Quadratice Equations and Systems of Quadraric Equations

We already learned that we should not fear the word "system". Don't worry, even when it comes to a system of quadratic equations, which are not linear, the solution is simple and straightforward.
Systems of quadratic or second-degree equations are quadratic equations written together, one next to the other or one below the other with a curly bracket, similar to what we already studied for systems of linear equations.
Generally, $X$ and $Y$ would not have been given to us, therefore, they are our unknowns.
When the problem states "given the system of equations" you should know that the result has to be $X$ and $Y$ that meet the conditions of said system.
The system can have one solution, two solutions, or even none.
The meaning of the solution of the system of equations is demonstrated by the points of intersection of the functions.
At those very points, the functions meet, they intersect.
These specific points exist in both the first equation and the second.

To solve a system of quadratic equations, we will use the comparison method between quadratic equations
When we have a system of quadratic equations and the $Y$ isolated in this way (with the same coefficient in both equations):

We will proceed in the following order:
We will verify that the unknown $Y$ is written the same way in both equations
We will compare the equations

We will find the $X$
We will gradually place an $X$ in one of the equations to solve for its $Y$.
We will neatly record the answers found.

Note
The coefficients of the unknown and the various parameters do not necessarily have to be the same, except for the unknown $Y$. When comparing equations, first verify that the unknown $Y$ is written in the same way in both equations
Let's see an example of solving a system of equations using the comparison method between quadratic equations:
Given the following system of equations:

We will proceed step by step:

1. We will verify that the unknown $Y$ is written the same way in both equations
   The unknown $Y$ indeed appears the same way in both equations, its coefficient is $1$.
2. We will compare the equations
   $x^2+2x-8=-x^2-x+12$
    
3. We will find the $X$
   We will solve this equation as an equation with $1$
   unknown:
   $x^2+2x-8=-x^2-x+12$
   
   We will transpose sides:
   $2x^2+3x-20=0$
   
   We will solve the quadratic equation (with the quadratic formula) and obtain:
   $X=2.5  ,-4$
    
4. We will gradually place an $X$ in one of the equations (the original) to solve for its $Y$
   Let's start with $X=2.5$
   We will place it in the selected equation within the system of equations (it doesn't matter which one, it's advisable to choose the one that seems easier):
   
   $​​​​​​​Y=-x^2-x+12$
   We will place $X=2.5$
   We will obtain:
   
   $Y=-(2.5)^2-2.5+12$
   $y= -6.25-2.5+12$
   $y=3.25$
   
   Let's write down this result: $( 3.25 ,  2.5)$
   
   Now let's move on to the second $X$ . 
   $X=-4$
   
   We will place it in the selected equation within the system of equations (it doesn't matter which one)
   $Y=-x^2-x+12$
   We will place $X=-4$ Remember to use parentheses to avoid confusion with the signs for addition and subtraction

We will obtain:
$Y=-(-4)^2-(-4)+12$
$Y=-16+4+12$
$y=0$
Let's write down this solution:

$( 0 ,  -4)$

5. Let's neatly record the answers we found.

$( 0 ,  -4) , (  3.25 ,  2.5)$
Now let's move on to the graphical method, so we can better understand the meaning of the solution to the system of equations.

The graphical solution of the system of quadratic equations are the points of intersection of the parabolas.
Therefore, when the parabolas are graphically represented, the solution of the system of equations will be their points of intersection.
If they do not intersect, it means that the system of equations has no solution.

If cut once, the system of equations will have one solution - the intersection point of the parabolas.

If they intersect twice, the system of equations will have two solutions - the intersection points of the parabolas.

Now, let's move on to the system of combined equations – that is, one linear and the other quadratic.

In case the system of equations is composed of a quadratic equation and a linear equation, in the following way:
$Y=ax^2+bx+c$
$y=mx+n$

Substitution method: It allows us to substitute the value of the variable y in an equation, thereby obtaining an equation with one variable.
Again, let's remember to neatly place the variable in parentheses to avoid confusion.

Note: Equations can be arranged in a different way, for example, the linear equation does not have to be represented in the same way seen previously and could have a coefficient that is not $Y$.

Let's see it in an example:

Given the system of equations (combined):
$Y=2x^2-5x+1$
$3x+2y=17$

We will notice that the equations are written differently from each other (in one the $Y$ is isolated and in the other it is not) and there is no problem with this.
To use the substitution method, we generally need to completely isolate one of the variables in the equations.
Now, we can place the value of $Y$ (the first quadratic equation) into the linear equation and obtain a linear equation with one unknown.
Remember to use parentheses to avoid confusion.
Let's see:

Let's remove the parentheses and we will get:
$3x+4x^2-10x+2=17$
Let's combine like terms and we will get:
$-7x+4x^2+2=17$
Let's arrange the equation, transpose terms and we will get:
$4x^2-7x-15=0$
Let's solve with the quadratic formula and we will get:
$x=3  ,-\frac{5}{4}$
Now, let's remember to place the found $X$ values to find the $Y$ values and thus be able to reach a complete result.
We will start with $x=3$ and place it in one of the original equations (the one that seems simpler - which is generally the linear one)
We will get:
$3 \times 3+2y=17$
$9+2y=17$
$2y=8$
$y=4$
Let's write this solution:$(4 ,  3)$
Now let's place the second $X$ we found, in one of the original equations and we will get: $x=-\frac{5}{4}$

