Trinomial Solution

Introduction to trinomial solution:

A polynomial with three terms is known as trinomial. One of the three terms is a constant and one of the remaining terms is a constant with a variable and another term is a constant with square of a variable. The constant is an optional. The roots of the trinomial are a solution for the given trinomial. The methods to solve a trinomial are by using quadratic formula or by using factoring method.


General form – Trinomial solution:


The general form of a trinomial is ax ^2+bx+c=0. In this x is a variable and a, b and c are constants.

The value of x is trinomial solution.

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Example problems – Trinomial solution:


Example 1 – Trinomial solution:

Solve the trinomial x ^2+8x+15=0.

Solution:

The given trinomial is x ^2+6x+9=0.

The given trinomial can be written as,

x ^2+3x+3x+9=0.

Take x as common from first two terms

x(x+3)+3x+9=0

Take 3 as common from last two terms

x(x+3)+3(x+3)=0

Take (x+3) as common

(x+3)(x+3)=0

The above product of monomials can be written as

x+3 = 0 and x+3=0

Now solve the above terms for x

So, `x=-3` and `x=-3`

The given trinomial’s solutions are -3 and -3.

So, the answer is -3.

Example 2 – Trinomial solution:

Solve the trinomial x ^2+12x+20=0.

Solution:

The given trinomial is x ^2+12x+20=0.

The given trinomial can be written as,

x ^2+10x+2x+20=0.

Take x as common from first two terms

x(x+10)+2x+20=0

Take 3 as common from last two terms

x(x+10)+2(x+10)=0

Take (x+10) as common

(x+10)(x+2)=0

The above product of monomials can be written as

x+10 = 0 and x+2=0

Now solve the above terms for x

So, `x=-10` and `x=-2`

The given trinomial’s solutions are -10 and -2.

Example 3 – Trinomial solution:

Solve the trinomial 2x ^2+20x+50=0.

Solution:

The given trinomial is 2x ^2+20x+50=0.

The given trinomial can be written as,

2x ^2+10x+10x+50=0.

Take 2x as common from first two terms

2x(x+5)+10x+50=0

Take 10 as common from last two terms

2x(x+5)+10(x+5)=0

Take (x+5) as common

(x+5)(2x+10)=0

The above product of monomials can be written as

x+5 = 0 and 2x+10=0

Now solve the above terms for x

So, `x=-5 ` and `x=-5`

The given trinomial’s solutions are -5 and -5.

So the solution is -5.

8th Grade Problem Solving

Introduction to 8th grade problem solving:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Here we are going to see about 8th grade problem solving and its example problems.                                                                                      Source: Wikipedia

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Example for 8th grade problem solving:


8th grade problem solving example: 1

Solve 3x + 3 = 12

Solution:

Given that 3x + 3 = 12

Subtract the -3 on both sides

3x + 3 – 3 = 12 – 3

3x = 9

Divide both side using 3

`(3x) / 3` = `9 / 3`

x = 3

The solution is x = 3.

8th grade problem solving example: 2

Solve 4x + 3y = 9

5x + 3y = 5

Solution:

Given that   4x + 3y = 9 ------------- (1)

5x + 3y = 5-------------- (2)

Subtract the first equation and second equation

4x + 3y = 9

(`-` )  5x + 3y = 5

_______________

-x = 4

x = - 4

x = - 4 take the value and substitute in equation (1).

4x + 3y = 9 ------------- (1)

x = -4

4(-4) + 3y = 9

-16 + 3y = 9

Add the +16 on both sides

+16 – 16 + 3y = 9 + 16

3y = 25

Divide using 3 on both sides

` (3y) / 3` = `25 / 3`

y = 8.33

The solution is x = -4

y = 8.33

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Example for 8th grade problem solving:


8th grade problem solving example: 3

Solve 6x + 4y = 2

7x + 5y = 3

Solution:

Given that   6x + 4y = 2 ------------- (1)

7x + 5y = 3 ------------- (2)

From these two equations we cannot cancel particular variable

So we need to change this equation at least one variable equal on both equations

(1)   * 5 =   30x + 20y = 10

(2)   * 4 =   28x + 20y = 12

From this equation we can cancel one variable.

30x + 20y = 10

( - ) 28x + 20y = 12

Subtract you can get the value

2x = - 2

Divide using 2 on both sides

x = - 2 / 2

x = -1

x = - 1 take the value and substitute in equation (1).

6x + 4y = 2 ------------- (1)

6(-1) + 4y = 2

-6 + 4y = 2

Add +6 on both sides

+6 – 6 + 4y = 2 + 6

4y = 8

Divide using 4 on both sides

`y / 4` = `8 / 4`

y = 2

The solution is x = -1

y = 2