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  • Let  be the roots of the equation <img alt="\mathrm{ x^{2}-\sqrt{2} x+\sqrt{6}=0}" src="/latex-

Let \mathrm{ \alpha, \beta} be the roots of the equation \mathrm{ x^{2}-\sqrt{2} x+\sqrt{6}=0} and \mathrm{ \frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1}be the roots of the equation\mathrm{ x^{2}+a x+b=0. } Then the roots of the equation \mathrm{ x^{2}-(a+b-2) x+(a+b+2)=0} are :

Option: 1

non-real complex numbers


Option: 2

real and both negative


Option: 3

real and both positive


Option: 4

real and exactly one of them is positive


Answers (1)

best_answer

\begin{aligned} & \mathrm{\alpha+\beta=\sqrt{2}, \quad \alpha \beta=\sqrt{6}}\\ &\text{Now}, \mathrm{a=-\left(\frac{1}{\alpha^2}+1+\frac{1}{\beta^2}+1\right)}\\ &\mathrm{=-\left(2+\frac{\alpha^2+\beta^2}{\alpha^2 \beta^2}\right) }\\ &\mathrm{=-\left(2+\frac{\left(\alpha+\beta \beta^2-2 \alpha \beta\right.}{\alpha^2 \beta^2}\right)} \\ &\mathrm{=-\left(2+\frac{2-2 \sqrt{6}}{6}\right) }\\ &=\mathrm{-\left(\frac{14-2 \sqrt{6}}{6}\right) }\end{aligned}\begin{aligned} \mathrm{b} &\mathrm{=\left(\frac{1}{\alpha^2}+1\right)\left(\frac{1}{\beta^2}+1\right)} \\ &\mathrm{=\frac{1}{\alpha^2 \beta^2}+\frac{1}{\alpha^2}+\frac{1}{\beta^2}+1 }\\ &\mathrm{=\frac{1}{6}+\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha^2 \beta^2}+1 }\end{aligned}

\begin{aligned} & \mathrm{=\frac{7}{6}+\frac{2-2 \sqrt{6}}{6}} \\ & \mathrm{+=\frac{3-2 \sqrt{6}}{6}} \\ \mathrm{\therefore a+b} &\mathrm{=-\frac{14}{6}+\frac{2 \sqrt{6}}{6}+\frac{9}{6}-\frac{2 \sqrt{6}}{6} }\\ &\mathrm{=-\frac{5}{6} }\\ &\mathrm{\therefore a+b+2 }=\mathrm{\frac{7}{6} }\\ &\mathrm{a+b-2 }= \mathrm{\frac{-17}{6}}\\ &\text{ Equation is } \mathrm{x^2+\frac{17}{6} x+\frac{7}{6}=0}\\ & \Rightarrow 6 x^2+17 x+7=0\\ &\text{ Both joots } \delta_t \text{ this equation are roal and negative }\end{aligned}

 

Posted by

Ajit Kumar Dubey

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