4 eso academics - unit 04 - equations and inequations

Unit 04 December

1. EQUATIONS.

An Equation is a statement using algebra that contains an unknown quantity

and an equals sign. The Solution of an equation is the set of values which, when

substituted for unknowns, make the equation a true statement. An Equation has

different Elements:

• Variables: The unknown quantities

• Member: The two expressions on either side of an equation.

• Term: Any of the addends of an equation.

• Degree: For a term with one variable, the degree is the variable's exponent. With

more than one variable, the degree is the sum of the exponents of the variables.

MATH VOCABULARY: Equation, Member, Term, Variable, Coefficient, Degree,

Constant, Solution.

Axel Cotón Gutiérrez Mathematics 4º ESO 4.4.1

Unit 04 December

2. FIRST-DEGREE AND SECOND DEGREE EQUATIONS.

2.1. FIRST-DEGREE EQUATIONS.

A First-Degree Equation is called a Linear Equation. The highest exponent of a

linear equation is 1. The standard form for a linear equation is:

𝒂𝒂,𝒃𝒃, 𝒄𝒄 ∈ ℤ 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂 ≠ 𝟎𝟎

3𝑥𝑥 + 2 = −10.

The solution of this equation is 𝑥𝑥 = −4

2.2. SECOND-DEGREE EQUATIONS.

A Second-Degree Equation is called a Quadratic Equation. The highest

exponent of a quadratic equation is 2. The standard form for a quadratic equation is:

𝒂𝒂,𝒃𝒃, 𝒄𝒄 ∈ ℤ 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂 ≠ 𝟎𝟎

5𝑥𝑥2 + 5𝑥𝑥 − 2 = 0

Quadratic equations can be solved using a special formula called the Quadratic

Formula:

The solutions to the quadratic equation are often called Roots, or sometimes

Zeroes.


Unit 04 December

Case 1:

𝑥𝑥2 − 6𝑥𝑥 + 5 = 0 ⟹ 𝑎𝑎 = 1; 𝑏𝑏 = −6 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐 = 5

𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐

2𝑎𝑎=−(−6) ± �(−6)2 − (4 ∙ 1 ∙ 5)

2 ∙ 1=

=6 ± √36 − 20

2=

6 ± √162

=6 ± 4

2=

3 ± 21

= 3 ± 2 = �3 + 2 = 53 − 2 = 1

This equation has two different solutions: 𝑥𝑥1 = 5 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥2 = 1

Case 2:

4𝑥𝑥2 + 4𝑥𝑥 + 1 = 0 ⟹ 𝑎𝑎 = 4; 𝑏𝑏 = 4 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐 = 1


2𝑎𝑎=−(+4) ± �(+4)2 − (4 ∙ 4 ∙ 1)

2 ∙ 4=

=−4 ± √16 − 16

8=−4 ± √0

8=−4 ± 0

8= �

−48

= −12

−48

= −12

Both solutions are equal. We say that this equation has a double solution or double

root.

Case 3:

𝑥𝑥2 − 2𝑥𝑥 + 5 = 0 ⟹ 𝑎𝑎 = 1; 𝑏𝑏 = −2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐 = 5


2𝑎𝑎=−(−2) ± �(−2)2 − (4 ∙ 1 ∙ 5)

2 ∙ 1=

=2 ± √4 − 20

2=

2 ± √−162

=

⎩⎪⎨

⎪⎧2 + √−16

2= ∄𝑥𝑥 ∈ ℝ

2 − √−162

= ∄𝑥𝑥 ∈ ℝ

This equation has no solution into the set of real numbers.


Unit 04 December

The previous examples show that the different types of solutions of the second-

degree equations depend on the value of 𝑫𝑫 = 𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄. This number is called the

Discriminant.

Sometimes you can find “Incomplete” quadratic equations, if 𝐛𝐛 = 𝟎𝟎 𝐨𝐨𝐨𝐨 𝐜𝐜 =

𝟎𝟎; 𝐨𝐨𝐨𝐨 𝐚𝐚 = 𝟎𝟎 𝐚𝐚𝐚𝐚𝐚𝐚 𝐛𝐛 = 𝟎𝟎. You can solve these equations in an easy way, without using the

quadratic formula.


Unit 04 December

MATH VOCABULARY: First-Degree Equation, Linear Equation, Second-Degree Equation,

Quadratic Equation, Quadratic Formula, Root, Zero, Double Root, Discriminant,

Incomplete Quadratic Equation.

