Saved Bookmarks
| 1. |
(C2 ab)x2-(a2 -bc)x+b2 ac=0 has equal roots prove that a=0 |
| Answer» We have,\xa0{tex}(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0{/tex}Here,\xa0{tex}A=(c^2-ab), B=-2(a^2-bc),C=b^2-ac{/tex}\xa0Now,\xa0{tex}B^2-4AC =0{/tex}{tex}4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0{/tex}{tex}4(a^2-bc)^2=4(c^2-ab)(b^2-ac){/tex}{tex}a^4-2a^2bc+b^2c^2=b^2c^2-ac^3-ab^3+a^2bc{/tex}{tex}a^4-2a^2bc+ac^3+ab^3-a^2bc=0{/tex}{tex}a^4-3a^2bc+ac^3+ab^3=0{/tex}{tex}a(a^3-3abc+c^3+b^3)=0{/tex}{tex}a=0 \\ or \\ a^3-3abc+c^3+b^3=0{/tex}{tex}a=0\\ or\\ a^3+b^3+c^3=3abc{/tex}Hence proved | |