If alpha, beta are the zeroes of x2-2x+3, find a quadratic equation whose roots are (alpha-1)(beta+1), (beta-1)(beta+1)
since the roots of the quadratic are 1±√2 i
(alpha-1) and (beta-1) are ±√2 i
If you want real coefficients, this will not produce the required complex conjugates.
I suspect a typo. So, fix that and then just plug and chug.
To find a quadratic equation with the given roots, let's expand the expressions (alpha-1)(beta+1) and (beta-1)(beta+1).
Expanding (alpha-1)(beta+1):
(alpha-1)(beta+1) = alpha * beta + alpha - beta - 1
Expanding (beta-1)(beta+1):
(beta-1)(beta+1) = beta^2 - beta + beta -1 = beta^2 - 1
We have two expressions now:
1. alpha * beta + alpha - beta - 1
2. beta^2 - 1
To find the quadratic equation, we know that if a, b, and c are the roots of a quadratic equation ax^2 + bx + c = 0, then the equation can be written as (x - a)(x - b)(x - c) = 0.
In this case, our roots are (alpha-1)(beta+1) and (beta-1)(beta+1).
Hence, the quadratic equation can be written as:
[(x - ((alpha-1)(beta+1))][(x - ((beta-1)(beta+1))] = 0
Expanding further:
[(x - (alpha * beta + alpha - beta - 1))][(x - (beta^2 - 1))] = 0
Simplifying:
(x - (alpha * beta + alpha - beta - 1))(x - beta^2 + 1) = 0
Expanding and combining like terms:
x^2 - (alpha * beta + alpha - beta - 1)x - beta^2x + (alpha * beta + alpha - beta - 1)beta^2 = 0
Finally, rearranging terms:
x^2 - (alpha * beta + alpha - beta - 1 + beta^2)x + (alpha * beta + alpha - beta - 1)beta^2 = 0
Therefore, the quadratic equation whose roots are (alpha-1)(beta+1) and (beta-1)(beta+1) is:
x^2 - (alpha * beta + alpha - beta - 1 + beta^2)x + (alpha * beta + alpha - beta - 1)beta^2 = 0