3. OTHER TYPES OF EQUATIONS.

3.1. BIQUADRATIC EQUATIONS.

Biquadratic Equations are quartic equations with no odd-degree terms. The

standard form for a biquadratic equation is:

Let´s see the case of fourth degree:

These equations can be written as a quadratic equation by making an

appropriate substitution called Change of variable:

To solve we have only to apply the Quadratic Formula:


Unit 04 December

The four roots of the equation will be:

Solve: 𝑥𝑥4 − 3𝑥𝑥2 − 4 = 0

𝑥𝑥2 = 𝑡𝑡

𝑡𝑡2 − 3𝑡𝑡 − 4 = 0

We apply quadratic formula:

𝑡𝑡 =−(−3) ± �(−3)2 − �4 ∙ 1 ∙ (−4)�

2 ∙ 1=

3 ± √252

= � 𝑡𝑡1 = 4𝑡𝑡2 = −1

⇒

⎩⎪⎨

⎪⎧ 𝑥𝑥1 = +√4 = 2

𝑥𝑥2 = −√4 = −2𝑥𝑥3 = +√−1 = ∄𝑥𝑥 ∈ ℝ𝑥𝑥4 = −√−1 = ∄𝑥𝑥 ∈ ℝ

MATH VOCABULARY: Biquadratic equations, Change of Variable.

3.2. EQUATIONS WHICH CAN BE SOLVED BY FACTORIZING.

We can solve an algebraic expression by factorizing if we can write it as the

product of factors.


Unit 04 December

𝑷𝑷𝟏𝟏(𝒙𝒙) ∙ 𝑷𝑷𝟐𝟐(𝒙𝒙) ∙ … ∙ 𝑷𝑷𝒂𝒂(𝒙𝒙) = 𝟎𝟎

We can do it extracting factors or with Ruffini´s rule as the unit 3.

Solve 𝑥𝑥4 − 5𝑥𝑥2 = 𝑥𝑥3 − 𝑥𝑥2 − 4x

Firstly we relocate the equation:

𝑥𝑥4 − 𝑥𝑥3 − 4𝑥𝑥2 + 4x = 0

We extract x as a common factor:

𝑥𝑥(𝑥𝑥3 − 𝑥𝑥2 − 4𝑥𝑥 + 4) = 0

Now we use Ruffini´s rule:

1 −1 −4 +4

1 +1 0 −4

1 0 −4 0

⇒ 𝑥𝑥(𝑥𝑥3 − 𝑥𝑥2 − 4𝑥𝑥 + 4) = 𝑥𝑥(𝑥𝑥 − 1)(𝑥𝑥2 − 4)

By using polynomial identities:

𝑥𝑥4 − 5𝑥𝑥2 = 𝑥𝑥3 − 𝑥𝑥2 − 4x = 𝑥𝑥(𝑥𝑥 − 1)(𝑥𝑥 + 2)(𝑥𝑥 − 2)

The solutions are:

⇒ �

𝑥𝑥1 = 0𝑥𝑥2 = 1𝑥𝑥3 = −2𝑥𝑥4 = 2


Unit 04 December

3.3. RATIONAL EQUATIONS.

Equations that contain rational expressions are called Rational Equations. We

solve rational equations by finding a common denominator. We can follow either of

two methods:

• We can rewrite the equation so that all terms have the common denominator and

we can solve for the variable with just the numerators.

• Or we can multiply both sides of the equation by the common denominator so

that all terms become polynomials instead of rational expressions.

An important step in solving rational equations is to reject any Extraneous

Solutions from the final answer. Extraneous solutions are solutions that don’t satisfy

the original form of the equation because they produce untrue statements or are

excluded values that make a denominator equal to 0. Therefore, we should always

check solutions in the original equation.

