If alpha and beata are zeros …
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Ram Kushwah 6 years, 7 months ago
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}
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