Question

If \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] are the two values of \[\lambda \] such that the roots \[\alpha \] and \[\beta \] of the quadratic equation, \[\lambda \left( {{x}^{2}}-x \right)+x+5=0\] satisfy \[\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+\dfrac{4}{5}=0\] , then \[\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}^{2}}+\dfrac{{{\lambda }_{2}}}{{{\lambda }_{1}}^{2}}\] is equal to
A.536
B.512
C.504
D.488

Answer 1

Hint:Transform \[\lambda \left( {{x}^{2}}-x \right)+x+5=0\] as \[\lambda {{x}^{2}}+x\left( 1-\lambda \right)+5=0\] . We know the formula of sum of roots, of the quadratic equation \[a{{x}^{2}}+bx+c=0\] , Sum of roots = \[\dfrac{-b}{a}\] and Product of roots = \[\dfrac{c}{a}\] . Apply this formula for the quadratic equation \[\lambda {{x}^{2}}+x\left( 1-\lambda \right)+5=0\] . Solve \[\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+\dfrac{4}{5}=0\] . Then using the formula, \[{{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \Rightarrow {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}}\] find the value of \[{{\alpha }^{2}}+{{\beta }^{2}}\] . Put the value of \[({{\alpha }^{2}}+{{\beta }^{2}})\] and \[\alpha \beta \] in the equation \[\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\dfrac{-4}{5}\] . Solving this, we get a quadratic equation in \[\lambda \] which has \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] as its roots. Using the formula of Sum of roots and Products of roots. Using the formula, \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] get the value of \[{{\lambda }_{1}}^{3}+{{\lambda }_{2}}^{3}\] . Now, using the formula, \[{{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \Rightarrow {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}}\] get the value of
\[{{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}\] and then solve it further.

