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  •  are roots of the equation <img alt="\lambda (x^2 -x ) + x + 5 = 0 ." src="https://learn.ca

\alpha , \beta are roots of the equation \lambda (x^2 -x ) + x + 5 = 0 . If \lambda _1 and \lambda _2 are the two values of \lambda for which the roots \alpha , \beta are connected by the relation \frac{\alpha}{\beta } + \frac{\beta}{\alpha} = 4 then the value of\frac{\lambda _1}{\lambda _2 } + \frac{\lambda _2}{\lambda _1} is

Option: 1

150


Option: 2

254 


Option: 3

180 


Option: 4

1022


Answers (1)

best_answer

As we have learnt in

 

Sum of Roots in Quadratic Equation -

\alpha +\beta = \frac{-b}{a}

- wherein

\alpha \: and\beta are root of quadratic equation

ax^{2}+bx+c=0

a,b,c\in C

 

 

Product of Roots in Quadratic Equation -

\alpha \beta = \frac{c}{a}

- wherein

\alpha \: and\ \beta are roots of quadratic equation:

ax^{2}+bx+c=0

a,b,c\in C

 

  

\alpha , \beta  are root of 

\lambda x^2 - ( \lambda -1 ) x +5 = 0 \\\\ \therefore \alpha + \beta = \frac{\lambda -1}{\lambda } \: \: and \: \: \alpha \beta = 5/ \lambda \\\\ \therefore

\frac{\alpha}{\beta } + \frac{\beta}{\alpha} = 4

  \frac{\alpha ^2 + \beta ^2 }{\alpha \beta } = 4 \\\\ ( \alpha + \beta )^2 = 6 \alpha \beta \\\\ \frac{(\lambda -1)^2}{\lambda } = 30 / \lambda \\\\ \lambda ^2 - 32 \lambda + 1 = 0 ....(1) \\\\ \lambda _1, \lambda _2 \: \: are \: \: roots\: \: of\: \: (1)\\\\ \lambda _1 + \lambda _2 = 32 \: \: and \: \: \lambda _1 \lambda _2 = 1 \\\\

\frac{\lambda _1}{\lambda _2 } + \frac{\lambda _2}{\lambda _1} = \frac{(\lambda _1+ \lambda _2 )^2-2 \lambda _1 \lambda _2}{\lambda _1 \lambda _2}  = \frac{(32)^2-2}{1} = 1022

 

Posted by

HARSH KANKARIA

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