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If \mathrm{ad} \neq \mathrm{bc}, then prove that the equation(a2 + b) x+ 2 (ac + bd)x+ (e + d) 0 has no real roots. |
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Answer» (a²+b²)x²+2(ac+bd)x+(c²+d²)=0a=(a²+b²) b=2(ac+bd) c=(c²+d²)putting a,b,c in formula[-b +/-√(b²-4ac)]/2a{-2(ac+bd) +/-√[4(ac+bd)²-4(a²+b²)(c²+d²)]}/2(a²+b²)[-2(ac+bd) +/-√4a²c²+4b²d²+8abcd-4a²c²-4a²d²-4b²c²-4b²d²]/2(a²+b²)-2(ac+bd) +/-√ - (4a²d²-8abcd+4b²c²)Hence root cant be negative.So, it has no real roots. |
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