ALGEBRA EQUATIONS - SOLVING POLYNOMIAL EQUATIONS OF HIGHER DEGREE WITH EXAMPLES
Please study introduction given to equations and Algebra Equations in Math Equations, if you have not already done so. We studied about equations of first degree in one and two variables and word problems on them in Linear Equations. We also covered equations of second degree, their solution, nature of roots, relationship between roots and coefficients, word problems in Quadratic Equations. In
Polynomials, we covered definition, degree, zeros of polynomials, remainder theorem, factor theorem. Applications of factor theorem and Synthetic Division to find the factors of polynomials were covered in Algebra Factoring. Please study all these before proceeding further. They are prerequisites here.
Here, we study Polynomial Equations of higher degree. In Algebra Equations, we cover Polynomial Equations.
Relation between the roots and coefficients of an nth degree equation : Algebra Equations
Consider the nth degree equation xn + p1x(n - 1) + p2x(n - 2) + p3x(n - 3) + .......... + pn = 0 Let α1, α2, α3,α4,..........αn be the roots of the equation. Then, s1 = sum of the roots = Σ α1 = (-1)1p1 s2 = sum of the products of the roots taken two at a time = Σ α1α2 = (-1)2p2 s3 = sum of the products of the roots taken three at a time = Σ α1α2α3 = (-1)3p3 s4 = sum of the products of the roots taken four at a time = Σ α1α2α3α4 = (-1)4p4 ................................................... .................................................... sn = sum of the products of the roots taken all at a time = α1α2α3α.4..........αn = (-1)npn
NOTE 1. The above relations are true only when the coefficient of xn is unity. In case it is not unity, the algebra equation shouldbe divided with the coefficient of xn to make it unitybefore writing the above relations between the roots and coefficients.
2. The number of relations we have is same as the degree of the equation.So there are n relations to find n roots. Though, n relations are sufficientto find n unknowns, it is practically difficult to find the roots with thehelp of the relations only.
Usually, we are given something more about the roots. If that additional data is not given, we can solve some Algebra Equations by trial and error, using remainder theorem or factor theorem, as we did in
3. For n = 3, we get a third degree or cubic algebra equation x3 + p1x2 + p2x + p3 = 0 Then s1 = α1 + α2 + α3 = -p1 s2 = α1α2 + α1α3 + α2α3 = p2 s3 = -p3
4. For n = 4, we get a fourth degree or biquadratic algebra equation x4 + p1x3 + p2x2 + p3x + p4 = 0 Then s1 = α1 + α2 + α3 + α4 = -p1 s2 = α1α2 + α1α3 + α1α4 + α2α3 + α2α4 + α3α4 = p2 s3 = α1α2α3 + α1α2α4 + α1α3α4 + α2α3α4 = -p3 s4 = α1α2α3α4 = p4
Revision of knowledge of Progressions :
In solving Algebra Equations, the following knowledge is useful.
Arithmetic Progression (A.P.) : Quantities are said to be in A.P., ifthe difference of any term and its preceding term is a constant. Harmonic Progression (H.P.) : Quantities are said to be in H.P., iftheir reciprocals are in A.P. Geometric Progression (G.P.) : Quantities are said to be in G.P., ifthe ratio of any term and its preceding term is a constant.
For a cubic equation, when the roots are (i) in A.P., they are taken as α - β, α, α + β (ii) in H.P., they are taken as 1⁄(α - β), 1⁄α, 1⁄(α + β) (iii) in G.P., they are taken as α⁄β, α, αβ
For a biquadratic equation, when the roots are (i) in A.P., they are taken as α - 3β, α - β, α + β, α + 3β (ii) in H.P., they are taken as 1⁄(α - 3β), 1⁄(α - β), 1⁄α, 1⁄(α + 3β) (iii) in G.P., they are taken as α⁄β3, α⁄β, αβ, αβ3
Progressive Learning of Math : Algebra Equations
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