Correct Answer - Option 4 : ad ≠ bc
Concept used:
We know, for no real roots, \(D = b^2 - 4ac < 0\)
Calculation:
Given equation is (a2 + b2) x2 + 2(ab + bd) x + c2 + d2 = 0
Here, a = (a2 + b2), b = 2(ab + bd), c = c2 + d2
\([(ac + bd)^2] - 4(a^2 + b^2) (c^2 + d^2) < 0\)
\(\Rightarrow [4(a^2 c^2 + b^2 d^2 + 2abcd)] - 4 (a^2 + b^2) (c^2 + d^2) < 0\)
\(\Rightarrow 4[(a^2 c^2 + b^2 d^2 + 2abcd) - (a^2 c^2 + a^2 d^2 + b^2 c^2 +b^2)] < 0\)
\(\Rightarrow 4[a^2 c^2 + b^2 d^2 + 2abcd - a^2 c^2 - a^2 d^2 - b^2 c^2 - b^2] < 0\)
\(\Rightarrow 4 [2abcd - b^2 c^2 - a^2 d^2] < 0\)
\(\Rightarrow -4 [a^2 d^2 + b^2 c^2 -2 abcd] < 0\)
\(\Rightarrow -4 [ad - bc]^2 < 0\)
Therefore,
(ad - bc) < 0
or ad < bc
∴ ad ≠ bc