$3 \times (--\frac{5}{4})+2y=17$

$-\frac{15}{4}+2y=17$
$2y=20.75$
$y=10.375$
Let's write this solution: $(10.375 ,  -\frac{5}{4})$

Let's neatly note down the solutions found $(4 ,  3) , (10.375 ,  -\frac{5}{4})$
These are the points of intersection of the line with the parabola.

Parabolas located on the same Cartesian plane can intersect more than once.
To find the intersection points between the parabolas, we should compare their equations, as we have learned, when $Y$ appears in the same way in both equations (with the same coefficient).
By comparing the equations, we will arrive at an equation with one unknown $X$.
We will solve for the value of $X$ and alternately place in each of the equations the value of $X$ to discover $Y$.
If we arrive at one solution, it means that the parabolas intersect only once.
If we arrive at two solutions, it means that the parabolas intersect twice.
If we find that there is no solution, it means that the parabolas do not intersect at all.

Useful Information
In some cases, as in the example, the parabolas may intersect when $X=5$
And the question will be what are the intersection points of the parabolas.
To find the intersection points, we simply place $X=5$ in one of the given parabola equations and thus discover the point.

We must convert verbal math problems into quadratic equations to solve them
Just as we have already done, equations are constructed based on the text given in the problem.
How will we do it?
The key to solving problems of this type is to describe the unknowns through a single unknown $X$.
This way instead of arriving at two equations with two unknowns we will arrive at one quadratic equation.
Another very important issue is to read and understand exactly what the problem is telling us. This way we can construct the equation correctly.
Let's see an example:
Find two numbers whose sum equals $11$ and the sum of these squared equals $61$
Solution:
First, let's define the variables with one unknown
First number: $X$
Second number:  $11-X$
How have we defined the second number?
We know that the sum of the two numbers we are looking for is $11$
Think about this: let's say the second number was $Y$,
we would write
$y+x=11$
and isolate the $Y$
We would get $Y=11-X$
Magnificent. Now let's move on to the second part of the problem: the sum of the numbers squared equals $61$ That is, if we square each of the numbers and then add them we will get $61$
Let's translate it into an equation:
$X^2+(11-x)^2=61$

Great! Now let's remove the parentheses using the distributive property of multiplication and we will arrive at a sorted quadratic equation:

$x^2+121-22x+x^2=61$

Let's combine terms and we will get:
$2x^2-22x+60=0$

Great! We can solve this quadratic equation easily with the quadratic formula. Let's solve and we will get to:

$X=6.5$

Pay attention! These are not the two numbers we are looking for! These are possible solutions for the first number.
Now we must find what the second number would be in each of these possibilities.
Let's begin:
When the first number is $5$,
the second is $11-X$ and therefore, $6$.
When the first number is $6$,
the second is $11-X$  and therefore $5$.

The quadratic inequality describes in which interval the function is positive and in which it is negative.
You have already studied all the steps to solve an inequality of this type.
Before we begin to learn how to solve the quadratic inequality, it is advisable to remember two important things:

1. Set of positivity and negativity of the function:
   Set of positivity - represents the $X$ in which the graph of the parabola is above the $X$ axis, with a positive $Y$ value.
   Set of negativity - represents the $X$ in which the graph of the parabola is below the $X$ axis, with a negative $Y$ value.
2. Dividing by a negative term - reverses the sign of the inequality.

Excellent, now we can begin.
When we have a quadratic inequality, we will act following these steps:

1. Transposition of terms and isolation of the quadratic equation until one side equals $0$. Remember that when we divide by a negative term, the inequality is reversed.
2. Drawing a parabola scheme - placing points of intersection with the $X$ axis and identifying the maximum and minimum of the parabola.
3. Calculation of the corresponding interval according to the exercise and the scheme.
   Interval of positivity: $0<$ quadratic equation
   Interval of positivity: $0>$ quadratic equation

Solve the following inequality:
$x^2+x-30<0$

Solution:
Let's proceed step by step:

1. Let's make sure that on one side we have $0$ In the case of the example, the equation is already in this form.
2. Let's draw a diagram.

Let's find the points of intersection with the $X$ axis
$x^2+x-30=0$
According to the quadratic formula, we will obtain:
$X=5,-6$

We will identify, according to the coefficient of $x^2$, that the parabola opens upwards.

Let's calculate the corresponding interval according to the exercise and the scheme. 
$x^2+x-30<0$

Since the exercise asks when the quadratic equation is less than $0$, they are asking about the interval of negativity, therefore, by observing the scheme we can determine that the quadratic equation is less than $0$ when :
$-6<x<5$
This is the solution to the inequality.