If a and b are distinct real numbers, show that the quadratic equation 2(a2 + b2) x2 + 2 (a + b) x + 1 = 0 has no real roots.
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2(a2 + b2) x2 + 2 (a + b) x + 1 = 0
Compare given equation with the general form of quadratic equation, which is ax2 + bx + c = 0
a = 2 (a2 + b2), b = 2(a + b), c = 1
Discriminant:
D = b2 – 4ac
=[2(a + b)]2 – 4. 2 (a2 + b2).1
= 4a2 + 4b2 + 8ab – 8a2 – 8b2
= – 4a2 – 4b2 + 8ab
= – 4(a2 + b2 – 2ab)
= – 4(a – b)2 < 0
Hence the equation has no real roots.