# Quadratic equations

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This solution deals with quadratic equations.

- Quadratic equations

## Step by Step Solution

### Rearrange:

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

y^2+y+13-(-5*y+5)=0

## Step by step solution :

## Step 1 :

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

1.1 Factoring y^{2}+6y+8

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

The middle term is, +6y its coefficient is 6 .

The last term, "the constant", is +8

Step-1 : Multiply the coefficient of the first term by the constant 1 • 8 = 8

Step-2 : Find two factors of 8 whose sum equals the coefficient of the middle term, which is 6 .

-8 | + | -1 | = | -9 | ||

-4 | + | -2 | = | -6 | ||

-2 | + | -4 | = | -6 | ||

-1 | + | -8 | = | -9 | ||

1 | + | 8 | = | 9 | ||

2 | + | 4 | = | 6 | That's it |

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 2 and 4

y^{2} + 2y + 4y + 8

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

y • (y+2)

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

4 • (y+2)

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

(y+4) • (y+2)

Which is the desired factorization

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

```
(y + 4) • (y + 2) = 0
```

## Step 2 :

#### Theory - Roots of a product :

2.1 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 :

2.2 Solve : y+4 = 0

Subtract 4 from both sides of the equation :

y = -4

#### Solving a Single Variable Equation :

2.3 Solve : y+2 = 0

Subtract 2 from both sides of the equation :

y = -2

### Supplement : Solving Quadratic Equation Directly

`Solving y`^{2}+6y+8 = 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 :

3.1 Find the Vertex of t = y^{2}+6y+8

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 "t" 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,Ay^{2}+By+C,the y -coordinate of the vertex is given by -B/(2A) . In our case the y coordinate is -3.0000

Plugging into the parabola formula -3.0000 for y we can calculate the t -coordinate :

t = 1.0 * -3.00 * -3.00 + 6.0 * -3.00 + 8.0

or t = -1.000

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

Root plot for : t = y^{2}+6y+8

Axis of Symmetry (dashed) {y}={-3.00}

Vertex at {y,t} = {-3.00,-1.00}

y -Intercepts (Roots) :

Root 1 at {y,t} = {-4.00, 0.00}

Root 2 at {y,t} = {-2.00, 0.00}

#### Solve Quadratic Equation by Completing The Square

3.2 Solving y^{2}+6y+8 = 0 by Completing The Square .

Subtract 8 from both side of the equation :

y^{2}+6y = -8

Now the clever bit: Take the coefficient of y , which is 6 , divide by two, giving 3 , and finally square it giving 9

Add 9 to both sides of the equation :

On the right hand side we have :

-8 + 9 or, (-8/1)+(9/1)

The common denominator of the two fractions is 1 Adding (-8/1)+(9/1) gives 1/1

So adding to both sides we finally get :

y^{2}+6y+9 = 1

Adding 9 has completed the left hand side into a perfect square :

y^{2}+6y+9 =

(y+3) • (y+3) =

(y+3)^{2}

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

y^{2}+6y+9 = 1 and

y^{2}+6y+9 = (y+3)^{2}

then, according to the law of transitivity,

(y+3)^{2} = 1

We'll refer to this Equation as Eq. #3.2.1

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

Note that the square root of

(y+3)^{2} is

(y+3)^{2/2} =

(y+3)^{1} =

y+3

Now, applying the Square Root Principle to Eq. #3.2.1 we get:

y+3 = √ 1

Subtract 3 from both sides to obtain:

y = -3 + √ 1

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

y^{2} + 6y + 8 = 0

has two solutions:

y = -3 + √ 1

or

y = -3 - √ 1

### Solve Quadratic Equation using the Quadratic Formula

3.3 Solving y^{2}+6y+8 = 0 by the Quadratic Formula .

According to the Quadratic Formula, y , the solution for Ay^{2}+By+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

__ __

- B ± √ B^{2}-4AC

y = ————————

2A

In our case, A = 1

B = 6

C = 8

Accordingly, B^{2} - 4AC =

36 - 32 =

4

Applying the quadratic formula :

-6 ± √ 4

y = —————

2

Can √ 4 be simplified ?

Yes! The prime factorization of 4 is

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).

√ 4 = √ 2•2 =

± 2 • √ 1 =

± 2

So now we are looking at:

y = ( -6 ± 2) / 2

Two real solutions:

y =(-6+√4)/2=-3+= -2.000

or:

y =(-6-√4)/2=-3-= -4.000

## Two solutions were found :

- y = -2
- y = -4