if alpha and bita are the zeros of polynomial p(x=3xsquare+2x+1,find the polynomial whose zeros are 1-alpha/1+alpha and 1-bita/1+bita
for ease of typing I will use a and b instead
for 3x^2 + 2x + 1= 0
the sum of the roots = a+b = -2/3
the product of the roots = ab = 1/3
then (1-a)/(1+a) + (1-b)/(1+v) , (sum of new roots)
= ((1-a)(1+b))/((1+a)(1+b))
= (2 - 2ab)/(1 + a+b + ab)
= (2 - 2(1/3))/(1 - 2/3 + 1/3)
= (4/3)/(2/3) = 2
(1-a)/(a+a) (1-b)/(1+b)
= (1 -a -b + ab)/(1+a+b+ab)
= (1 + 2/3 + 1/3)/(1 -2/3 + 1/3)
= 2/(2/3)
= 3
so new equation is
x^2 - 2x + 3 = 0
check my arithmetic
To find the polynomial whose zeros are 1 - (alpha / (1 + alpha)) and 1 - (bita / (1 + bita)), we can first determine the zeros of the polynomial p(x) = 3x^2 + 2x + 1.
Given that alpha and bita are the zeros of p(x), we have:
p(alpha) = 0
p(bita) = 0
Now, let's find the values of alpha and bita.
For p(alpha) = 0:
3(alpha)^2 + 2(alpha) + 1 = 0
For p(bita) = 0:
3(bita)^2 + 2(bita) + 1 = 0
Now, let's solve these equations to find the values of alpha and bita.
For p(alpha) = 0:
3(alpha)^2 + 2(alpha) + 1 = 0
Simplifying:
3(alpha)^2 + 2(alpha) = -1
For p(bita) = 0:
3(bita)^2 + 2(bita) + 1 = 0
Simplifying:
3(bita)^2 + 2(bita) = -1
To find the zeros of the new polynomial, we substitute 1 - (alpha / (1 + alpha)) and 1 - (bita / (1 + bita)) for x in p(x) = 3x^2 + 2x + 1.
So, the new polynomial with the given zeros is p(x) = 3[(1 - (alpha / (1 + alpha)))^2] + 2[(1 - (alpha / (1 + alpha)))] + 1.
To find the polynomial whose zeros are given by 1 - α/(1 + α) and 1 - β/(1 + β), we can make use of Vieta's formulas. Vieta's formulas state that for a polynomial
p(x) = ax^2 + bx + c,
if α and β are the roots of the polynomial with α ≠ β, then the sum of the roots is -b/a and the product of the roots is c/a.
Given that the zeros of the polynomial p(x) = 3x^2 + 2x + 1 are α and β, we can apply Vieta's formulas to find their sum and product.
Sum of the roots = -b/a = -2/3
Product of the roots = c/a = 1/3
Let's construct a new polynomial with the desired roots.
Let's say the new polynomial is q(x) = Ax^2 + Bx + C.
The sum of the roots of q(x) is (1 - α/(1 + α)) + (1 - β/(1 + β)). Simplifying this expression gives:
1 - α/(1 + α) + 1 - β/(1 + β) = 2 - (α/(1 + α) + β/(1 + β))
Now, substitute the value of the sum of the roots of p(x) into the expression:
2 - (α/(1 + α) + β/(1 + β)) = 2 - (-2/3) = 2 + 2/3 = 8/3
The sum of the roots of the new polynomial q(x) is 8/3.
The product of the roots of q(x) is obtained similarly:
(1 - α/(1 + α))(1 - β/(1 + β)) = 1 - (α/(1 + α) + β/(1 + β)) + (α/(1 + α))(β/(1 + β))
Substitute the values from the product of the roots of p(x):
1 - (α/(1 + α) + β/(1 + β)) + (α/(1 + α))(β/(1 + β)) = 1 - (1/3) + (1/3)(1/3) = 5/9
The product of the roots of the new polynomial q(x) is 5/9.
So, the new polynomial q(x) is given by:
q(x) = A(x - 1 - α/(1 + α))(x - 1 - β/(1 + β))
Using the values of the sum and product of the roots:
q(x) = A(x - 8/3)(x - 5/9)
Now, we can multiply this out to get the specific polynomial:
q(x) = A(x^2 - (8/3)x - (5/9)x + (8/3)(5/9))
= A(x^2 - (29/9)x + (40/27))
Therefore, the polynomial whose zeros are given by 1 - α/(1 + α) and 1 - β/(1 + β) is q(x) = x^2 - (29/9)x + (40/27).