Q3 of 227 Page 110

Show that the roots of the equation x2 + 2(a + b) x + 2 (a2 + b2) = 0 are unreal.

x2 + 2(a + b) x + 2 (a2 + b2) = 0


Compare this equation with ax2 + bx + c = 0


∴ a = 1, b = 2(a + b) and c = 2(a2 + b2)


b2 – 4ac = (2a + 2b)2 – 4(1)[2(a2 + b2)]


= 4a2 + 8ab + 4b2 – 8a2 – 8b2


= 8ab – 4a2 – 4b2


= – 4a2 + 8ab – 4b2


= –4(a2 – 2ab + b2)


= –4 (a – b)2


Since squared quantity is always positive.


Hence, (a – b)2 ≥ 0


Now, it is given a ≠ b, so (a – b)2 > 0


So, D = –4(a – b)2 will be negative.


Hence the equation has no real roots.


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