Equations

Basic concepts

• Variable: They're the last letters of the alphabet, they show us unknown quantities

• Constants: They never change, their value is fixed like constant, rational numbers or sometimes we can use letters like a and b

• Members: Everything, before the equal is the first member, and everything after the equal, will be our second member

• Terms: Each one of the elements of the equation separated by a sign

• Degree: It's the exponent of greater value than we've in the equation

• Solution: It's the final value of our variable

• Linear equations: They're those that only have one unknown and their highest degree is 1. They describe a straight line

• Second-degree equations or quadratic equations: They will be second-degree equations and knowing how to factor will help us a lot to solve them

Equations: First-degree equations

In order to solve first-degree equations, we must make use of transposition. The transposition is that one operation is converted into another by passing it to the other side of equality

• Transposition of the addition: Goes to subtract the 2nd member

• Transposition of the subtraction: Goes to add to the 2nd member

• Transposition of multiplication: Goes to divide the 2nd member

• Transposition of the division: Goes to multiply the 2nd member

The purpose of the equations is to find the value of our variables

1. We must identify our variable

2. We should leave our variables on the first member (left side of equality) and constants on the second member (right side of equality)

3. We will solve the equation using whatever transpositions we need

Application problems: First-degree equations

Although this type of problem could be solved "by trial and error", it's easier, faster, and more recommendable to learn to do it by solving first degree equations because it's the basis for being able to solve more complex equations of higher degrees

Linear equations systems

When we have two unknowns or more, we are talking about a system of equations. According to the number of unknowns, we will have at least that same number of equations to solve the system

Solution of systems of equations:

• Equalization: We equalize one equation with another to obtain the result of a variable

• Substitution: We isolate a variable to substitute it in another equation

• Elimination: It's commonly called addition and subtraction because we will eliminate the same variable in two equations at the same time to obtain the result of the other

• Graphic: We will omit it because it's not used anymore. Points are placed, lines are drawn and where they cross, that will be the solution. It's a nice but time-consuming method, it's not used because we currently have better tools

• Cramer: It's solved from determinants and matrices. This method is taught in the linear algebra course

To solve a linear equation, we must first identify which method best fits our system of equations

Substitution: It's chosen when there's a variable in one of the equations without a coefficient

• We list the two equations, the first being the one with the variable without a coefficient

• We isolate the variable without a coefficient from equation number 1 and we will number the resulting equation as 3

• In equation 2, we substitute the variable that we solved for earlier. In this way, we will obtain the equivalent of the other variable

• Having the value of one of the variables, we can substitute its value in any of the initial equations to obtain the value of the missing one

To confirm that we've the correct result, we can substitute the values ​​of the variables in the initial equations and check that the equalities hold

Troubleshooting Methods

Elimination: We will eliminate a variable with the help of arithmetic operations

• We choose the variable that in one of the equations appears with a positive sign and in the other with a negative sign. Or that variable that appears in one of the two equations without a coefficient

• Of the variable that we choose, we must identify what its coefficients are. The entire equation 1 will be multiplied by the coefficient of the variable in the second equation and the entire second equation will be multiplied by the coefficient of the variable in the first equation

• We will have equations 3 and 4 as a result, which will have the same value in the variable we chose but with opposite signs. This allows us to eliminate the variable from both equations

• We add the variable left over from the two equations and also their constants. From there we can isolate this remaining variable and find its result

• We substitute the variable we found into any of our initial equations and thus obtain the result of the missing variable

Equalization: We've to establish equality between the first equation and the second

• We choose one of the variables and isolate it in both equations

• We equate the right part of the equal of both equations

• We clear the variable that remained in the resulting equation, finding its value

• We substitute the value of the found variable into one of the initial equations to find the value of the missing variable

Discriminants in the equations

A quadratic equation is one that has variables of second-degree. Before solving these equations it's very useful that we know the discriminant which is equal to the coefficient of the second term squared minus 4 times the coefficient of the first term times the third

• If the discriminant is positive, the equation has two solutions

• If the discriminant is 0, the equation has only one solution

• If the discriminant is negative, it has no real solutions, only imaginary ones

The discriminant is useful to us because it saves us by quickly telling us if an equation has a solution or not

Complete equations of second degree

The general formula is the best tool to avoid making mistakes and although it can be confusing, we can learn it after a while. It helps us to find the roots or solutions to our quadratic equations (of the second degree)

We can solve all quadratic equations using the general formula or by factoring with square trinomials

Incomplete second degree equations: Type 2 and 3

When the independent term is equal to 0, it's easily solved with common term factorization. When we get the two factors it's just a matter of setting each one equal to 0 to find the X's

When we don't have a linear term, we solve by finding X with the formula: plus-minus square root of minus cover a, where c is our independent term and a is the coefficient of our variable squared

