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| 1. |
If alpha and beata are zeros of kx^2+5x+2such that 1/alpha ^2+1/beata^2=17/4.find the value of k |
| Answer» let f(x)=kx2+5x+2{tex}\\begin{array}{l}\\mathrm\\alpha\\;\\;\\mathrm{are}\\;\\mathrm\\beta\\;\\mathrm{are}\\;\\mathrm{zer}o\\;\\mathrm{of}\\;\\mathrm f(\\mathrm x)\\\\\\mathrm{so}\\;\\alpha+\\mathrm\\beta=-\\frac5{\\mathrm k},\\mathrm{αβ}=\\frac2{\\mathrm k}\\end{array}{/tex}{tex}\\begin{array}{l}\\frac1{\\alpha^2}+\\frac1{\\beta^2}\\\\=\\frac{\\alpha^2+\\beta^2}{\\alpha^2\\beta^2}=\\frac{{\\displaystyle\\left(\\alpha+\\beta\\right)^2}{\\displaystyle-}{\\displaystyle2}{\\displaystyle\\alpha}{\\displaystyle\\beta}}{\\alpha^2\\beta^2}\\\\=\\frac{\\displaystyle25/k^2-4/k}{\\displaystyle\\left({\\displaystyle\\frac2k}\\right)^2}=\\frac{\\displaystyle\\frac{25-4k}{k^2}}{\\displaystyle\\frac4{k^2}}=\\frac{25-4k}4=\\frac{17}4\\\\\\\\\\end{array}{/tex}{tex}\\begin{array}{l}4\\ast(25-4k)=17\\ast4\\\\100-16k=68\\\\16k=100-68\\\\16k=32\\\\k=2\\\\\\\\\\\\\\end{array}{/tex} | |