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If one root of the x square-x-k=0 be the square of other than k is equal to the |
| Answer» The given equation is:x2\xa0- x + k = 0\xa0As it is given that\xa03 is root of the equation x2\xa0- x + k = 0, therefore we have:(3)2\xa0- 3 + k = 0{tex}\\Rightarrow{/tex}\xa09 - 3 + k = 0{tex}\\Rightarrow{/tex}\xa06 + k = 0{tex}\\Rightarrow{/tex}\xa0k = - 6For p = - 6 the other given equation becomes:x2\xa0- 6(2x - 6 + 2) + p = 0{tex}\\Rightarrow{/tex}\xa0x2\xa0-12x + 24 + p = 0This is the form of ax2\xa0+ bx + c = 0,where a = -1, b = -12 and c = 24 + pSince the given equation has equal roots, therefore D = 0.i.e., b2\xa0- 4ac = 0{tex}\\Rightarrow{/tex}\xa0(-12)2\xa0- 4(1)(24 + p) = 0{tex}\\Rightarrow{/tex}\xa0144 - 96 - 4p = 0{tex}\\Rightarrow{/tex}\xa048 - 4p = 0{tex}\\Rightarrow{/tex}\xa04p = 48{tex}\\Rightarrow{/tex} p = 12Hence, the value of p is 12. | |