If alpha and beta are the zeroes of the quadratic polynomial ax^2+bx+c...
Solution:
Given quadratic polynomial ax^2 + bx + c
Zeroes of the polynomial are α and β
Sum of zeroes = α + β = -b/a
Product of zeroes = αβ = c/a
(i) α^2β^2 = (αβ)^2 = (c/a)^2
(ii) α/β and β/α
α/β = αβ/β^2 = c/aβ
β/α = βα/α^2 = c/aα
(iii) 1/α^3 and 1/β^3
1/α^3 = 1/(α^2 α) = (αβ/a)^3
1/β^3 = 1/(β^2 β) = (αβ/a)^3
(iv) α^2/β and β^2/α
α^2/β = αβ/β^2 = c/aβ
β^2/α = βα/α^2 = c/aα
If alpha and beta are the zeroes of the quadratic polynomial ax^2+bx+c...
Given that ax^2+bx+c, by which we can get alpha+beta=(-b/a) and alpha*beta=(c/a) then (1) alpha^2+beta^2=(alpha+beta)^2 -2*alpha*beta=(b^2-2ac)/a^2 (2)alpha/beta+beta/alpha=(alpha^2+beta^2)/alpha*beta=[(b^2-2ac)/a^2]/c/a by solving it we get (b^2-2ac)/ac (3) 1/alpha^3+1/beta^3=(beta^3+alpha^3)/(alpha*beta)^3=[(alpha+beta)(alpha^2+beta^2-alpha*beta)]/(alpha*beta)^3 by putting the value of (alpha+beta),(alpha^2+beta^2)&(alpha*beta) we get the answer that is( -b^3+3abc)/c^3 (4) alpha^2/beta+beta^2/alpha=(alpha^3tbeta^3)/alpha*beta=[(alpha+beta)(alpha^2+beta^2-alpha*beta)]/(alpha*beta) again by putting the value of (alpha+beta),(alpha^2+beta^2)&(alpha*beta)as I already give in 1st question; we can get the answer which is (-b^3+3abc)/ac^2