if alpha and beta are the zeroes of a quadratic polynomial 2x^2 - 3x - 5 form a polynomial whose zeroes are (i)alpha^2 and beta^2 (ii)1/alpha^2 and 1/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).
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) $$\frac{1}{\alpha^2} \;\;and\;\; \frac{1}{\beta^2}$$
$$\\Roots\;are\;\frac{1}{6.25}=\frac{4}{25}=0.16\;\;and\;\;1\\\\
so\;\;\frac{-b}{a}=1.16\;\;and\;\; \frac{c}{a}=0.16\\
Let a=1\\
$So one solution would be $ x^2-1.16x+0.16$$