If AD is not equal to …

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Sia ? 6 years, 9 months ago
Given, (a2 + b2)x2 + 2(ac +bd)x + (c2 + d2) = 0
The given equation is of the form {tex}Ax^2+Bx+C=0{/tex}
{tex}A=a^2+b^2,\ \ B=2(ac+bd),\ \ C=c^2+d^2{/tex}
D = b2 - 4ac
= {tex}[(2(ac + bd))^2 - 4 (a^2 + b^2) (c^2 + d^2)]{/tex}
= {tex}[4(a^2c^2 + 2abcd + b^2d^2] - 4[a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2]{/tex}
= 4a2c2 + 8abcd + 4b2d2 - 4a2c2 - 4a2d2 - 4b2c2 - 4b2d2
= 8abcd - 4a2d2 - 4b2c2
= -4[a2d2 + b2c2 - 2abcd]
= -4[ad - bc]2 < 0
Hence, the equation has no real roots.
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