Asked by Kirandoon | 20th Dec, 2008, 07:58: PM
B2 - 4AC = 4(ab+cd)2 - 4 (a2+c2 )(b2+d2)
=4 ( a2b2 +c2d2+2abcd - a2b2-a2d2-b2c2-c2d2)
=4 ( 2abcd -a2d2-b2c2)
= - 4 (ad-bc)2
which is always negative.
so roots of this equation will be imaginary.
Answered by | 20th Dec, 2008, 08:10: PM
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