Quadratice Equations and Systems of Quadraric Equations

🏆Practice equations and systems of quadratic equations

Quadratic Equations System

To find the solution of the system of quadratic equations (points of intersection of the functions) we will act as follows:

  1. We will verify that the unknown YY is written the same way in both equations
  2. We will compare the equations
  3. Let's find the XX
  4. Gradually place an XX in one of the equations to solve for its YY.
  5. Let's neatly record the answers we find.
Start practice

Test yourself on equations and systems of quadratic equations!

einstein

Look at the rectangle in the figure.

\( x>0 \)

The area of the rectangle is:

\( x^2-13 \).

Calculate x.

x-4x-4x-4x+1x+1x+1x²-13

Practice more now

Quadratic Equations System

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, XX and YY 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 XX and YY 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.

Algebraic Solution

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 YY isolated in this way (with the same coefficient in both equations):

1 - Algebraic solution Y=ax²+bx+c

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

We will find the XX
We will gradually place an XX in one of the equations to solve for its YY.
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 YY. When comparing equations, first verify that the unknown YY 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:

2 - the following system of equations Y=X²+2X-8 and Y=-X²-X+12

We will proceed step by step:

  1. We will verify that the unknown YY is written the same way in both equations
    The unknown YY indeed appears the same way in both equations, its coefficient is 11.
  2. We will compare the equations
    x2+2x8=x2x+12x^2+2x-8=-x^2-x+12
     
  3. We will find the XX
    We will solve this equation as an equation with 11
    unknown:
    x2+2x8=x2x+12x^2+2x-8=-x^2-x+12

    We will transpose sides:
    2x2+3x20=02x^2+3x-20=0

    We will solve the quadratic equation (with the quadratic formula) and obtain:
    X=2.5 ,4X=2.5  ,-4
     
  4. We will gradually place an XX in one of the equations (the original) to solve for its YY
    Let's start with X=2.5X=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=x2x+12​​​​​​​Y=-x^2-x+12
    We will place X=2.5X=2.5
    We will obtain:

    Y=(2.5)22.5+12Y=-(2.5)^2-2.5+12
    y=6.252.5+12y= -6.25-2.5+12
    y=3.25y=3.25

    Let's write down this result: (3.25, 2.5) ( 3.25 ,  2.5) 

    Now let's move on to the second XX
    X=4X=-4

    We will place it in the selected equation within the system of equations (it doesn't matter which one)
    Y=x2x+12Y=-x^2-x+12
    We will place X=4X=-4 Remember to use parentheses to avoid confusion with the signs for addition and subtraction

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

(0, 4) ( 0 ,  -4) 

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

(0, 4),( 3.25, 2.5)( 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.

Join Over 30,000 Students Excelling in Math!
Endless Practice, Expert Guidance - Elevate Your Math Skills Today
Test your knowledge

Graphical Solution

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.

For example

3 - Graphic solution

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

For example

4 - the intersection point of the parabolas

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

For example

5 - System of equations with two solutions

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

Solution of a system of equations when one of them is 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=ax2+bx+cY=ax^2+bx+c
y=mx+ny=mx+n

We will use the substitution method

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 YY.

Let's see it in an example:

Example of using the substitution method

Given the system of equations (combined):
Y=2x25x+1Y=2x^2-5x+1
3x+2y=173x+2y=17

We will notice that the equations are written differently from each other (in one the YY 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 YY (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:

6 - Example of using the substitution method

Let's remove the parentheses and we will get:
3x+4x210x+2=173x+4x^2-10x+2=17
Let's combine like terms and we will get:
7x+4x2+2=17-7x+4x^2+2=17
Let's arrange the equation, transpose terms and we will get:
4x27x15=04x^2-7x-15=0
Let's solve with the quadratic formula and we will get:
x=3 ,54x=3  ,-\frac{5}{4}
Now, let's remember to place the found XX values to find the YY values and thus be able to reach a complete result.
We will start with x=3x=3 and place it in one of the original equations (the one that seems simpler - which is generally the linear one)
We will get:
3×3+2y=173 \times 3+2y=17
9+2y=179+2y=17
2y=82y=8
y=4y=4
Let's write this solution:(4, 3) (4 ,  3)
Now let's place the second XX we found, in one of the original equations and we will get: x=54x=-\frac{5}{4}

