Prove that equation x²(a²+b²) + 2x(ac+bd)+(c²+d²)=0 …
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Posted by Abhishek Jain 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
According to question{tex}{/tex}, the given equation is
x{tex}^2{/tex}(a2 + b2)+ 2x (ac + bd) +(c2 + d2)= 0 .
{tex}{/tex}Let D be the discriminant of this equation.
Therefore,D = 4(ac + bd)2 - 4(a2 + b2)(c2 + d2)
{tex} \Rightarrow{/tex} D = 4 [(ac+ bd)2 - (a2 + b2) (c2 + d2)]
{tex} \Rightarrow{/tex} D = 4 [a2c2 + b2d2 + 2ac.bd - a2c2 - a2d2 - b2c2 - b2d2]
{tex} \Rightarrow{/tex}D = 4 [2ac.bd - a2d2 - b2c2] = - 4[a2d2 + b2c2 - 2ad.bc] = -4(ad - bc){tex}^2{/tex}
It is given that ad {tex} \neq{/tex} bc.
{tex} \Rightarrow{/tex} ad - bc {tex} \neq{/tex} 0
{tex} \Rightarrow{/tex} (ad - bc){tex}^2{/tex} > 0
{tex} \Rightarrow{/tex} - 4 (ad - bc){tex}^2{/tex} < 0
{tex} \Rightarrow{/tex} D < 0.
Therefore, given equation has no real root.
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