### Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

n*(n+1)/2-(1000)=0

## Step by step solution :

## Step 1 :

` n + 1 Simplify ————— 2 `

#### Equation at the end of step 1 :

` (n + 1) (n • ———————) - 1000 = 0 2 `

## Step 2 :

#### Equation at the end of step 2 :

` n • (n + 1) ——————————— - 1000 = 0 2 `

## Step 3 :

#### Rewriting the whole as an Equivalent Fraction :

3.1Subtracting a whole from a fraction

Rewrite the whole as a fraction using 2 as the denominator :

` 1000 1000 • 2 1000 = ———— = ———————— 1 2 `

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

#### Adding fractions that have a common denominator :

3.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

` n • (n+1) - (1000 • 2) n`^{2} + n - 2000 —————————————————————— = ————————————— 2 2

#### Trying to factor by splitting the middle term

3.3Factoring n^{2} + n - 2000

The first term is, n^{2} its coefficient is 1.

The middle term is, +n its coefficient is 1.

The last term, "the constant", is -2000

Step-1 : Multiply the coefficient of the first term by the constant 1•-2000=-2000

Step-2 : Find two factors of -2000 whose sum equals the coefficient of the middle term, which is 1.

-2000 | + | 1 | = | -1999 | ||

-1000 | + | 2 | = | -998 | ||

-500 | + | 4 | = | -496 | ||

-400 | + | 5 | = | -395 | ||

-250 | + | 8 | = | -242 | ||

-200 | + | 10 | = | -190 |

For tidiness, printing of 14 lines which failed to find two such factors, was suppressed

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

#### Equation at the end of step 3 :

` n`^{2} + n - 2000 ————————————— = 0 2

## Step 4 :

#### When a fraction equals zero :

`4.1 When a fraction equals zero ...`

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

` n`^{2}+n-2000 ————————— • 2 = 0 • 2 2

Now, on the left hand side, the 2 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape:

n^{2}+n-2000=0

#### Parabola, Finding the Vertex:

4.2Find the Vertex ofy = n^{2}+n-2000Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting "y" because the coefficient of the first term,1, is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,An^{2}+Bn+C,the n-coordinate of the vertex is given by -B/(2A). In our case the n coordinate is -0.5000Plugging into the parabola formula -0.5000 for n we can calculate the y-coordinate:

y = 1.0 * -0.50 * -0.50 + 1.0 * -0.50 - 2000.0

or y = -2000.250

#### Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = n^{2}+n-2000

Axis of Symmetry (dashed) {n}={-0.50}

Vertex at {n,y} = {-0.50,-2000.25}

n-Intercepts (Roots) :

Root 1 at {n,y} = {-45.22, 0.00}

Root 2 at {n,y} = {44.22, 0.00}

#### Solve Quadratic Equation by Completing The Square

4.3Solvingn^{2}+n-2000 = 0 by Completing The Square.Add 2000 to both side of the equation :

n^{2}+n = 2000

Now the clever bit: Take the coefficient of n, which is 1, divide by two, giving 1/2, and finally square it giving 1/4

Add 1/4 to both sides of the equation :

On the right hand side we have:

2000+1/4or, (2000/1)+(1/4)

The common denominator of the two fractions is 4Adding (8000/4)+(1/4) gives 8001/4

So adding to both sides we finally get:

n^{2}+n+(1/4) = 8001/4

Adding 1/4 has completed the left hand side into a perfect square :

n^{2}+n+(1/4)=

(n+(1/2))•(n+(1/2))=

(n+(1/2))^{2}

Things which are equal to the same thing are also equal to one another. Since

n^{2}+n+(1/4) = 8001/4 and

n^{2}+n+(1/4) = (n+(1/2))^{2}

then, according to the law of transitivity,

(n+(1/2))^{2} = 8001/4

We'll refer to this Equation as Eq. #4.3.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(n+(1/2))^{2} is

(n+(1/2))^{2/2}=

(n+(1/2))^{1}=

n+(1/2)

Now, applying the Square Root Principle to Eq.#4.3.1 we get:

n+(1/2)= √ 8001/4

Subtract 1/2 from both sides to obtain:

n = -1/2 + √ 8001/4

Since a square root has two values, one positive and the other negative

n^{2} + n - 2000 = 0

has two solutions:

n = -1/2 + √ 8001/4

or

n = -1/2 - √ 8001/4

Note that √ 8001/4 can be written as

√8001 / √4which is √8001 / 2

### Solve Quadratic Equation using the Quadratic Formula

4.4Solvingn^{2}+n-2000 = 0 by the Quadratic Formula.According to the Quadratic Formula,n, the solution forAn^{2}+Bn+C= 0 , where A, B and C are numbers, often called coefficients, is given by :

-B± √B^{2}-4AC

n = ————————

2A In our case,A= 1

B= 1

C=-2000 Accordingly,B^{2}-4AC=

1 - (-8000) =

8001Applying the quadratic formula :

-1 ± √ 8001

n=——————

2Can √ 8001 be simplified ?

Yes!The prime factorization of 8001is

3•3•7•127

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 8001 =√3•3•7•127 =

±3 •√ 889

√ 889 , rounded to 4 decimal digits, is 29.8161

So now we are looking at:

n=(-1±3• 29.816 )/2

Two real solutions:

n =(-1+√8001)/2=(-1+3√ 889 )/2= 44.224

or:

n =(-1-√8001)/2=(-1-3√ 889 )/2= -45.224

## Two solutions were found :

- n =(-1-√8001)/2=(-1-3√ 889 )/2= -45.224
- n =(-1+√8001)/2=(-1+3√ 889 )/2= 44.224