3×(54)+2y=173 \times (--\frac{5}{4})+2y=17

154+2y=17-\frac{15}{4}+2y=17
2y=20.752y=20.75
y=10.375y=10.375
Let's write this solution: (10.375, 54)(10.375 ,  -\frac{5}{4})

Let's neatly note down the solutions found (4, 3),(10.375, 54)(4 ,  3) , (10.375 ,  -\frac{5}{4})
These are the points of intersection of the line with the parabola.

Do you know what the answer is?

Intersection between two parabolas

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 YY 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 XX.
We will solve for the value of XX and alternately place in each of the equations the value of XX to discover YY.
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=5X=5
And the question will be what are the intersection points of the parabolas.
To find the intersection points, we simply place X=5X=5 in one of the given parabola equations and thus discover the point.

Word Problems

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 XX.
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 1111 and the sum of these squared equals 6161
Solution:
First, let's define the variables with one unknown
First number: XX
Second number:  11X   11-X     
How have we defined the second number?
We know that the sum of the two numbers we are looking for is 1111
Think about this: let's say the second number was YY,
we would write
 y+x=11  y+x=11  
and isolate the YY
We would get Y=11XY=11-X
Magnificent. Now let's move on to the second part of the problem: the sum of the numbers squared equals 6161 That is, if we square each of the numbers and then add them we will get 6161
Let's translate it into an equation:
X2+(11x)2=61X^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:

x2+12122x+x2=61x^2+121-22x+x^2=61

Let's combine terms and we will get:
2x222x+60=02x^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=6.5X=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 55,
the second is 11X11-X and therefore, 66.
When the first number is 66,
the second is 11X11-X  and therefore 55.

Check your understanding

Quadratic Inequality

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 XX in which the graph of the parabola is above the XX axis, with a positive YY value.
    Set of negativity - represents the XX in which the graph of the parabola is below the XX axis, with a negative YY 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 00. Remember that when we divide by a negative term, the inequality is reversed.
  2. Drawing a parabola scheme - placing points of intersection with the XX 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<0< quadratic equation
    Interval of positivity: 0>0> quadratic equation

Let's look at an example

Solve the following inequality:
x2+x30<0x^2+x-30<0

Solution:
Let's proceed step by step:

  1. Let's make sure that on one side we have 00 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 XX axis
x2+x30=0 x^2+x-30=0 
According to the quadratic formula, we will obtain:
X=5,6X=5,-6

We will identify, according to the coefficient of x2x^2 , that the parabola opens upwards.

Let's draw a diagram

7 - We will identify based on the coefficient of x^2 that the parabola is the minimum

Let's calculate the corresponding interval according to the exercise and the scheme. 
x2+x30<0x^2+x-30<0

Since the exercise asks when the quadratic equation is less than 00, they are asking about the interval of negativity, therefore, by observing the scheme we can determine that the quadratic equation is less than 00 when :
6<x<5-6<x<5
This is the solution to the inequality.


Examples and exercises with solutions of systems of quadratic equations

Exercise #1

Look at the rectangle in the figure.

x>0

The area of the rectangle is:

x213 x^2-13 .

Calculate x.

x-4x-4x-4x+1x+1x+1x²-13

Video Solution

Answer

x=3 x=3

Exercise #2

Consider the following relationships between the variables x and y:

x2+4=6y x^2+4=-6y

y2+9=4x y^2+9=-4x

Which answer is correct?

Video Solution

Answer

(x+2)2+(y+3)2=0 (x+2)^2+(y+3)^2=0

Do you think you will be able to solve it?
Start practice