# 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.

Great Deals on School & Homeschool Curriculum Books

## 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.

## Get The Best Grades With the Least Amount of Effort : Solving Algebra Equations

Here is a collection of proven tips,
tools and techniques to turn you into
a super-achiever - even if you've never
thought of yourself as a "gifted" student.

The secrets will help you absorb, digest
and remember large chunks of information
quickly and easily so you get the best grades
with the least amount of effort.

If you apply what you read from the above
collection, you can achieve best grades without
giving up your fun, such as TV, surfing the net,
playing video games or going out with friends!

Know more about the Speed Study System.

## 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.

## Research-based personalized Math Help tutoring program : Solving Algebra Equations

Here is a resource for Solid Foundation in
Math Fundamentals from Middle thru High School.
You can check your self by the

### FREE TRIAL.

Are you spending lot of money for math tutors to your
child and still not satisfied with his/her grades ?

Do you feel that more time from the tutor and
more personalized Math Help to identify and fix
the problems faced by your child will help ?

Here is a fool proof solution I strongly recommend
and that too With a minuscule fraction of the amount
you spent on tutors with unconditional 100% money
back Guarantee, if you are not satisfied.

### SUBSCRIBE, TEST, IF NOT SATISFIED, RETURN FOR FULL REFUND

It is like having an unlimited time from an excellent Tutor.

It is an Internet-based math tutoring software program
that identifies exactly where your child needs help and
then creates a personal instruction plan tailored to your
child’s specific needs.

If your child can use a computer and access
the Internet, he or she can use the program.
And your child can access the program anytime
from any computer with Internet access.

### Unique program to help improve math skills quickly and painlessly.

There is an exclusive, Parent Information Page provides YOU
with detailed reports of your child’s progress so you can
monitor your child’s success and give them encouragement.
These Reports include

• Time spent using the program
• Assessment results
• Personalized remediation curriculum designed for your child
• Details the areas of weakness where your child needs additional help
• Provides the REASONS WHY your child missed a concept
• List of modules accessed and amount of time spent in each module
• Quiz results
• Creates reports that can be printed and used to discuss issues with your child’s teachers
These reports are created and stored in a secure section
of the program, available exclusively to you, the parent.
The section is accessed by a password that YOU create and use.
No unauthorized users can access this information.

### Personalized remediation curriculum designed for your child

Thus The features of this excellent Tutoring program are

• Using detailed testing techniques
• Identifing exactly where a student needs help.
• Its unique, smart system pinpointing precise problem areas -
• slowly and methodically guiding the student
• raising to the necessary levels to fix the problem.

### Not a “one-size-fits-all” approach!

Its research-based results have proven that
it really works for all students! in improving
math skills and a TWO LETTER GRADE INCREASE in
math test scores!,if they invest time in using
the program.

Proven for More than 10,000 U.S. public school
students who increased their math scores.

## 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.

## Progressive Learning of Math : Solving Algebra Equations

Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts
from their existing knowledge.
These provide many pages of practice that gradually
increases in difficulty and provide constant review.

These also provide teachers and parents with lessons
on how to work with the child on the concepts.

The series is low to reasonably priced and include

Elementary Math curriculum

and

Algebra Curriculum.

## 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