# What is drug plan 22n 33n 4-5

## Quadratic equations

### Reformatting the input:

Changes made to your input should not affect the solution:

(1): "n2" was replaced by "n ^ 2".

### Step by step solution:

### Step 1:

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

((5 • 7n^{2}) + 22n) + 3 = 0

### Step 2:

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

2.1 Factoring 35n^{2}+ 22n + 3

The first term is, 35n^{2} its coefficient is 35.

The middle term is, + 22n its coefficient is 22.

The last term, "the constant", is +3

Step-1: Multiply the coefficient of the first term by the constant 35 • 3 = 105

Step-2: Find two factors of 105 whose sum equals the coefficient of the middle term, which is 22.

-105 | + | -1 | = | -106 | ||

-35 | + | -3 | = | -38 | ||

-21 | + | -5 | = | -26 | ||

-15 | + | -7 | = | -22 | ||

-7 | + | -15 | = | -22 | ||

-5 | + | -21 | = | -26 | ||

-3 | + | -35 | = | -38 | ||

-1 | + | -105 | = | -106 | ||

1 | + | 105 | = | 106 | ||

3 | + | 35 | = | 38 | ||

5 | + | 21 | = | 26 | ||

7 | + | 15 | = | 22 | That's it |

Step-3: Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 7 and 15

35n^{2} + 7n + 15n + 3

Step-4: Add up the first 2 terms, pulling out like factors:

7n • (5n + 1)

Add up the last 2 terms, pulling out common factors:

3 • (5n + 1)

Step-5: Add up the four terms of step 4:

(7n + 3) • (5n + 1)

Which is the desired factorization

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

(5n + 1) • (7n + 3) = 0### Step 3:

#### Theory - Roots of a product:

3.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

#### Solving a Single Variable Equation:

3.2 Solve: 5n + 1 = 0

Subtract 1 from both sides of the equation:

5n = -1

Divide both sides of the equation by 5:

n = -1/5 = -0.200

#### Solving a Single Variable Equation:

3.3 Solve: 7n + 3 = 0

Subtract 3 from both sides of the equation:

7n = -3

Divide both sides of the equation by 7:

n = -3/7 = -0.429

### Supplement: Solving Quadratic Equation Directly

Solving 35n^{2}+ 22n + 3 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by completing the square and by using the quadratic formula

#### Parabola, Finding the Vertex:

4.1 Find the Vertex of y = 35n^{2}+ 22n + 3

Parabolas 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, 35, 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 -0.3143

Plugging into the parabola formula -0.3143 for n we can calculate the y -coordinate:

y = 35.0 * -0.31 * -0.31 + 22.0 * -0.31 + 3.0

or y = -0.457

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

Root plot for: y = 35n^{2}+ 22n + 3

Axis of Symmetry (dashed) {n} = {- 0.31}

Vertex at {n, y} = {-0.31, -0.46}

n -Intercepts (roots):

Root 1 at {n, y} = {-0.43, 0.00}

Root 2 at {n, y} = {-0.20, 0.00}

#### Solve Quadratic Equation by Completing The Square

4.2 Solving 35n^{2}+ 22n + 3 = 0 by completing the square.

Divide both sides of the equation by 35 to have 1 as the coefficient of the first term:

n^{2}+ (22/35) n + (3/35) = 0

Subtract 3/35 from both side of the equation:

n^{2}+ (22/35) n = -3/35

Now the clever bit: Take the coefficient of n, which is 22/35, divide by two, giving 11/35, and finally square it giving 121/1225

Add 121/1225 to both sides of the equation:

On the right hand side we have:

-3/35 + 121/1225 The common denominator of the two fractions is 1225 Adding (-105/1225) + (121/1225) gives 16/1225

So adding to both sides we finally get:

n^{2}+ (22/35) n + (121/1225) = 16/1225

Adding 121/1225 has completed the left hand side into a perfect square:

n^{2}+ (22/35) n + (121/1225) =

(n + (11/35)) • (n + (11/35)) =

(n + (11/35))^{2}

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

n^{2}+ (22/35) n + (121/1225) = 16/1225 and

n^{2}+ (22/35) n + (121/1225) = (n + (11/35))^{2}

then, according to the law of transitivity,

(n + (11/35))^{2} = 16/1225

We'll refer to this Equation as Eq. # 4.2.1

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

Note that the square root of

(n + (11/35))^{2} is

(n + (11/35))^{2/2} =

(n + (11/35))^{1} =

n + (11/35)

Now, applying the Square Root Principle to Eq. # 4.2.1 we get:

n + (11/35) = √ 16/1225

Subtract 11/35 from both sides to obtain:

n = -11/35 + √ 16/1225

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

n^{2} + (22/35) n + (3/35) = 0

has two solutions:

n = -11/35 + √ 16/1225

or

n = -11/35 - √ 16/1225

Note that √ 16/1225 can be written as

√ 16 / √ 1225 which is 4/35

### Solve Quadratic Equation using the Quadratic Formula

4.3 Solving 35n^{2}+ 22n + 3 = 0 by the Quadratic Formula.

According to the Quadratic Formula, n, the solution for An^{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 = 35

B = 22

C = 3

Accordingly, B.^{2} - 4AC =

484 - 420 =

64

Applying the quadratic formula:

-22 ± √ 64

n = ——————

70

Can √ 64 be simplified?

Yes! The prime factorization of 64 is

2•2•2•2•2•2

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).

√ 64 = √ 2•2•2•2•2•2 =2•2•2•√ 1 =

± 8 • √ 1 =

± 8

So now we are looking at:

n = (-22 ± 8) / 70

Two real solutions:

n = (- 22 + √64) / 70 = (- 11 + 4) / 35 = -0,200

or:

n = (- 22-√64) / 70 = (- 11-4) / 35 = -0.429

### Two solutions were found:

- n = -3/7 = -0.429
- n = -1/5 = -0.200

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