Show that the equation $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ has no real roots, when $a \neq b$

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Solution

$D = b^{2} - 4 a c = 0$
$= 2 {[2 (a + b)]}^{2} - 2 [2 (a^{2} + b^{2})] [1]$
$= 4 (a + b)^{2} - 8 (a^{2} + b^{2})$
$= 4 a^{2} + 4 b^{2} + 8 a b - 8 a^{2} - 8 b^{2}$
$= - 4 a^{2} - 4 b^{2} + 8 a b = - 4 (a - b)^{2}$
must be less than zero
Hence no real roots.

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