If m is chosen in the quadratic equation
(m2 + 1)
x2 -- 3x + (m2 + 1)2 = 0
such that the sum of its
roots is greatest, then the absolute difference of
the cubes of its roots is :-
Solutions · 1
Official solution
Given quadratic equation
(m2 + 1)
x2 -- 3x + (m2 + 1)2 = 0
Let roots of the equation \[\alpha \] and \[\beta \].
\[ \therefore \] Sum of roots = \[\alpha \] + \[\beta \] = \[{3 \over {{m^2} + 1}}\]
Product of roots = \[\alpha \] β= m2 + 1{3 \over {{m^2} + 1}}\[ is maximum when m = 0
Hence equation becomes x2 -- 3x + 1 = 0
\]\alpha + \beta = 3,αβ=1∣α−β∣=(α+β)2−4αβ=(3)2−4.1=5α3−β3=(α−β)(α2+β2+αβ)=5(α+β)2−αβ=5(9−1)=85