# SOLVING ALGEBRA EQUATIONS OF HIGHER DEGREE WITH EXAMPLES

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 15**x**^{3} - 23**x**^{2} + 9**x** - 1 = 0, the roots of which are in H.P.

Solution to Example 1 of Solving Algebra Equations :

The given Algebra Equation is 15**x**^{3} - 23**x**^{2} + 9**x** - 1 = 0

Dividing both sides of the equation by 15, we get

**x**^{3} - (23⁄15)**x**^{2} + (9⁄15)**x** - 1⁄15 = 0

Comparing this with **x**^{3} + **p**_{1}**x**^{2} + **p**_{2}**x** + **p**_{3} = 0, we get

**p**_{1} = -23⁄15; **p**_{2} = (9⁄15); **p**_{3} = -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 = -**p**_{1} ⇒ 1⁄(α - β) + 1⁄α + 1⁄(α + β) = 23⁄15.....(i)

sum of the product of the roots taken two at a time = **p**_{2} = (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 = -**p**_{3} = 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 - β^{2} = 5⇒ β^{2} = 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 8**x**^{3} - 14**x**^{2} + 7**x** - 1 = 0, the roots of which are in G.P.

Solution to Example 2 of Solving Algebra Equations :

The given Algebra Equation is 8**x**^{3} - 14**x**^{2} + 7**x** - 1 = 0

Dividing both sides of the equation by 8, we get

**x**^{3} - (7⁄4)**x**^{2} + (7⁄8)**x** - 1⁄8 = 0

Comparing this with **x**^{3} + **p**_{1}**x**^{2} + **p**_{2}**x** + **p**_{3} = 0, we get

**p**_{1} = -7⁄4; **p**_{2} = 7⁄8; **p**_{3} = -1⁄8

By data the roots are in G.P.

Let the roots be α⁄β, α, αβ

We know, sum of roots = -**p**_{1} = 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 = **p**_{2} = 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 = -**p**_{3} = 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 + 2β^{2} = 7β⇒ 2β^{2} - 5β + 2 = 0⇒ 2β^{2} - 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

- If one root of the Algebra Equation 24
**x**^{3} - 14**x**^{2} - 63**x** + 45 = 0 is double another root, solve the equation. - Solve the Algebra Equation 18
**x**^{3} + 81**x**^{2} +121**x** + 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

- 3⁄4, 3⁄2, -5⁄3
- -3⁄2, -4⁄3, -5⁄3