Material: Complete equations

COMPLETE EQUATION

There is no relationship between the number of solutions of the equation and the type of equation (complete or incomplete)

The equation is complete when all three coefficients a, b, and c are different from 0

If b, c≠0, the equation is said to be complete and its solutions are given by the formula

$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

Examples of complete equations:

1. $x^2+x+1=0$

$$\begin{split} \frac {-1 \pm \sqrt {1^2 - 4}}{2} &=\frac {-1 \pm \sqrt {-3}}{2} \end{split}$$

${x}_1=\frac {-1 + \sqrt {-3}}{2}; {x}_2=\frac {-1 - \sqrt {-3}}{2}$

1. $2x^2-x+3=0$

$$\begin{split} \frac {1 \pm \sqrt {1 - 24}}{2} &=\frac {1 \pm \sqrt {-23}}{4} \end{split}$$

${x}_1=\frac {1 + \sqrt {-23}}{4}; {x}_2=\frac {1 - \sqrt {-23}}{4}$

1. $-x^2+5x-3=0$

$$\begin{split} \frac {-5 \pm \sqrt {25 - 12}}{-2} &=\frac {-5 \pm \sqrt {13}}{-2} \\ &=\frac {5 \pm \sqrt {13}}{2} \end{split}$$

${x}_1=\frac {5 + \sqrt {13}}{2}; {x}_2=\frac {5 - \sqrt {13}}{2}$

Challenges

1. $x^2-3x-10=0;{x}_1=5,{x}_2=-2$

$$\begin{split} \frac {3 \pm \sqrt {9 + 40}}{2} &=\frac {3 \pm \sqrt {49}}{2} \\ &=\frac {3 \pm 7}{2} \\ \end{split}$$

${x}_1=\frac {3 + 7}{2}=\frac {10}{2}=5; {x}_2=\frac {3 - 7}{2}=\frac {-4}{2}=-2$

1. $x^2-2x-15=0;x1{x}_1=5,{x}_2=-3$

$$\begin{split} \frac {2 \pm \sqrt {4 + 60}}{2} &=\frac {2 \pm \sqrt {64}}{2} \\ &=\frac {2 \pm 8}{2} \end{split}$$

${x}_1=\frac {2 + 8}{2}=\frac {10}{2}=5; {x}_2=\frac {2 - 8}{2}=\frac {-6}{2}=-3$

1. $5x^2-10x-15=0;{x}_1=3,{x}_2=-1$

$$\begin{split} \frac {10 \pm \sqrt {100 + 300}}{10} &=\frac {10 \pm \sqrt {400}}{10} \\ &=\frac {10 \pm 20}{10} \end{split}$$

${x}_1=\frac {10 + 20}{10}=\frac {30}{10}=3; {x}_2=\frac {10 - 20}{10}=\frac {-10}{10}=-1$

1. $3x^2-2x-1=0;{x}_1=1,{x}_2=-\left(\frac{1}{3}\right)$

$$\begin{split} \frac {2 \pm \sqrt {4 + 14}}{6} &=\frac {2 \pm \sqrt {16}}{6} \\ &=\frac {2 \pm 4}{6} \end{split}$$

${x}_1=\frac {2 + 4}{6}=\frac {6}{6}=1; {x}_2=\frac {2 - 4}{6}=\frac {-2}{6}=-\left(\frac{1}{3}\right)$

1. $x^2+8x+16=0;{x}_1=-4$

$$\begin{split} \frac {-8 \pm \sqrt {64-64}}{2} &=\frac {-8}{2} \\ &=-4 \end{split}$$

${x}_1=-4$

1. $10x^2-7x+1=0;{x}_1=\frac{1}{2},{x}_2=\frac{1}{5}$

$$\begin{split} \frac {7 \pm \sqrt {49-40}}{20} &=\frac {7 \pm \sqrt {9}}{20} \\ &=\frac {7 \pm 3}{20} \end{split}$$

${x}_1=\frac {7 + 3}{20}=\frac {10}{20}=\frac {1}{2}; {x}_2=\frac {7 - 3}{20}=\frac {4}{20}=\frac {2}{10}=\frac {1}{5}$

1. $x^2-2x+5=0;{x}_1=1+2i,{x}_2=1-2i$

$$\begin{split} \frac {2 \pm \sqrt {4-20}}{2} &=\frac {2 \pm \sqrt {-16}}{2} \\ &=\frac {2 \pm 4\sqrt {-1}}{2}\\ &=\frac {2 \pm 4i}{2} \end{split}$$

${x}_1=\frac {2 + 4i}{2}=1+2i; {x}_2=\frac {2-4i}{2}=1-2i$