Easy one .

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I will try my best to provide a nice and elegant solution

First of all for clarity let $\lambda=a$

then we will get $a(x^2+x)+x+5=0 \\ ax^2+x(a+1)+5=0$

Now by Vieta we have $\alpha+\beta=\dfrac{-(a+1)}{a} \\ \alpha\beta=\dfrac{5}{a}$

According to question $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=4 \\ \dfrac{\alpha^2+\beta^2}{\alpha\beta}=4$

Now $\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=4$

$( \alpha+\beta)^2=(\dfrac{a^2+1+2a}{a^2}) \\ \therefore \dfrac{a^2+1+2a-\dfrac{10}{a}}{\dfrac{5}{a}}=4$

$\Rightarrow \dfrac { \left( \dfrac { a^{ 2 }+1+2a }{ a^{ 2 } } -\dfrac { 10 }{ a } \right) }{ \dfrac { 5 }{ a } } =4\\ \Rightarrow \dfrac { \left( \dfrac { a^{ 2 }+1-8a }{ a^{ 2 } } \right) }{ \dfrac { 5 }{ a } } =4\\ \Rightarrow \dfrac { a^{ 2 }+1-8a }{ a^{ 2 } } \times \dfrac { a }{ 5 } =4\\ \Rightarrow \dfrac { a^{ 2 }+1-8a }{ 5a } =4\\ \Rightarrow a^{ 2 }-28a+1=0$

Now let roots of this eqn. be $\zeta$ and $\eta$ (here $\lambda_1=\zeta,\lambda_2=\eta$ )

By vieta we have

$\Rightarrow \zeta +\eta =28\\ \Rightarrow \zeta \eta =1\\ \therefore (\zeta +\eta )^{ 2 }=784\\ \Rightarrow \zeta ^{ 2 }+\eta ^{ 2 }=784-2\times 1^{ 2 }\\ \Rightarrow \zeta ^{ 2 }+\eta ^{ 2 }=782\\ \therefore \dfrac { \zeta ^{ 2 }+\eta ^{ 2 } }{ \zeta \eta } =\dfrac { 782 }{ 1 }$

$\therefore$ the answer is $\huge\Rightarrow\boxed{\color{royalblue}{782}}$

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Upvote if you like it