Solve:

𝑥𝑥𝑥𝑥 − 3

+3

𝑥𝑥 + 1= −1

𝐿𝐿𝐶𝐶𝐶𝐶(𝑥𝑥 − 3, 𝑥𝑥 + 1) = (𝑥𝑥 − 3)(𝑥𝑥 + 1)

𝑥𝑥(𝑥𝑥 + 1)(𝑥𝑥 − 3)(𝑥𝑥 + 1) +

3(𝑥𝑥 − 3)(𝑥𝑥 − 3)(𝑥𝑥 + 1) =

−1(𝑥𝑥 − 3)(𝑥𝑥 + 1)(𝑥𝑥 − 3)(𝑥𝑥 + 1)

𝑥𝑥(𝑥𝑥 + 1) + 3(𝑥𝑥 − 3) = −1(𝑥𝑥 − 3)(𝑥𝑥 + 1)

𝑥𝑥2 + 𝑥𝑥 + 3𝑥𝑥 − 9 = −𝑥𝑥2 − 𝑥𝑥 + 3𝑥𝑥 + 3

2𝑥𝑥2 + 2𝑥𝑥 − 12 = 𝑥𝑥2 + 𝑥𝑥 − 6 = 0


Unit 04 December

𝑥𝑥 =−1 ± √1 + 24

2= � 𝑥𝑥1 = 2

𝑥𝑥2 = −3

We have to check the answers:


+3

𝑥𝑥 + 1= −1

𝑥𝑥=2��

22 − 3

+3

2 + 1=

2−1

+33

= −1 ⟹ 𝑖𝑖𝑡𝑡´𝑠𝑠 𝑎𝑎 𝑠𝑠𝑜𝑜𝑜𝑜𝑜𝑜𝑡𝑡𝑖𝑖𝑜𝑜𝑎𝑎


+3

𝑥𝑥 + 1= −1

𝑥𝑥=−3��

−3−3 − 3

+3

−3 + 1=−3−6

+3−2

= −1 ⟹ 𝑖𝑖𝑡𝑡´𝑠𝑠 𝑎𝑎 𝑠𝑠𝑜𝑜𝑜𝑜𝑜𝑜𝑡𝑡𝑖𝑖𝑜𝑜𝑎𝑎

MATH VOCABULARY: Rational equations, Extraneous Solutions.

3.4. RADICAL EQUATIONS.

A Radical Equation is an equation in which the variable is contained inside a

radical symbol (in the radicand). To solve it we:

• Isolate the radical on one side of the equation.

• If the radical is a square root, square each side of the equation. (If the radical is

not a square root, raise each side to a power equal to the index of the root).

• Simplify and solve as you would any equations.

• Substitute answers back into the original equation to make sure that your

solutions are valid (there could be some extraneous roots that do not satisfy the

original equation and that you must throw out).

Solve: √𝑥𝑥 + 5 − √8 − 𝑥𝑥 = −1

√𝑥𝑥 + 5 = −1 + √8 − 𝑥𝑥

Squaring both members:


Unit 04 December

�√𝑥𝑥 + 5�2

= �−1 + √8 − 𝑥𝑥�2

𝑥𝑥 + 5 = 1 − 2√8 − 𝑥𝑥 + 8 − 𝑥𝑥

2𝑥𝑥 − 4 = −2√8 − 𝑥𝑥

𝑥𝑥 − 2 = −√8 − 𝑥𝑥

2 − 𝑥𝑥 = √8 − 𝑥𝑥

Squaring again:

(2 − 𝑥𝑥)2 = �√8 − 𝑥𝑥�2

4 − 4𝑥𝑥 + 𝑥𝑥2 = 8 − 𝑥𝑥

𝑥𝑥2 − 3𝑥𝑥 − 4 = 0

𝑥𝑥 =3 ± √9 + 16

2= � 𝑥𝑥1 = 4

𝑥𝑥2 = −1

We have to check the answers:

√𝑥𝑥 + 5 − √8 − 𝑥𝑥 = −1𝑥𝑥=4��√4 + 5 − √8 − 4 = 3 − 2 = 1 ≠ −1

⟹ 𝐼𝐼𝑡𝑡 𝑖𝑖𝑠𝑠 𝑎𝑎𝑜𝑜𝑡𝑡 𝑎𝑎 𝑠𝑠𝑜𝑜𝑜𝑜𝑜𝑜𝑡𝑡𝑖𝑖𝑜𝑜𝑎𝑎

√𝑥𝑥 + 5 − √8 − 𝑥𝑥 = −1𝑥𝑥=−1��√−1 + 5 −�8 − (−1) = 2 − 3 = −1

⟹ 𝑖𝑖𝑡𝑡´𝑠𝑠 𝑎𝑎 𝑠𝑠𝑜𝑜𝑜𝑜𝑜𝑜𝑡𝑡𝑖𝑖𝑜𝑜𝑎𝑎

MATH VOCABULARY: Radical equations, To Isolate.


