a - b)²

Since square quantity is always positive

So, (a - b)² ≥ 0

So, D = -4(a - b)² will be negative.

Hence the equation has no real roots.