Prove that equation x²(a²+b²) + 2x(ac+bd)+(c²+d²)=0 has no real roots, if ad≠bc

Prove that equation x²(a²+b²) + 2x(ac+bd)+(c²+d²)=0 has no real r

1.	Prove that equation x²(a²+b²) + 2x(ac+bd)+(c²+d²)=0 has no real roots, if ad≠bc
Answer» According to question{tex}{/tex}, the given equation isx{tex}^2{/tex}(a2 + b2)+ 2x (ac + bd) +(c2 + d2)= 0 .\xa0{tex}{/tex}Let D be the discriminant of this equation.Therefore,D = 4(ac + bd)2 - 4(a2 + b2)(c2 + d2){tex} \\Rightarrow{/tex}\xa0D = 4 [(ac+ bd)2 - (a2 + b2) (c2 + d2)]{tex} \\Rightarrow{/tex}\xa0D = 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}\xa0bc.{tex} \\Rightarrow{/tex} ad - bc\xa0{tex} \\neq{/tex}\xa00{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.