CUBIC EQUATION - SOLVING POLYNOMIAL EQUATIONS OF THIRD DEGREE WITH EXAMPLES

Please study Method of Solving Cubic Equation 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.

Solved Example 1 : Cubic Equation

Solve the following problem on Cubic Equation.

If one root of the Algebra Equation 2x^{3} + 3x^{2} - 8x + 3 = 0 is double another root, solve the equation.

Solution to Example 1 of Cubic Equation : The given Algebra Equation is 2x^{3} + 3x^{2} - 8x + 3 = 0 Dividing both sides of the equation by 2, we get x^{3} + (3⁄2)x^{2} - 4x + 3⁄2 = 0 Comparing this with x^{3} + p_{1}x^{2} + p_{2}x + p_{3} = 0, we get p_{1} = 3⁄2; p_{2} = -4; p_{3} = 3⁄2

By data one root is double the other root. So if one root is α, then the other root is 2α. Let the third root be β We know, sum of roots = -p_{1}⇒ α + 2α + β = -3⁄2 ⇒ 3α + β = -3⁄2 ⇒ 6α + 2β = -3...........(i)

sum of the product of the roots taken two at a time = p_{2} = -4 ⇒ (α)(2α) + (α)(β) + (2α)(β) = -4 ⇒ α(2α + β + 2β) = -4 ⇒ α(2α + 3β) = -4.............(ii)

product of the roots = -p_{3} = -3⁄2 ⇒ (α)(2α)(β) = -3⁄2 ⇒ 4(α)^{2}(β) = -3..........(iii)

From (i), β = (-3 - 6α)⁄2⇒ 3β = (-9 - 18α)⁄2 = -9⁄2 - 9α Using this value of 3β in (ii), we get α(2α + 3β) = -4⇒ α(2α - 9⁄2 - 9α) = -4 ⇒ α( -9⁄2 - 7α) = -4⇒ 7α^{2} + (9⁄2)α - 4 = 0 ⇒ 14α^{2} + 9α - 8 = 0⇒ 14α^{2} + 16α - 7α - 8 = 0 ⇒ 2α(7α + 8) -1(7α + 8) = 0⇒ (7α + 8)(2α - 1) = 0 ⇒ α = 1⁄2 or -8⁄7

For α = -8⁄7 and β = 27⁄14, L.H.S. of (iii) = 4(α)^{2}(β) = 4( -8⁄7)^{2}(27⁄14) = (4)(64)(27)⁄(49)(14) ≠ -3 ∴ α = -8⁄7 and β = 27⁄14, do not satisfy (iii) and is not a valid solution. α = 1⁄2, β = -3 satisfy equation (iii) and is a valid solution.

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Solve the Algebra Equation 4x^{3} + 20x^{2} - 23x + 6 = 0 two of the roots being equal..

Solution to Example 2 of Cubic Equation :

The given Algebra Equation equation is 4x^{3} + 20x^{2} - 23x + 6 = 0 Dividing both sides of the equation by 4, we get x^{3} + 5x^{2} - (23⁄4)x + 3⁄2 = 0 Comparing this with x^{3} + p_{1}x^{2} + p_{2}x + p_{3} = 0, we get p_{1} = 5; p_{2} = - (23⁄4); p_{3} = 3⁄2

By data two roots are equal. So if one root is α, then the other root is α. Let the third root be β We know, sum of roots = -p_{1}⇒ α + α + β = -5 ⇒ 2α + β = -5 ...........(i)

sum of the product of the roots taken two at a time = p_{2} = - (23⁄4) ⇒ (α)(α) + (α)(β) + (α)(β) = -(23⁄4) ⇒ α(α + 2β) = -(23⁄4) ⇒ 4α(α + 2β) = -23.............(ii)

product of the roots = -p_{3} = -3⁄2 ⇒ (α)(α)(β) = -3⁄2 ⇒ 2(α)^{2}(β) = -3..........(iii)

From (i), β = (-5 - 2α)⇒ 2β = (-10 - 4α) Using this value of 2β in (ii), we get 4α(α + 2β) = -23⇒ 4α(α - 10 - 4α) = -23 ⇒ 4α( -10 - 3α) = -23⇒ 12α^{2} + 40α - 23 = 0 ⇒ 12α^{2} + 46α - 6α - 23 = 0 ⇒ 2α(6α + 23) -1(6α + 23) = 0⇒ (6α + 23)(2α - 1) = 0 ⇒ α = 1⁄2 or -23⁄6

For α = -23⁄6 and β = 8⁄3, L.H.S. of (iii) = 2(α)^{2}(β) = 2(-23⁄6)^{2}(8⁄3) ≠ -3 ∴ α = -23⁄6 and β = 8⁄3, do not satisfy (iii) and is not a valid solution. α = 1⁄2, β = -6 satisfy equation (iii) and is a valid solution.

∴ the roots of the given Cubic Equation are α, α, β = 1⁄2, 1⁄2, -6. Ans.

Exercise : Cubic Equation

Solve the following problems on Cubic Equation.

Solve the Algebra Equation 8x^{3} - 36x^{2} - 18x + 81 = 0, the roots of which are in A.P.

Solve the Algebra Equation 6x^{3} - 11x^{2} + 6x - 1 = 0, the roots of which are in H.P.

For Answers to the above problems on Cubic Equation see at the bottom of the page.

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