Q-1-A-quadratic-equation-x-2-3x-4-0-has-roots-and-without-solving-a-write-down-the-values-of-2-2-b-find-the-quadratic-equation-with-integral-coefficients-whose-roots-are-1-2-and-1-

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Question Number 35363 by Rio Mike last updated on 18/May/18

$$Q_1.Aquadraticequation{x}^2-3x+4=0$$$$hasroots\alpha and\beta .withoutsolving$$$$a)writedownthevaluesof{\alpha }^2+{\beta }^2$$$$b)findthequadraticequation$$$$withintegralcoefficients,whose$$$$rootsare\frac{1}{{\alpha }^2}and\frac{1}{{\beta }^2}$$$$Q_2.thefirsttermofaGPis32$$$$andthesumtoinfinityis48.$$$$findthecommonratioandthe{8}^{th}$$$$termoftheprogression$$

Answered by Rasheed.Sindhi last updated on 18/May/18

$${\mathrm{Q}}_1$$$$\alpha \mathrm{\&}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2-3\mathrm{x}+4=0$$$$\alpha +\beta =\frac{-(-3)}{1}=3,\alpha \beta =\frac{4}{1}=4$$$$\left(a\right){\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$$$$={\left(3\right)}^2-2\left(4\right)=9-8=1$$$$\left(b\right)\mathrm{Thwe}\mathrm{equation}\mathrm{whose}\mathrm{roots}\mathrm{are}\frac{1}{{\alpha }^2}\mathrm{\&}\frac{1}{{\beta }^2}$$$$\mathrm{Sum}-\mathrm{of}-\mathrm{roots}=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{{\left(\alpha \beta \right)}^2}$$$$=\frac{1}{{\left(4\right)}^2}=1/16$$$$\mathrm{Product}-\mathrm{of}-\mathrm{roots}=\frac{1}{{\alpha }^2}\times \frac{1}{{\beta }^2}=\frac{1}{{\left(\alpha \beta \right)}^2}$$$$=\frac{1}{{\left(4\right)}^2}=\frac{1}{16}$$$$\mathrm{The}\mathrm{new}\mathrm{equation}:$$$${\mathrm{x}}^2-\left(\mathrm{sum}\right)\mathrm{x}+\mathrm{product}=0$$$${\mathrm{x}}^2-\left(1/16\right)\mathrm{x}+\left(1/16\right)=0$$$${16\mathrm{x}}^2-\mathrm{x}+1=0$$

Commented by MJS last updated on 18/May/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{typing}\mathrm{faster}\mathrm{than}\mathrm{me};-)$$

Commented by Rasheed.Sindhi last updated on 18/May/18

$$\mathrm{Only}\mathrm{for}\mathrm{this}\mathrm{time}.\mathrm{Always}\mathrm{you}\mathrm{beat}\mathrm{me}!$$

Answered by Rasheed.Sindhi last updated on 18/May/18

$${\mathrm{Q}}_2$$$$a=32$$$$S_{\mathrm{\infty }}=\frac{a}{1-r}=48$$$$\frac{32}{1-r}=48$$$$32=48-48r$$$$r=\frac{48-32}{48}=\frac{1}{3}$$$$t_n={ar}^{n-1}$$$$t_8=32\times {\left(1/3\right)}^{8-1}=\frac{32}{3^7}=\frac{32}{2187}$$

Answered by MJS last updated on 18/May/18

$$(x-\alpha )(x-\beta )=x^2-(\alpha +\beta )x+\alpha \beta$$$$\mathrm{we}\mathrm{know}\mathrm{that}\alpha \beta =4\mathrm{and}\alpha +\beta =3\Rightarrow$$$${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta =3^2-2\times 4=1$$$$(x-\frac{1}{{\alpha }^2})(x-\frac{1}{{\beta }^2})=x^2-\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}x+\frac{1}{{\alpha }^2{\beta }^2}=$$$$=x^2-\frac{1}{16}x+\frac{1}{16}$$$$16x^2-x+1=0$$

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