Complete step-by-step answer:
According to the question, we have
\[\lambda \left( {{x}^{2}}-x \right)+x+5=0\] …………………..(1)
Transforming the above quadratic equation in form of \[a{{x}^{2}}+bx+c=0\] , we get
\[\begin{align}
  & \lambda \left( {{x}^{2}}-x \right)+x+5=0 \\
 & \Rightarrow \lambda {{x}^{2}}-\lambda x+x+5=0 \\
\end{align}\]
\[\Rightarrow \lambda {{x}^{2}}+x\left( 1-\lambda \right)+5=0\] …………………(2)
We know the formula of sum of roots, of the quadratic equation \[a{{x}^{2}}+bx+c=0\] ,
Sum of roots = \[\dfrac{-b}{a}\] …………………..(3)
Product of roots = \[\dfrac{c}{a}\] ………………………(4)
Similarly, for the quadratic equation \[\lambda {{x}^{2}}+x\left( 1-\lambda \right)+5=0\] , we have \[\alpha \] and \[\beta \] as its roots,
Using equation (3), we get
Sum of roots, \[\alpha +\beta =\dfrac{-\left( 1-\lambda \right)}{\lambda }\] ………………..(5)
Product of roots, \[\alpha \beta =\dfrac{5}{\lambda }\] …………………(6)
According to the information provided in the question, we also have
\[\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+\dfrac{4}{5}=0\]
 \[\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\dfrac{-4}{5}\] …………………….(7)
We know the formula, \[{{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \Rightarrow {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}}\] …………(8)
From equation (5), equation (6), and equation (8), we get
\[\begin{align}
  & {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}} \\
 & \Rightarrow {{\left\{ \left( \dfrac{-(1-\lambda )}{\lambda } \right) \right\}}^{2}}-2.\dfrac{5}{\lambda }={{\alpha }^{2}}+{{\beta }^{2}} \\
 & \Rightarrow \dfrac{1+{{\lambda }^{2}}-2\lambda -10\lambda }{{{\lambda }^{2}}}={{\alpha }^{2}}+{{\beta }^{2}} \\
\end{align}\]
\[\Rightarrow \dfrac{{{\lambda }^{2}}-12\lambda +1}{{{\lambda }^{2}}}={{\alpha }^{2}}+{{\beta }^{2}}\] ………………….(9)
Now, from equation (6), equation (7), and equation (9), we get
\[\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\dfrac{-4}{5}\]
\[\begin{align}
  & \dfrac{\dfrac{{{\lambda }^{2}}-12\lambda +1}{{{\lambda }^{2}}}}{\dfrac{5}{\lambda }}=\dfrac{-4}{5} \\
 & \Rightarrow \dfrac{{{\lambda }^{2}}-12\lambda +1}{5\lambda }=\dfrac{-4}{5} \\
 & \Rightarrow {{\lambda }^{2}}-12\lambda +1=-4\lambda \\
 & \Rightarrow {{\lambda }^{2}}-12\lambda +4\lambda +1=0 \\
\end{align}\]
\[\Rightarrow {{\lambda }^{2}}-8\lambda +1=0\] ………………..(10)
The above equation is quadratic in \[\lambda \] whose roots are \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] .
Using equation (3) and equation (4), we can get the sum and products of roots.
Sum of roots, \[{{\lambda }_{1}}+{{\lambda }_{2}}=8\] ………………..(11)
Product of roots, \[{{\lambda }_{1}}{{\lambda }_{2}}=1\] …………………(12)
We have to find the value of \[\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}^{2}}+\dfrac{{{\lambda }_{2}}}{{{\lambda }_{1}}^{2}}\] …………….(13)
On solving equation (13), we get
\[\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}^{2}}+\dfrac{{{\lambda }_{2}}}{{{\lambda }_{1}}^{2}}\]
\[=\dfrac{{{\lambda }_{1}}^{3}+{{\lambda }_{2}}^{3}}{{{\lambda }_{2}}^{2}{{\lambda }_{1}}^{2}}\]
\[=\dfrac{{{\lambda }_{1}}^{3}+{{\lambda }_{2}}^{3}}{{{({{\lambda }_{1}}{{\lambda }_{2}})}^{2}}}\] …………………..(14)
We know the formula, \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] .
Now, replacing a by \[{{\lambda }_{1}}\] and b by \[{{\lambda }_{2}}\] in the above formula, we get
\[{{\lambda }_{1}}^{3}+{{\lambda }_{2}}^{3}=\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)\left( {{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}-{{\lambda }_{1}}{{\lambda }_{2}} \right)\] …………………(15)
 From equation (14) and equation (15), we have
\[=\dfrac{{{\lambda }_{1}}^{3}+{{\lambda }_{2}}^{3}}{{{({{\lambda }_{1}}{{\lambda }_{2}})}^{2}}}\]
\[=\dfrac{\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)\left( {{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}-{{\lambda }_{1}}{{\lambda }_{2}} \right)}{{{({{\lambda }_{1}}{{\lambda }_{2}})}^{2}}}\] …………………..(16)
We know the formula, \[{{\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)}^{2}}={{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}+2{{\lambda }_{1}}{{\lambda }_{2}}\Rightarrow {{\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)}^{2}}-2{{\lambda }_{1}}{{\lambda }_{2}}={{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}\] .
Now, using equation (11) and equation (12), we get
\[\begin{align}
  & {{\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)}^{2}}-2{{\lambda }_{1}}{{\lambda }_{2}}={{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2} \\
 & \Rightarrow {{(8)}^{2}}-2.1={{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2} \\
\end{align}\]
\[\Rightarrow 62={{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}\] …………………..(17)
Now, from equation (11), equation (12), equation (16), and equation (17), we get
\[=\dfrac{\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)\left( {{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}-{{\lambda }_{1}}{{\lambda }_{2}} \right)}{{{({{\lambda }_{1}}{{\lambda }_{2}})}^{2}}}\]
\[\begin{align}
  & =\dfrac{\left( 8 \right)\left( 62-1 \right)}{{{(1)}^{2}}} \\
 & =\dfrac{8.61}{1} \\
 & =488 \\
\end{align}\]
So, the value of \[\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}^{2}}+\dfrac{{{\lambda }_{2}}}{{{\lambda }_{1}}^{2}}\] is 488.
Hence, the correct option is (D).

Note: This question involves a little bit longer method which leads to calculation mistakes. As each step matters so, approach this question step by step. Also, for this question one can miss the point that \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] are the roots of the equation \[{{\lambda }^{2}}-8\lambda +1=0\] .