SOLVING ALGEBRA EQUATIONS OF HIGHER DEGREE WITH EXAMPLES

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Please study
Method of Solving Algebra Equations
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.

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Example 1 : Solving Algebra Equations

Solve the following problem on Solving Algebra Equations

Solve the Algebra Equation 15x3 - 23x2 + 9x - 1 = 0, the roots of which are in H.P.

Solution to Example 1 of Solving Algebra Equations :
The given Algebra Equation is 15x3 - 23x2 + 9x - 1 = 0
Dividing both sides of the equation by 15, we get
x3 - (23⁄15)x2 + (9⁄15)x - 1⁄15 = 0
Comparing this with x3 + p1x2 + p2x + p3 = 0, we get
p1 = -23⁄15; p2 = (9⁄15); p3 = -1⁄15

By data the roots are in H.P.
Let the roots be 1⁄(α - β), 1⁄α, 1⁄(α + β)
We know, sum of roots = -p1 ⇒ 1⁄(α - β) + 1⁄α + 1⁄(α + β) = 23⁄15.....(i)

sum of the product of the roots taken two at a time = p2 = (9⁄15)
⇒ 1⁄{(α - β)(α)} + 1⁄{(α - β)(α + β)} + 1⁄{(α)(α + β)} = (9⁄15)
⇒ {(α + β) + α + (α - β)}⁄{(α - β)(α)(α + β)} = (9⁄15)
⇒ 3α⁄{(α - β)(α)(α + β)} = (9⁄15).......(ii)

product of the roots = -p3 = 1⁄15
⇒ 1⁄{(α - β )(α)(α + β)} = 1⁄15.......(iii)

(ii) ÷ (iii) gives 3α = (9⁄15)⁄(1⁄15) = 9⇒ &alpha = 3
From (iii), (α - β )(α)(α + β) = 15
Using the value of α here, we get
(3 - β )(3)(3 + β) = 15⇒ 9 - β2 = 5⇒ β2 = 4⇒ β = ±2

The values of α and β are found from (ii) and (iii).
Let us verify whether they satisfy (i) or not.

First consider &alpha = 3 and β = +2
L.H.S. of (i) = 1⁄(α - β) + 1⁄α + 1⁄(α + β) = 1⁄(3 - 2) + 1⁄&3 + 1⁄(3 + 2) = 1 + 1⁄&3 + 1⁄5
= (15 + 5 + 3)⁄15 = 23⁄15 = R.H.S. of (i) [satisfied.]
Now consider &alpha = 3 and β = -2L.H.S. of (i) = 1⁄(α - β) + 1⁄α + 1⁄(α + β) = 1⁄(3 + 2) + 1⁄&3 + 1⁄(3 - 2) = 1⁄5 + 1⁄&3 + 1
= (3 + 5 + 15)⁄15 = 23⁄15 = R.H.S. of (i) [satisfied.]
So, &alpha = 3, β = ±2 satisfy equation (i).∴ The required roots are 1⁄(α - β), 1⁄α, 1⁄(α + β)
= 1⁄(3 - 2), 1⁄&3, 1⁄(3 + 2) (or) 1⁄(3 + 2), 1⁄&3, 1⁄(3 - 2) = 1, 1⁄&3, 1⁄5 (or) 1⁄5, 1⁄&3, 1
As we can see, both sets are same.

Thus, the roots of the given Algebra Equation are
1, 1⁄&3, 1⁄5. Ans.

Example 1 of Solving Algebra Equations is thus solved.

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Example 2 : Solving Algebra Equations

Solve the following problem on Solving Algebra Equations

Solve the Algebra Equation 8x3 - 14x2 + 7x - 1 = 0, the roots of which are in G.P.

Solution to Example 2 of Solving Algebra Equations :
The given Algebra Equation is 8x3 - 14x2 + 7x - 1 = 0
Dividing both sides of the equation by 8, we get
x3 - (7⁄4)x2 + (7⁄8)x - 1⁄8 = 0
Comparing this with x3 + p1x2 + p2x + p3 = 0, we get
p1 = -7⁄4; p2 = 7⁄8; p3 = -1⁄8

By data the roots are in G.P.
Let the roots be α⁄β, α, αβ
We know, sum of roots = -p1 = 7⁄4
⇒ α⁄β + α + αβ = 7⁄4⇒ α(1⁄β + 1 + β) = 7⁄4.........(i)

sum of the product of the roots taken two at a time = p2 = 7⁄8
⇒ (α⁄β)α + (α⁄β)(αβ) + α(αβ) = 7⁄8⇒ (α)2(1⁄β + 1 + β) = 7⁄8.......(ii)

product of the roots = -p3 = 1⁄8
⇒ (α⁄β)(α)(αβ) = 1⁄8⇒ α = 1⁄2.........(iii)

(ii) ÷ (i) gives α = 1⁄2 [same as (iii)]

Using the value of α = 1⁄2 in (i), we get
(1⁄2)(1⁄β + 1 + β) = 7⁄4⇒ (1⁄β + 1 + β) = 7⁄2
Multiplying both sides with 2β, we get
2 + 2β + 2β2 = 7β⇒ 2β2 - 5β + 2 = 0⇒ 2β2 - 4β - β + 2 = 0
⇒ 2β(β - 2) -1(β - 2) = 0 ⇒ (β - 2)(2β - 1) = 0⇒ β = 2 or 1⁄2

∴ The required roots are α⁄β, α, αβ
= (1⁄2)⁄2, (1⁄2), (1⁄2)2 (or) (1⁄2)⁄(1⁄2), (1⁄2), (1⁄2)(1⁄2) = 1⁄4, 1⁄2, 1 (or) 1, 1⁄2, 1⁄4As we can see, both sets are same.

Thus, the roots of the given Algebra Equation are
1⁄4, 1⁄2, 1. Ans.

Example 2 of Solving Algebra Equations is thus solved.

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Exercise : Solving Algebra Equations

Solve the following problems on Solving Algebra Equations

  1. If one root of the Algebra Equation 24x3 - 14x2 - 63x + 45 = 0 is double another root, solve the equation.
  2. Solve the Algebra Equation 18x3 + 81x2 +121x + 60 = 0, given a root is equal to half the sum of the remaining roots.

For Answers see at the bottom of the page.

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Answers to Exercise : Solving Algebra Equations

Answers to problems on Solving Algebra Equations

  1. 3⁄4, 3⁄2, -5⁄3
  2. -3⁄2, -4⁄3, -5⁄3