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### My Notes

#### Max: 2000 Characters

The solution is 0<x<4.

But I think the solution should be x<0 x>4.

Can someone please assist? Thanks.

The solution is 0<x<4.

But I think the solution should be x<0 x>4.

Can someone please assist? Thanks.

Hi happyapple123,

\(x^2 - 4x <0 => x (x - 4) < 0\)

This means either \(x < 0\) or \((x - 4) < 0\) but not both, because if both are negative then multiplication of two negative numbers is positive so the equation will not be true.

Now, if

\(x < 0\) then \((x - 4)\) is also less than 0, which makes both negative, so this cannot be true.

for example,

if \(x = -1\), then \((x - 4) => (-1 - 4) = -5 < 0\)

=> so, \((-1) * (-5) = 5 > 0\), so equation doesn't agree.

Hence, only \((x - 4) < 0\) but \(x > 0\) which means

\(0 < x < 4\)

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Basal Area: Definition & Formula

There are different ways to describe a stand of trees, one of which is basal area. In this lesson you'll learn about basal area, how to calculate it for a given tree, and why it's an important forestry assessment tool.

Using Quadratic Formulas in Real Life Situations

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## Quadratic equations

### Step by step solution :

### Step 1 :

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

1.1 Factoring x^{2}-4x-2

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

The middle term is, -4x its coefficient is -4 .

The last term, "the constant", is -2

Step-1 : Multiply the coefficient of the first term by the constant 1 • -2 = -2

Step-2 : Find two factors of -2 whose sum equals the coefficient of the middle term, which is -4 .

-2 | + | 1 | = | -1 | |

-1 | + | 2 | = | 1 |

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

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

x^{2}- 4x - 2 = 0

### Step 2 :

#### Parabola, Finding the Vertex :

2.1 Find the Vertex of y = x^{2}-4x-2

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, 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,Ax^{2}+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 2.0000

Plugging into the parabola formula 2.0000 for x we can calculate the y -coordinate :

y = 1.0 * 2.00 * 2.00 - 4.0 * 2.00 - 2.0

or y = -6.000

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

Root plot for : y = x^{2}-4x-2

Axis of Symmetry (dashed) {x}={ 2.00}

Vertex at {x,y} = { 2.00,-6.00}

x -Intercepts (Roots) :

Root 1 at {x,y} = {-0.45, 0.00}

Root 2 at {x,y} = { 4.45, 0.00}

#### Solve Quadratic Equation by Completing The Square

2.2 Solving x^{2}-4x-2 = 0 by Completing The Square .

Add 2 to both side of the equation :

x^{2}-4x = 2

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

Add 4 to both sides of the equation :

On the right hand side we have :

2 + 4 or, (2/1)+(4/1)

The common denominator of the two fractions is 1 Adding (2/1)+(4/1) gives 6/1

So adding to both sides we finally get :

x^{2}-4x+4 = 6

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

x^{2}-4x+4 =

(x-2) • (x-2) =

(x-2)^{2}

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

x^{2}-4x+4 = 6 and

x^{2}-4x+4 = (x-2)^{2}

then, according to the law of transitivity,

(x-2)^{2} = 6

We'll refer to this Equation as Eq. #2.2.1

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

Note that the square root of

(x-2)^{2} is

(x-2)^{2/2} =

(x-2)^{1} =

x-2

Now, applying the Square Root Principle to Eq. #2.2.1 we get:

x-2 = √ 6

Add 2 to both sides to obtain:

x = 2 + √ 6

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

x^{2} - 4x - 2 = 0

has two solutions:

x = 2 + √ 6

or

x = 2 - √ 6

### Solve Quadratic Equation using the Quadratic Formula

2.3 Solving x^{2}-4x-2 = 0 by the Quadratic Formula .

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

__ __

- B ± √ B^{2}-4AC

x = ————————

2A

In our case, A = 1

B = -4

C = -2

Accordingly, B^{2} - 4AC =

16 - (-8) =

24

Applying the quadratic formula :

4 ± √ 24

x = —————

2

Can √ 24 be simplified ?

Yes! The prime factorization of 24 is

2•2•2•3

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).

√ 24 = √ 2•2•2•3 =

± 2 • √ 6

√ 6 , rounded to 4 decimal digits, is 2.4495

So now we are looking at:

x = ( 4 ± 2 • 2.449 ) / 2

Two real solutions:

x =(4+√24)/2=2+√ 6 = 4.449

or:

x =(4-√24)/2=2-√ 6 = -0.449

### Two solutions were found :

- x =(4-√24)/2=2-√ 6 = -0.449
- x =(4+√24)/2=2+√ 6 = 4.449

## Linear equations with one unknown

### Step by step solution :

### Step 1 :

### Step 2 :

#### Pulling out like terms :

2.1 Pull out like factors :

x^{2} - 4x = x • (x - 4)

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

x • (x - 4) = 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 : x = 0

Solution is x = 0

#### Solving a Single Variable Equation :

3.3 Solve : x-4 = 0

Add 4 to both sides of the equation :

x = 4

### Two solutions were found :

- x = 4
- x = 0

## 4x solve 2

.

Quadratic Equation: Solve 4x^2 + 2x = 0.

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