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| 1. |
If ad #bc then prove the equation (a²+ b²) x² + 2 (ac+bd)x+(c²+d²) =0 has no real roots |
| Answer» D=b²-2ab =2²(ac+bd)² - 4(a²+b²)(c²+d²) =4(a²c²+b²d²+2acbd) - 4(a²c²+a²d²+b²c²+b²d²) =4(a²c²+b²d²+2acbd - a²c²+a²d²+b²c²+b²d²) =4(2acbd - a²d² - b²c²) = -4(a²d²+b²c² - 2acbd) = -4{ (ad)² + (bc)² - 2(ad)(bc) } = -4 (ad - bc)² : - Hence, the value of the Discriminant will be negative. : - So, the equation will have no real roots. Good Bye ! ? | |