If a and b are distinct real numbers, show that the quadratic equation

The given equation is $2\left(a^{2}+b^{2}\right) x^{2}+2(a+b) x+1=0$.

$\therefore D=[2(a+b)]^{2}-4 \times 2\left(a^{2}+b^{2}\right) \times 1$

$=4\left(a^{2}+2 a b+b^{2}\right)-8\left(a^{2}+b^{2}\right)$

$=4 a^{2}+8 a b+4 b^{2}-8 a^{2}-8 b^{2}$

$=-4 a^{2}+8 a b-4 b^{2}$

$=-4\left(a^{2}-2 a b+b^{2}\right)$

$=-4(a-b)^{2}<0$

Hence, the given equation has no real roots.