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

11 and 12 Grade Math 

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