ALGEBRA PUZZLES - SOLVING POLYNOMIAL EQUATIONS OF HIGHER DEGREE WITH EXAMPLES

Please study Method of Solving Equations before Algebra Puzzles if you have not already done so. There we gave introduction to Algebra Equations of higher degree and discussed the method of solving them.

We developed the relations between the roots and coefficients of the equation and gave Formulas for the same. That knowledge is a prerequisite here. So please study them before proceeding further.

Using (i) and (ii), we get (α - β)^{2} = (α + β)^{2} - 4αβ= (-2)^{2} - 4(±3) = 4 - 12 or 4 + 12 = -8 or 16 Taking positive value, α - β = ±4.......(iii) (i) + (iii) gives 2α = 2 or -6⇒ α = 1 or -3 using these in (i), we get β = -2 - α = -2 -1 or -2 +3 = -3 or 1 These are the same as α values. ∴ the roots are 1, 1, -3, -3.

These roots are found using s_{1} and s_{4}. Let us verify whether these satisfy s_{2} and s_{3}. s_{2} = (1)(1) + (1)(-3) + (1)(-3) + (1)(-3) + (1)(-3) + (-3)(-3) = 10 - 12 = -2 [satisfied.] s_{3} = (1)(1)(-3) + (1)(1)(-3) + (1)(-3)(-3) + (1)(-3)(-3) = 18 - 6 = 12 [satisfied.]

Thus the roots of the given Algebra Equation are 1, 1, -3, -3. Ans.

Thus Example 1 on Algebra Puzzles is solved.

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Solve the Algebra Equation 16x^{4} - 64x^{3} + 56x^{2} + 16x - 15 = 0, the roots of which are in A.P.

Solution to Example 2 on Algebra Puzzles : The given Algebra Equation is 16x^{4} - 64x^{3} + 56x^{2} + 16x - 15 = 0 Dividing both sides of the equation by 16, we get x^{4} - 4x^{3} + (7⁄2)x^{2} + x - 15⁄16 = 0 Comparing this with x^{4} + p_{1}x^{3} + p_{2}x^{2} + p_{3}x + p_{4} = 0, we get p_{1} = -4; p_{2} = 7⁄2; p_{3} = 1; p_{4} = -15⁄16

By data the roots are in A.P. Let the roots be α - 3β, α - β, α + β, α + 3β We know, s_{1} = -p_{1} = 4⇒ α - 3β + α - β + α + β + α + 3β = 4⇒ 4α = 4⇒ α = 1......(i)

Using the value of α = 1, the roots become 1 - 3β, 1 - β, 1 + β, 1 + 3β

When β = +(1⁄2), the roots become -1⁄2, 1⁄2, 3⁄2, 5⁄2. When β = -(1⁄2), the roots become 5⁄2, 3⁄2, 1⁄2, -1⁄2. As we can see, both sets are same.

Let us verify whether the set of roots satisfy values of s_{3} and s_{4}We know, s_{3} = -p_{3} = -1 and s_{4} = p_{4} = -15⁄16But, s_{3} = (-1⁄2)(1⁄2)(3⁄2) + (-1⁄2)(1⁄2)(5⁄2) + (-1⁄2)(3⁄2)(5⁄2) + (1⁄2)(3⁄2)(5⁄2) = (-3⁄8) + (-5⁄8) + (-15⁄8) + (15⁄8) = -1 [satisfied.]s_{4} = (-1⁄2)(1⁄2)(3⁄2)(5⁄2) = -15⁄16 [satisfied.]

∴ The roots of the given Algebra Equation are -1⁄2, 1⁄2, 3⁄2, 5⁄2. Ans.

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