if alpha and beta are the zeroes of a quadratic polynomial 2x^2 - 3x - 5 form a polynomial whose zeroes are alpha^2 and beta^2

The polynomial can be factored into: (x+1)(2x-5), so we have α = -1 and β = 5/2

(i)  The polynomial with zeros α² and β² can be factored as (x - α²)(x - β²).   This can be multiplied out to get 

x² - (α² + β²)x + α²β²    or    x² - (1 + 25/4)x + 25/4    or    x² -  (29/4)x + 25/4  

Multiply all the terms by 4 to get   4x² -  29x + 25

See if you can now do part (ii).

If alpha and beta are the zeroes of a quadratic polynomial   $$2x^2 - 3x - 5$$   form a polynomial whose zeroes are

The roots for this is  2.5 an -1         These are     $$\alpha\;\;and\;\;\beta$$

(i)       $$\alpha^2=6.25\;\;and \;\;\beta^2 =1$$

In any quadratic    ax^2+bx+c   the sum of the roots is given by  -b/a = 7.25    

and the product of the roots is c/a = 6.25

So one solution would be       $$x^2-7.25x+6.25$$

(ii)      $$\frac{1}{\alpha^2} \;\;and\;\; \frac{1}{\beta^2}$$        

    $$\\Roots\;are\;\frac{1}{6.25}=\frac{4}{25}=0.16\;\;and\;\;1\\\\ so\;\;\frac{-b}{a}=1.16\;\;and\;\; \frac{c}{a}=0.16\\ Let a=1\\ $So one solution would be $ x^2-1.16x+0.16$$