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| 1. |
Alfa,bita n gama r zeroes of polynomial x³+px²+qx+2Such that ab+1=0find 2p+q+5. |
| Answer» P(x) =\xa0x3\xa0+ px2\xa0+ qx + 2Here, a = 1, b = p, c = q, d = 2Now,\xa0{tex}\\alpha + \\beta + \\gamma = \\frac { - b } { a }{/tex}= -p.......(i){tex}\\alpha \\beta + \\beta \\gamma + \\alpha \\gamma = \\frac { c } { a }{/tex}= q{tex}\\Rightarrow \\alpha \\beta + \\gamma ( \\beta + \\alpha ){/tex}\xa0= q..............(ii)and\xa0{tex}\\alpha \\cdot \\beta \\cdot \\gamma = \\frac { - d } { a }{/tex}= -2{tex}\\Rightarrow \\quad \\alpha . \\beta \\cdot \\gamma{/tex}= -2.......................(iii)Also,\xa0{tex}\\alpha \\beta + 1 = 0 \\Rightarrow \\alpha \\beta = - 1{/tex}Therefore, (iii) becomes -1{tex}\\times \\gamma = - 2 \\Rightarrow \\gamma = 2{/tex}Substituting in (i), we get{tex}\\alpha + \\beta + 2 = - p \\Rightarrow \\alpha + \\beta{/tex}\xa0=- p - 2Substituting these value in (ii), we get-1 + 2(-p - 2) = q{tex}\\Rightarrow{/tex}\xa0-1 - 2p - 4 = q{tex}\\Rightarrow{/tex}\xa02p + q + 5 = 0 | |