Task 2:
Non-Linear System of Equations
Create a system of equations that includes one linear equation and one quadratic equation.
Part 1: Show all work to solving your system of equations algebraically.
Part 2: Graph your system of equations, and show the solution graphically to verify your solution.
Part 1:
Let's create a system of equations that includes one linear equation and one quadratic equation.
Linear Equation:
y = 2x + 1
Quadratic Equation:
y = x^2 - 4
To solve this system of equations algebraically, we can substitute the value of y from the linear equation into the quadratic equation and solve for x.
Substituting y = 2x + 1 into the quadratic equation:
2x + 1 = x^2 - 4
Rearranging the equation:
x^2 - 2x - 5 = 0
This is a quadratic equation. Let's solve it by factoring or using the quadratic formula.
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -2, and c = -5.
x = (-(-2) ± √((-2)^2 - 4(1)(-5))) / (2(1))
x = (2 ± √(4 + 20)) / 2
x = (2 ± √24) / 2
x = (2 ± 2√6) / 2
x = 1 ± √6
So we have two possible solutions for x: x = 1 + √6 and x = 1 - √6.
Now let's substitute these values of x back into the linear equation to find the corresponding values of y.
For x = 1 + √6:
y = 2(1 + √6) + 1
y = 2 + 2√6 + 1
y = 3 + 2√6
For x = 1 - √6:
y = 2(1 - √6) + 1
y = 2 - 2√6 + 1
y = 3 - 2√6
So the algebraic solution to the system of equations is:
(x, y) = (1 + √6, 3 + 2√6) and (1 - √6, 3 - 2√6)
Part 2:
To graph the system of equations, we will plot the linear equation and the quadratic equation on the same graph.
The linear equation is a straight line, and the quadratic equation is a parabola.
Graphing y = 2x + 1:
Using the slope-intercept form, we can see that the line intersects the y-axis at y = 1 and has a slope of 2.
Plotting this line on the graph, we have:
Graph:
https://www.desmos.com/calculator/odgrdjba9s
Graphing y = x^2 - 4:
This is a parabola that opens upward and its vertex is at the point (0, -4).
We can also find the x-intercepts (zeros) of the quadratic equation by setting y = 0 and solving for x:
0 = x^2 - 4
x^2 = 4
x = ± √4
x = ±2
Plotting this parabola on the graph, we have:
Graph:
https://www.desmos.com/calculator/fond6cbcpw
The solution to the system of equations can be found by identifying the point(s) of intersection between the linear equation and the quadratic equation. From the graph, we can see that the two equations intersect at two points: (1 + √6, 3 + 2√6) and (1 - √6, 3 - 2√6), confirming our algebraic solution.
Sure! Let's start by creating a system of equations that includes one linear equation and one quadratic equation:
Linear Equation:
y = 2x + 3
Quadratic Equation:
y = x^2 - 4x + 4
Part 1: Solving the system of equations algebraically:
To solve this system of equations algebraically, we need to find the values of x and y that satisfy both equations simultaneously.
Step 1: Set the equations equal to each other:
2x + 3 = x^2 - 4x + 4
Step 2: Rearrange the equation to bring all terms to one side to simplify:
x^2 - 6x + 1 = 0
Step 3: Solve the quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Here, a = 1, b = -6, and c = 1, so plugging the values into the formula:
x = (-(-6) ± √((-6)^2 - 4(1)(1))) / (2(1))
x = (6 ± √(36 - 4)) / 2
x = (6 ± √32) / 2
x = (6 ± 4√2) / 2
Step 4: Simplify the expression for x:
x = 3 ± 2√2
Now that we have the values of x, let's substitute them into either equation to find the corresponding values of y.
For x = 3 + 2√2:
y = 2(3 + 2√2) + 3
y = 6 + 4√2 + 3
y = 9 + 4√2
For x = 3 - 2√2:
y = 2(3 - 2√2) + 3
y = 6 - 4√2 + 3
y = 9 - 4√2
So, the solution to the system of equations is:
(3 + 2√2, 9 + 4√2) and (3 - 2√2, 9 - 4√2)
Part 2: Graphical representation:
To graph the system of equations and verify the solution, we will plot the linear equation and the quadratic equation on the same coordinate plane.
The linear equation, y = 2x + 3, is a straight line with a slope of 2 and a y-intercept of 3.
The quadratic equation, y = x^2 - 4x + 4, represents a parabola.
We can use any graphing tool or software to plot these equations and see where they intersect. By finding the points of intersection, we can confirm the solution we obtained algebraically.
Once the graph is generated, we can visually inspect the points of intersection. The coordinates of these points should match the solutions we found earlier algebraically: (3 + 2√2, 9 + 4√2) and (3 - 2√2, 9 - 4√2).
By comparing the graphical and algebraic solutions, we can verify the accuracy of our algebraic solution.
y=x
y=x^2
x^2=x
x^2-x=0
x(x-1)=0
x=0,1