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More Lessons for NYSED Regents Exam
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High School Math based on the topics required for the Regents Exam conducted by NYSED.
The following diagrams show the types of solutions for Linear-Quadratic Systems. Scroll down the page for more examples and solutions on how to solve Linear-Quadratic Systems.
Linear Quadratic Systems Part 1
This lesson shows how to solve linear-quadratic systems
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Linear Quadratic Systems Part 2
This lesson shows how to solve linear - quadratic systems.
Linear and Quadratic Systems
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The graphical form of linear equation is a straight line.
Graphical form of a quadratic equation is a parabola.
If we solve both linear and quadratic equation, then we will get the point of intersection of parabola and a straight line.
Example 1 :
Solve :
y = x2 + 3x - 5
y = x + 3
Solution :
y = x2 + 3x - 5 ----(1)
y = x + 3 ----(2)
By equating (1) and (2), we get
x + 3 = x2 + 3x - 5
x2 + 3x - x - 3 - 5 = 0
x2 + 2x - 8 = 0
(x + 4)(x - 2) = 0
x + 4 = 0 or x - 2 = 0
x = -4 or x = 2
To get corresponding y values, we can apply each x values one by one either in first or second equation.
If x = -4, then y = -4 + 3 = -1.
If x = 2, then y = 2 + 3 = 5.
So, the given parabola and straight line will intersect at the points (-4, -1) and (2, 5).
By observing the graph given below, we can understand what is point of intersection.
Example 2 :
Solve :
y = x2 - 4x + 6
y = x + 2
Solution :
y = x2 - 4x + 6 ----(1)
y = x + 2 ----(2)
By equating (1) and (2), we get
x + 2 = x2 - 4x + 6
x2 - 4x - x + 6 - 2 = 0
x2 - 5x + 4 = 0
(x - 1)(x - 4) = 0
x - 1 = 0 or x - 4 = 0
x = 1 or x = 4
To get corresponding y values, we can apply each x values one by one either in first or second equation.
If x = 1, then y =1 + 2 = 3.
If x = 4, then y = 4 + 2 = 6.
So, the given parabola and straight line will intersect at the points (1, 3) and (4, 6).
Example 3 :
Solve :
y = x2 - 10x + 14
y = 7x - 16
Solution :
y = x2 - 10x + 14 ----(1)
y = 7x - 16 -----(2)
By equating (1) and (2), we get
x2 - 10x + 14 = 7x - 16
x2 - 10x - 7x + 14 + 16 = 0
x2 - 17x + 30 = 0
(x - 12)(x - 5) = 0
x - 12 = 0 and x - 5 = 0
x = 12 and x = 5
To get corresponding y values, we can apply each x values one by one either in first or second equation.
In (2), if x = 12, then y = 7(12) - 16 y = 84 - 16 y = 68 | In (2), if x = 5, then y = 7(5) - 16 y = 35 - 16 y = 19 |
So, the given parabola and straight line will intersect at the points (12, 68) and (5, 19).
Example 4 :
Solve :
y = x2 - 24
y = x - 12
Solution :
y = x2 - 24 ----(1)
y = x - 12 ----(2)
By equating (1) and (2), we get
x - 12 = x2 - 24
x2 - x - 24 + 12 = 0
x2 - x - 12 = 0
(x - 4)(x + 3) = 0
x - 4 = 0 or x + 3 = 0
x = 4 or x = -3
To get corresponding y values, we can apply each x values one by one either in first or second equation.
In (2), if x = 4, then y = 4 - 12 y = -8 | In (2), if x = -3, then y = -3 - 12 y = -15 |
So, the given parabola and straight line will intersect at the points (4, -8) and (-3, -15).
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