Saved Bookmarks
| 1. |
Prove that x2+4x+5 has no real zeroes |
| Answer» f(x) = x4\xa0+ 4x2\xa0+ 5= (x2)2\xa0+ 4x2\xa0+ 5Let x2\xa0=n,Then, f(x) = n2\xa0+ 4n + 5,Here a=1,b=4,c=5The discriminant(D) = {tex}\\text{b}^2-4\\mathrm{ac}=\\;(4)^2-4\\times1\\times5=16-20=-4{/tex}Since the discriminant is negative so this polynomial has no zerosHence, f(x) = x4\xa0+ 4x2\xa0+ 5\xa0has no zero. | |