InterviewSolution
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InterviewSolution
| 1. |
If alpha and beta are zeros of x2-px+q prove that alpha2/beta2+ beta2/alpha2=p4/q2-4p2/q+2 |
| Answer» Here\xa0α and β are the zeros of polynomial f(x) = x2 - px + qSo a=1,b=-p,c=qSum of the zeroes α + β={tex}-\\frac ba{/tex}\xa0= pProduct of the zeroes αβ=q\xa0{tex}\\frac{{{\\alpha ^2}}}{{{\\beta ^2}}} + \\frac{{{\\beta ^2}}}{{{\\alpha ^2}}}{/tex}{tex}= \\frac{{{\\alpha ^4} + {\\beta ^4}}}{{{\\alpha ^2}{\\beta ^2}}}{/tex}{tex}=\\frac{\\left(\\mathrm\\alpha^2+\\mathrm\\beta^2\\right)^2-2\\left(\\mathrm{αβ}\\right)^2}{\\left(\\mathrm{αβ}\\right)^2}=\\frac{\\{(\\mathrm\\alpha+\\mathrm\\beta)^2-2\\mathrm{αβ}\\}^2-2\\left(\\mathrm{αβ}\\right)^2}{\\left(\\mathrm{αβ}\\right)^2}{/tex}{tex}=\\frac{(\\mathrm p^2-2\\mathrm q)^2-2\\mathrm q^2}{\\mathrm q^2}=\\frac{\\mathrm p^4+4\\mathrm q^2-4\\mathrm p^2\\mathrm q-2\\mathrm q^2}{\\mathrm q^2}=\\frac{\\mathrm p^4+2\\mathrm q^2-4\\mathrm p^2\\mathrm q}{\\mathrm q^2}{/tex}{tex}=\\frac{\\mathrm p4}{\\mathrm q^2}-\\frac{4\\mathrm p^2\\mathrm q}{\\mathrm q^2}+\\frac{2\\mathrm q^2}{\\mathrm q^2}=\\frac{\\mathrm p4}{\\mathrm q^2}-\\frac{4\\mathrm p^2}{\\mathrm q}+2=\\mathrm{RHS}{/tex}Hence, proved. | |