Transformation of Roots
Fundamental Theorem of Symmetric Polynomials
If you have an expression involving \(\alpha\) and \(\beta\) where swapping \(\alpha\) and \(\beta\) doesn't change the expression (that's what we call 'symmetric'), then you can always rewrite this expression using just the sum of \(\alpha\) and \(\beta\) and the product of \(\alpha\) and \(\beta\).
Example 1:
Consider the expression \(\alpha^3 + \beta^3  3\alpha  3\beta + \alpha\beta\). This is a symmetric polynomial because if you swap \(\alpha\) and \(\beta\), the expression remains unchanged.
This expression can be rewritten in terms of the sum of the roots \(\alpha + \beta\) and the product of the roots \(\alpha\beta\). We can use known identities such as \(\alpha^3 + \beta^3 = (\alpha + \beta)^3  3\alpha\beta(\alpha + \beta)\) to rewrite the expression:
The entire expression is now in terms of \(\alpha + \beta\) and \(\alpha\beta\), which correspond to the coefficients of a quadratic equation in which \(\alpha\) and \(\beta\) are roots.
Example 2:
Consider the expression \(\alpha^2/\beta + \beta^2/\alpha\). This is a symmetric expression because swapping \(\alpha\) and \(\beta\) does not change its value. To express this in terms of the sum and product of \(\alpha\) and \(\beta\), first find a common denominator to combine the terms:
Now, use the identity \(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2  \alpha\beta + \beta^2)\) and the fact that \(\alpha^2 + \beta^2 = (\alpha + \beta)^2  2\alpha\beta\). This allows the expression to be rewritten in terms of \(\alpha + \beta\) and \(\alpha\beta\), showcasing the symmetry in the expression.
Use the following identities to easily convert symmetric expression in sum and product form: To express the given expressions in terms of \(\alpha + \beta\) and \(\alpha\beta\):

\(\alpha^2 + \beta^2 = (\alpha + \beta)^2  2\alpha\beta\).

\(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2  \alpha\beta + \beta^2)\).

\(\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2  2(\alpha\beta)^2\).

\(\alpha^5 + \beta^5 = (\alpha^3 + \beta^3)(\alpha^2 + \beta^2)  \alpha^2\beta^2(\alpha + \beta)\)
Now, apply the previously discussed identities for \(\alpha^3 + \beta^3\) and \(\alpha^2 + \beta^2\) and simplify the result using the values of \(\alpha + \beta\) and \(\alpha\beta\). This approach allows for the expression of \(\alpha^5 + \beta^5\) in terms of the sum and product of the roots, \(\alpha\) and \(\beta\).
Transformation of roots
To understand transformation of roots, consider the following example:
Example 3
Find the quadratic equation whose roots are \(\alpha + k\) and \(\beta + k\), where \(\alpha\) and \(\beta\) are roots of the given equation \(ax^2 + bx + c = 0\).
Solution:
So we are transforming roots of \(ax^2 + bx + c = 0\), \(\alpha\) and \(beta\) to new roots \(\alpha + k\) and \(\beta + k\). To form the corresponding quadratic equation, we use the sum and product of the new roots:

Sum of the new roots: The sum is \(S=(\alpha + k) + (\beta + k) = \alpha + \beta + 2k\). Since \(\alpha + \beta = \frac{b}{a}\), the new sum is \(S=\frac{b}{a} + 2k\).

Product of the new roots: The product is \(P=(\alpha + k)(\beta + k) = \alpha\beta + k(\alpha + \beta) + k^2\). Since \(\alpha\beta = \frac{c}{a}\), the new product is \(P=\frac{c}{a} + k(\frac{b}{a}) + k^2\).
The new quadratic equation having roots \(\alpha + k\) and \(\beta + k\) is formed using these new sum and product values:
Simplifying this equation will give the desired quadratic equation.
Observe that in the above problem we are shifting the roots forward by \(k\) units
Transformation via substitution.

Substitute \( y = x + k \): This means when \( x = \alpha \), \( y = \alpha + k \), and similarly, when \( x = \beta \), \( y = \beta + k \). Essentially, we're shifting each root by \( k \) units.

Replace \( x \) with \( y  k \) in the original quadratic equation \( ax^2 + bx + c = 0 \): \( a(y  k)^2 + b(y  k) + c = 0 \)

Simplify the Equation: Expand and simplify the equation to find the new quadratic in terms of \( y \). \( a(y  k)^2 + b(y  k) + c = 0 \)
\( \implies a(y^2  2yk + k^2) + b(y  k) + c = 0 \)
\( \implies ay^2  2ayk + ak^2 + by  bk + c = 0 \)
\( \implies ay^2 + (b  2ak)y + (ak^2  bk + c) = 0 \)

Roots of the New Equation: The roots of this new quadratic equation in \( y \) will be \( y = \alpha + k \) and \( y = \beta + k \), since these values of \( y \) correspond to the original roots \(\alpha\) and \(\beta\) shifted by \( k \).
This method effectively finds the quadratic equation whose roots are the original roots shifted by \( k \), and is a useful technique in algebra for modifying and analyzing quadratic equations as we will see in more detail in polynomials chapter.
Example 4
Given that \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2  px + q = 0\) and neither of them equals 1, find the quadratic equation whose roots are \(\frac{\alpha}{1  \alpha}\) and \(\frac{\beta}{1  \beta}\).
Solution:

Transformation of Roots:
 The transformation we want is \( x \to \frac{x}{1  x} \). Let's denote the transformed variable as \( y \), so \( y = \frac{x}{1  x} \).

Solve for \( x \):
 From \( y = \frac{x}{1  x} \), we get \( x = \frac{y}{1 + y} \).

Substitute in the Original Equation:
 Substitute \( x = \frac{y}{1 + y} \) into \(x^2  px + q = 0\): \( \left(\frac{y}{1 + y}\right)^2  p\left(\frac{y}{1 + y}\right) + q = 0 \)

Simplify the Equation:
 Expand and simplify the equation in terms of \( y \).
\( \left(\frac{y}{1 + y}\right)^2  p\left(\frac{y}{1 + y}\right) + q = 0 \)
\( \implies \frac{y^2}{(1 + y)^2}  \frac{py}{1 + y} + q = 0 \)
\( \implies y^2  py(1 + y) + q(1 + y)^2 = 0 \)
\( \implies y^2  py^2  py + q(1 + 2y + y^2) = 0 \)
\( \implies y^2  py^2  py + q + 2qy + qy^2 = 0 \)
\( \implies (1  p + q)y^2 + (2q  p)y + q = 0 \)

New Quadratic Equation:
 \((1  p + q)y^2 + (2q  p)y + q = 0 \)
You are advised to verify this answer by using the alternative method of calculating the sum and product of roots.