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Question: If \[\;\lambda \in R\] is such that the sum of the cubes of the roots of the equation, \[{x^2} + (2-...

If   λ∈R\;\lambda \in R is such that the sum of the cubes of the roots of the equation, x2+(2−λ)x+(10−λ)=0{x^2} + (2-\lambda )x + (10-\lambda ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is
A.2020
B.  25\;2\sqrt 5
C.  27\;2\sqrt 7
D.  42\;4\sqrt 2

Explanation

Solution

First we have to define what the terms we need to solve the problem are.
First, we check the question what the terms given
This question involves a simple concept of quadratic equation and its roots.
Here in given equation we have to get sum of roots and products of roots and then by using the formula
α,β=−b±b2−4ac2a\alpha ,\beta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step by step answer:
Let a quadratic equation is
ax2  + bx+c=0a{x^2}\; + {\text{ }}bx + c = 0has root of α\alpha and β\beta
Now, given equation is
x2+(2−λ)x+(10−λ)=0{x^2} + (2-\lambda )x + (10-\lambda ) = 0
now by comparing this equation with ax2  +bx+c=0a{x^2}\; + bx + c = 0
we have a,b,ca,b,c value
a=1,b=  (2−λ),c=(10−λ)a = 1,b = \;(2-\lambda ),c = (10-\lambda )
let us put the values in this equation.
By quadratic formula, the root of this equation is:
α,β=−b±b2−4ac2a\alpha ,\beta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Given equationx2+(2−λ)x+(10−λ)=0{x^2} + (2-\lambda )x + (10-\lambda ) = 0
a=1,b=(2−λ),c=(10−λ)a = 1,b = (2-\lambda ),c = (10-\lambda )
let us put the values in this equation.
So, we onsimplifying,
α,β=((λ−2)±(2−λ)2−4(10−λ))2\alpha ,\beta = \dfrac{{((\lambda - 2) \pm \sqrt {{{(2 - \lambda )}^2} - 4(10 - \lambda )} )}}{2}
We substituting the formula for above equation,
Again,we onsimplifying,
α,β=((λ−2)±(λ2−36))2\alpha ,\beta = \dfrac{{((\lambda - 2) \pm \sqrt {({\lambda ^2} - 36)} )}}{2} and hence α=((λ−2)+(λ2−36))2\alpha = \dfrac{{((\lambda - 2) + \sqrt {({\lambda ^2} - 36)} )}}{2} β=((λ−2)−(λ2−36))2\beta = \dfrac{{((\lambda - 2) - \sqrt {({\lambda ^2} - 36)} )}}{2}
The givendifference of the magnitude roots is clearly ∣λ(λ2−36)∣|\sqrt {\lambda ({\lambda ^2} - 36)} |
We have,
the sum of the cubes of root equation,
α3+β3=(λ−2)34+(3(λ−2)(λ2−36))4{\alpha ^3} + {\beta ^3} = \dfrac{{{{(\lambda - 2)}^3}}}{4} + \dfrac{{(3(\lambda - 2)({\lambda ^2} - 36))}}{4}
So, we onsimplifying,
=((λ−2)(λ2−4λ−104))4= \dfrac{{((\lambda - 2)({\lambda ^2} - 4\lambda - 104))}}{4}
Again,we onsimplifying,
=(λ−2)(λ2−λ−26)= (\lambda - 2)({\lambda ^2} - \lambda - 26)
This function attains its minimum value at λ=4\lambda = 4
We know that,
Thus, the given difference of the magnitude rootsis clearly ∣i20∣\left| {i\sqrt {20} } \right|
Which is equal to, ∣i20∣=25\left| {i\sqrt {20} } \right| = 2\sqrt 5

Hence, the correct answer is option (B)   25\;2\sqrt 5

Note:
The sum of the cubes of the roots of the equation, x2+(2−λ)x+(10−λ)=0{x^2} + (2-\lambda )x + (10-\lambda ) = 0 is minimum, then the magnitude of the difference of the roots of this equation we find the quadratic equations and its roots. In basic formula of quadratic equation, we used all quadratic equation sums. First mainly we find out aa value, bb value and cc value and find the answer.