If a and b are distinct real numbers, show that the quadratic equation $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ has no real roots.

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Solution

The equation is $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0 A = 2 (a^{2} + b^{2}) B = 2 (a + b) C = 1$

Calculating Determinant, $D = B^{2} - 4 A C (2 (a + b))^{2} - 4 (2 (a^{2} + b^{2})) (1) = 4 a^{2} + 4 b^{2} + 8 a b - 8 a^{2} - 8 b^{2} = - 4 a^{2} - 4 b^{2} + 8 a b = - 4 (a^{2} + b^{2} - 2 a b) = - 4 (a - b)^{2}$

Since squared quantity is always positive,

hence $(a - b)^{2} \geq 0$

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

Hence equation has no real roots.

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