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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