# Symmetric Functions of Roots of a Quadratic Equation

Let α and β be the roots of the quadratic equation ax$$^{2}$$ + bx + c = 0, (a ≠ 0), then the expressions of the form α + β, αβ, α$$^{2}$$ + β$$^{2}$$, α$$^{2}$$ - β$$^{2}$$, 1/α^2 + 1/β^2 etc. are known as functions of the roots α and β.

If the expression doesn’t change on interchanging α and β, then it is known as symmetric. In other words, an expression in α and β which remains same when α and β are interchanged, is called symmetric function in α and β.

Thus $$\frac{α^{2}}{β}$$ + $$\frac{β^{2}}{α}$$ is a symmetric function while α$$^{2}$$ - β$$^{2}$$ is not a symmetric function. The expressions α + β and αβ are called elementary symmetric functions.

We know that for the quadratic equation ax$$^{2}$$ + bx + c = 0, (a ≠ 0), the value of α + β = -$$\frac{b}{a}$$ and αβ = $$\frac{c}{a}$$. To evaluate of a symmetric function of the roots of a quadratic equation in terms of its coefficients; we always express it in terms of α + β and αβ.

With the above information, the values of other functions of α and β can be determined:

(i) α$$^{2}$$ + β$$^{2}$$ = (α + β)$$^{2}$$ - 2αβ

(ii) (α - β)$$^{2}$$ = (α + β)$$^{2}$$ - 4αβ

(iii) α$$^{2}$$ - β$$^{2}$$ = (α + β)(α - β) = (α + β) √{(α + β)^2 - 4αβ}

(iv) α$$^{3}$$ + β$$^{3}$$ = (α + β)$$^{3}$$ - 3αβ(α + β)

(v) α$$^{3}$$ - β$$^{3}$$ = (α - β)(α$$^{2}$$ + αβ + β$$^{2}$$)

(vi) α$$^{4}$$ + β$$^{4}$$ = (α$$^{2}$$ + β$$^{2}$$)$$^{2}$$ - 2α$$^{2}$$β$$^{2}$$

(vii) α$$^{4}$$ - β$$^{4}$$ = (α + β)(α - β)(α$$^{2}$$ + β$$^{2}$$) = (α + β)(α - β)[(α + β)$$^{2}$$ - 2αβ]

Solved example to find the symmetric functions of roots of a quadratic equation:

If α and β are the roots of the quadratic ax$$^{2}$$ + bx + c = 0, (a ≠ 0), determine the values of the following expressions in terms of a, b and c.

(i) $$\frac{1}{α}$$ + $$\frac{1}{β}$$

(ii) $$\frac{1}{α^{2}}$$ + $$\frac{1}{β^{2}}$$

Solution:

Since, α and β are the roots of ax$$^{2}$$ + bx + c = 0,
α + β = -$$\frac{b}{a}$$ and αβ = $$\frac{c}{a}$$

(i) $$\frac{1}{α}$$ + $$\frac{1}{β}$$

= $$\frac{α + β}{αβ}$$ = -b/a/c/a = -b/c

(ii) $$\frac{1}{α^{2}}$$ + $$\frac{1}{β^{2}}$$

= α^2 + β^2/α^2β^2

= (α + β)$$^{2}$$ - 2αβ/(αβ)^2

= (-b/a)^2 – 2c/a/(c/a)^2 = b^2 -2ac/c^2