Unit 04 December

4. INEQUATIONS.

An Inequation, also known as an Inequality, is an algebraic expression

containing an inequality sign (<, >,≤,≥).

2𝑥𝑥 − 4 > −2; 𝑥𝑥2 + 5 ≤ 0

The solution to an inequation is the set of values that make true the inequality.

We will see two methods for solving inequations: graphical method and algebraic

method.

4.1. GRAPHICAL METHOD.

We have to draw the graph of the inequation as if they were equations and

then set the inequalities.

Case 1:

2x + 4 > 0

It you draw the graph of the line 𝑦𝑦 = 2𝑥𝑥 + 4, the solution to the inequation will be the

set of values of 𝑥𝑥 such that 𝑦𝑦 > 0


Unit 04 December

Looking at the graph, you can see that when 𝑥𝑥 > −2, then 𝑦𝑦 > 0.

That is, the solution to the inequation 2x + 4 > 0 is the interval (−2,∞).

Case 2:

−2x + 7 >𝑥𝑥2− 3

The ordinate of the line 𝑦𝑦 = −2x + 7 is greater than or equal to the ordinate of the

line 𝑦𝑦 = 𝑥𝑥2− 3 for the values of 𝑥𝑥 less than or equal to 4.

The solution to the inequation is the interval (−∞, 4)

Case 3:

−𝑥𝑥2 + 4𝑥𝑥 > 2𝑥𝑥 − 3

The ordinate of the parabola 𝑦𝑦 = −𝑥𝑥2 + 4𝑥𝑥 is greater than or equal to the ordinate of

the line 𝑦𝑦 = 2𝑥𝑥 − 3 for the values of 𝑥𝑥 between −1 and 3.

The solution to the inequation is the interval (−3,4)


Unit 04 December

In general, to solve an inequation with one variable, 𝒇𝒇(𝒙𝒙) ≤ 𝒈𝒈(𝒙𝒙) or

𝒇𝒇(𝒙𝒙) ≤ 𝒈𝒈(𝒙𝒙) :

• Represent the graphs of the functions 𝒚𝒚 = 𝒇𝒇(𝒙𝒙) and 𝒚𝒚 = 𝒈𝒈(𝒙𝒙) , and get the points

where they intersect.

• Find the intervals that make true the inequality.

4.2. ALGEBRAIC METHOD.

Solving inequations is similar to solving equations. We add, subtract, multiply or

divide both sides until we get the variable by itself. There is one important difference

though. With inequations we can add or subtract the same thing to both sides and the

inequality stays the same. We can multiply or divide both sides by a positive number

and the inequality stays the same. But if we multiply by a negative number, then the

sign of the inequality Reverses.

Solve:


Unit 04 December

−2x − 7 <𝑥𝑥2

+ 3

−4𝑥𝑥 − 14 < 𝑥𝑥 + 6

−4𝑥𝑥 − 𝑥𝑥 < 6 + 14

−5𝑥𝑥 < 20 ⇒ 𝑥𝑥 >20−5

= −4

The solution is the interval (−4,∞)

Solve:

𝑥𝑥2 − 5𝑥𝑥 − 6 ≤ −𝑥𝑥 − 1

𝑥𝑥2 − 4𝑥𝑥 − 5 ≤ 0

We find the roots of the polynomial 𝑥𝑥2 − 4𝑥𝑥 − 5

𝑥𝑥1 = −1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥2 = 5

We plot these roots on the number line and find what is the sign of the number value

of the polynomial 𝑥𝑥2 − 4𝑥𝑥 − 5 in each of the intervals in which this line has been split.

We take any value of these intervals, for instance:

If 𝐼𝐼𝐼𝐼 𝑥𝑥 = −2 ⇒ (−2)2 − 4(−2) − 5 = 7 > 0


Unit 04 December

If 𝐼𝐼𝐼𝐼 𝑥𝑥 = 0 ⇒ (0)2 − 4(0) − 5 = 7 − 5 < 0

If 𝐼𝐼𝐼𝐼 𝑥𝑥 = 6 ⇒ (6)2 − 4(6) − 5 = 7 > 0

The solution to the inequation is the interval [−1,5]


4 eso academics - unit 04 - equations and inequations

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