# CUBIC EQUATION - SOLVING POLYNOMIAL EQUATIONS OF THIRD DEGREE WITH EXAMPLES

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 2x3 + 3x2 - 8x + 3 = 0 is double another root, solve the equation.

Solution to Example 1 of Cubic Equation :
The given Algebra Equation is 2x3 + 3x2 - 8x + 3 = 0
Dividing both sides of the equation by 2, we get
x3 + (3⁄2)x2 - 4x + 3⁄2 = 0
Comparing this with x3 + p1x2 + p2x + p3 = 0, we get
p1 = 3⁄2; p2 = -4; p3 = 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 = -p1⇒ 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 = p2 = -4
⇒ (α)(2α) + (α)(β) + (2α)(β) = -4
⇒ α(2α + β + 2β) = -4 ⇒ α(2α + 3β) = -4.............(ii)

product of the roots = -p3 = -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⇒ 7α2 + (9⁄2)α - 4 = 0
⇒ 14α2 + 9α - 8 = 0⇒ 14α2 + 16α - 7α - 8 = 0
⇒ 2α(7α + 8) -1(7α + 8) = 0⇒ (7α + 8)(2α - 1) = 0
⇒ α = 1⁄2 or -8⁄7

when α = 1⁄2, β = (-3 - 6α)⁄2 = (-3 - 3)⁄2 = -3
when α = -8⁄7, β = {-3 - 6(-8⁄7)}⁄2 = (-21 + 48)⁄14 = 27⁄14

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.

∴ The roots of the given Algebra Equation are α, 2α, β
= 1⁄2, 1, -3. Ans. Great Deals on School & Homeschool Curriculum Books

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## Solved Example 2 : Cubic Equation

Solve the following problem on Cubic Equation.

Solve the Algebra Equation 4x3 + 20x2 - 23x + 6 = 0 two of the roots being equal..

Solution to Example 2 of Cubic Equation :

The given Algebra Equation equation is 4x3 + 20x2 - 23x + 6 = 0
Dividing both sides of the equation by 4, we get
x3 + 5x2 - (23⁄4)x + 3⁄2 = 0
Comparing this with x3 + p1x2 + p2x + p3 = 0, we get
p1 = 5; p2 = - (23⁄4); p3 = 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 = -p1⇒ α + α + β = -5
⇒ 2α + β = -5 ...........(i)

sum of the product of the roots taken two at a time = p2 = - (23⁄4)
⇒ (α)(α) + (α)(β) + (α)(β) = -(23⁄4)
⇒ α(α + 2β) = -(23⁄4) ⇒ 4α(α + 2β) = -23.............(ii)

product of the roots = -p3 = -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⇒ 12α2 + 40α - 23 = 0
⇒ 12α2 + 46α - 6α - 23 = 0
⇒ 2α(6α + 23) -1(6α + 23) = 0⇒ (6α + 23)(2α - 1) = 0
⇒ α = 1⁄2 or -23⁄6

when α = 1⁄2, β = (-5 - 2α) = (-5 - 1) = -6
when α = -23⁄6, β = -5 - 2(-23⁄6) = (-15 + 23)⁄3 = 8⁄3

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.

1. Solve the Algebra Equation 8x3 - 36x2 - 18x + 81 = 0, the roots of which are in A.P.
2. Solve the Algebra Equation 6x3 - 11x2 + 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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## Answers to Exercise : Cubic Equation

Answers to Problems on Cubic Equation :

1. -3⁄2, 3⁄2, 9⁄2
2. 1, 1⁄2, 1⁄3

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