Tag Archives: second

Solving the equations of the second degree

Solving the equations of the second degree is quite an important skill in mathematics. Many life situations can be modelled with the second degree equations

The objective of this unit is to show you some of the most important technique for factor equations of the second degree. See below:

Important note:

Before starting this unit, it is very important to study, review or what factoring is

Technique # 1:

Solve equations of the second degree by factoring
This technique shows you how to solve the equations of the form ax2 + bx + c = 0 when a = 1 or more

When a = 1, it’s quite simple!

When one is equal to 1, it is a bit tedious. You need to do the trials and errors

Technique # 2:

Address by completing the square
This technique shows you how to fill the place to solve second degree equations

Particular attention to all stages and to learn more!

The evidence even of the quadratic formula is entirely based on this technique.

Make sure that you understand, or you can fight a lot to understand the proof of the quadratic formula

Technique # 3:

Solve using the quadratic formula
Technique # 1 (a) of the limits in the resolution of the equations of the second degree. When the answers are not whole numbers, but the real numbers, it is very difficult or almost impossible to find solutions.

With this formula, you can solve the equations of the second degree and it does not matter how complex the equation is or how weird is the answer.

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Solve equations of the second degree by factoring

Solve equations of the second degree by factoring could be several times the most simple and fastest solve the quadratic equation as long as you know how factor.

I urge you to study or review the following important unit: factoring

Example # 1: Solving the equations of the second degree by factoring

Solve x 2 + 3 x + 2 = 0.

First, you need to factor x 2 + 3 x + 2

Given that the coefficient of x 2 is 1 (x 2 = 1x 2), you can factor in looking for factors of the last Parliament (latter term is 2) which adds to the coefficient of the second term (3 x, factor is 3)

2 = 1 × 2

-2 =-1 × 2

Given that 1 + 2 = 3, and 3 is the coefficient of the second term, x 2 + 3 x + 2 = (x + 2) × (x + 1)

x 2 + 3 x + 2 = 0 gives:

(x + 2) × (x + 1) = 0

(x + 2) × (x + 1) = 0 when x + 2 = 0 or x + 1 = 0.

x + 2 = 0 when x = – 2

x + 1 = 0 when x = – 1

Let us now check x =-2 and x =-1 are solutions of 2 x + 3 x + 2 = 0

(-2) 2 + 3 ×-2 + 2 = 4 + 6 + 2 = 3 + 6 = 0

(-1) 2 + 3 ×-1 + 2 = 1 + 3 + 2 = 3 + 3 = 0

If instead you were resolution x-2 + 3 x + 2 = 0, you’ll:

x-2 + 3 x + 2 = 0.

( x + -2) × ( x + -1) = 0

-(x + 2) × (x +-1) = 0 when x +-2 = 0 or x +-1 = 0.

-x + 2 = 0 when x = 2

-x + 1 = 0 when x = 1

Make sure that they are indeed solutions

Example # 2: Resolution of the equations of the second degree by factoring

Solve x 2 + x-30 = 0.

First, you need to factor x 2 + x – 30

-30 = 30 ×-1

-30-15 × 2 =

-30 = 6 ×-5

-30 =-30 × 1

-30-15 × 2 =

-30 =-6 × 5

Since only 6 +-5 = 1 and 1 are the coefficient of the second term (x = 1x), x 2 + x – 30 = (x + 6) × (x – 5)

x 2 + x-30 = 0 gives:

(x + 6) × (x – 5) = 0

(x + 6) × (x – 5) = 0 when x + 6 = 0 or x – 5 = 0.

x + 6 = 0 when x = – 6

x – 5 = 0 when x = 5

Let us now check x =-6 and x = 5 are indeed x 2 + x-30 = 0 solutions

(-6) 2 +-6 – 30 = 36-6-30 = 30-30 = 0

2 (5) + 5 + 5-30-30 = 25 = 30-30 = 0.

If instead you were resolution x 2 + x – 30 = 0, you’ll:

x 2 + – x – 30 = 0.

(x – 6) × (x + 5) = 0

(x – 6) × (x + 5) = 0 when x – 6 = 0 or x + 5 = 0.

x – 6 = 0 when x = 6

x + 5 = 0 when x = – 5

Make sure that they are indeed solutions

Solve equations of the second degree by factoring can become very difficult. See below:

Example # 3: Solving the equations of the second degree by factoring

6 x 2 + 27 x + 30 = 0.

First factor 6 x 2 + 27 x + 30

6 x 2 + x + 30 = 27 (3 x +?) × (2 x +?) or (x 6 +?) × (x +?)

Now, the last factor term 30

30 = 30 × 1

30 = 15 × 2

30 = 6 × 5

To obtain 27 x, you should try to multiply cross below. There are 6 of them. Cross multiply and add!

6 x x
30 1

x 6 + 30 x = x 36

6 x x
15 2

x 12 + 15 x = x 27

6 x x
6          5

x 30 + 6 x = x 36

3 x 2 x
30 1

3 x + x 60 = x 63

3 x 2 x
15 2

x 6 + 30 x = x 36

3 x 2 x
6           5

15 x + 12 x = x 27

You have two choices in bold!

6 x 2 + 27 x + 30 = 0.

(6 x + 15) × (x + 2) = 0

x 6 + 15 = 0.

x 6 = – 15

x = – 15/6

x + 2 = 0.

x = – 2

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