Solving Polynomials using roots

if α and β are roots of x^2 + 5x + 7 = 0 form the quadratic equations whose roots are (α^2-β^2) and (α^2+β^2)

i got that α+β=-5 and αβ=7
which equate the two roots
(α-β)^2 = 25 - 14 --> = 11
(α+β)^2= 25+14 = 25 + 14 --> =39
which equates to the quadratic
x^2+50x+429=0 --> But according to the book is wrong

the answer is
x^2+22x+75=0

We have (α-β)^2 = 25 - 28 = -3 and (α+β)^2 = 25
Then (α-β)^2 + (α+β)^2 = -3 + 25 = 22
and (α-β)^2 (α+β)^2 = (-3)* 25 = -75
Then the required quadratic equation should be x^2 -22x -75 = 0.
There must be some mistakes somewhere!

The equation whose roots are (α^2-β^2) and (α^2+β^2) is

x^2 - {(α^2-β^2) + (α^2+β^2)}x + {(α^2-β^2)(α^2+β^2)} = 0 or

x^2 - 2α^2x + (α^4 - β^4) = 0 .

From the given equation we have ,

α + β = - 5 and αβ = 7 ;

(α - β)^2 = (α + β)^2 - 4αβ = 25 - 28 = - 3 => (α - β) = ± √3 i.

also,

α = (- 5) + sqrt(25 - 28)/2 = - (5/2) + i√3/2 ;

β = - (5/2) - i√3/2 .

α^2 = 25/4 - (5√3/2)i - 3/4 = (22/4) - (5√3/2)i ;

β^2 = (25/4) + (5√3/2)i - 3/4 = (22/4) + (5√3/2)i ;

α^4 - β^4 = (α^2 + β^2)(α^2 - β^2)

= {(22/4) - (5√3/2)i + (22/4) + (5√3/2)i}{(22/4) - (5√3/2)i - (22/4) - (5√3/2)i

= - 11 X (5√3)i = - 55√3 i ;

the equation is

x^2 - (11 - 5√3 i)x - 55√3 i = 0 .

This polynomial has no real roots. The roots are alpha= (1/2)(-5+sqrt(3)i) and beta=(1/2)(-5-sqrt(3)i).
alpha^2+beta^2=11 and alpha^2-beta^2=-5sqrt(3)i.

The polynomial with these roots would be
f(x)=(x-11)(x+5sqrt(3)i).

Hope this helps