If `alpha` and `beta` are the roots of the equation `x^2+ax+b=0` and `alpha^4` and `beta^4` are the roots of the equation `x^2-px+q=0` then the roots of `x^2-4bx+2b^2-p=0` are always

Options

A. both non-real
B. both positive
C. both negative
D. positive and negative

Answer

D

Explanation

We have, `alpha + beta = - a, alpha beta = b, alpha^(4) + beta^(4) = p and alpha^(4) beta^(4) = q`.
Now, `alpha^(4)+beta^(4) = (alpha^(2) + beta^(2))^(2) - 2 alpha^(2) beta^(2)`
`rArr" "alpha^(4)+beta^(4)=[(alpha+beta)^(2)-2alpha beta]^(2)-2(alpha beta)^(2)`
`rArr" "alpha^(4) + beta^(4)=(a^(2)-2b)^(2)-2b^(2)" "[because alpha + beta = -a, alpha beta = b]`
`rArr" "alpha^(4) + beta^(4) = (a^(2))^(2) - 4a^(2)b + 2b^(2)`
`rArr" "p=(a^(2))^(2_ - 4b (a^(2))+ 2b^(2)" "[because alpha^(4) + beta^(4) = p]`
`rArr" "(a^(2))^(2) - 4b(a^(2))+2b^(2) - p = 0" "...(i)`
`rarr" "a^(2)` is a root of the equation `x^(2) - 4bx + 2b^(2) - p = 0`
Let `gamma` be the other root of `x^(2) - 4bx + 2b^(2) - p = 0`
`a^(2) + gamma = 4b rArr gamma = 4b - a^(2)" "...(ii)`
Since `alpha, beta` are real roots of `x^(2) + ax + b = 0`. Therefore,
Disc `ge 0 rArr a^(2) - 4b - a^(2) le 0 rArr gamma lt 0" "["Using (ii)"]`
Let D be the discriminant of `x^(2) - 4bx + 2b^(2) - p = 0`. Then,
`D = 16b^(2) - 4 (2b^(2) - p) = 16b^(2) + 4 {(a^(2))^(2) - 4b(a^(2))}" "["Using (i)"]`
`rArr" "D = 4 {(a^(2))^(2) - 4 ba^(2) + 4b^(2)} = 4 (a^(2) - 2b)^(2) ge 0`
Hence, the roots of the given equation are real and are of opposite signs.