Quadratic Simultaneous Equations
In this article, we will discuss:
Here is one more link to practice a few extra questions: Maths Genie Quadratic Simultaneous Equations Questions
ax² + bx + c = 0 (quadratic equation)
dx + ey + f = 0 (linear equation)
Step #1: Rearrange the linear equation to solve for one of the variables.
For example, if the linear equation is 2x – 3y = 7,
solve for y:
y = (2x – 7)/3.
Step #2: Substitute the value of y into the quadratic equation, resulting in a quadratic equation in terms of x.
For example, if the quadratic equation is x2 – 5x + 6 = 0, substitute (2x – 7)/3 for y to get
(2x – 7)2/9 – 5(2x – 7)/3 + 6 = 0.
Step #3: Simplify the quadratic equation and solve for x. In the example above,
we can simplify the equation to
4x2 – 38x + 65 = 0.
Factor the quadratic equation to get
(2x – 5)(2x – 13) = 0.
Therefore, x = 5/2 or x = 13/2.
Step #4: Substitute the value of x back into the linear equation to solve for y.
In the example above, if x = 5/2, then y = (2(5/2) – 7)/3 = -1/3.
If x = 13/2, then y = (2(13/2) – 7)/3 = 5/3.
Therefore, the solution for the quadratic simultaneous equations
x2 – 5x + 6 = 0 and 2x – 3y = 7 is (5/2, -1/3) and (13/2, 5/3).
y = mx + b
For the linear equation,
ax + by + c = 0,
we can solve for y to get:
y = (-ax – c)/b
For the quadratic equation, y = ax2 + bx + c, we can use a graphing calculator or plotting points to create a graph.
Solved Example:
Question 1: How can we solve the system of equations consisting of a quadratic equation (3x2 – 4x – 5 = 0) and a linear equation (2x – 7y + 1 = 0)?
Solution:
To solve the system of equations, we will follow these steps:
2x – 7y + 1 = 0
-7y = -2x – 1
y = (2/7)x + 1/7
(1.5, 1.1) and (-0.6, -0.4).
For the point (1.5, 1.1):
x ≈ 1.5
y ≈ 1.1
For the point (-0.6, -0.4):
x ≈ -0.6
y ≈ -0.4
These solutions represent the values of x and y that satisfy both the quadratic and linear equations in the system.
Question 1. Solve the simultaneous equations:
x2 - 3x - 4 = 0
y2 - 4y - 5 = 0
Answer : ( , )Equation 1:
(x - 4)(x + 1) = 0
Equation 2:
(y - 5)(y + 1) = 0
Equation 1:
(x - 4) = 0 ⇒ x = 4
(x + 1) = 0 ⇒ x = -1
Equation 2:
(y - 5) = 0 ⇒ y = 5
(y + 1) = 0 ⇒ y = -1
Question 2: Solve the simultaneous equations
x2 - 9 = 0
y2 - 4y - 21 = 0
Answer : ( , )Solution:
Equation 1:
(x - 3)(x + 3) = 0
Equation 2:
(y - 7)(y + 3) = 0
Equation 1:
(x - 3) = 0 ⇒ x = 3
(x + 3) = 0 ⇒ x = -3
Equation 2:
(y - 7) = 0 ⇒ y = 7
(y + 3) = 0 ⇒ y = -3
Quadratic simultaneous equations involve two equations, one of which is a quadratic equation.
The goal is to find the values of the variables that satisfy both equations.
There are two main methods to solve quadratic simultaneous equations: factorisation and graphing.
Factorisation involves factoring both equations and setting each factor equal to zero to find the solutions.
Graphing the two equations allows us to visually determine the points where they intersect, which represent the solutions.
Understanding the basics of quadratic simultaneous equations and the methods used to solve them is important for students.
Being proficient in solving these equations can boost confidence when encountering similar problems in exams or real-life situations.
Question 1: Solve the simultaneous equations
2x + y = 7
x2 - y2 = 8
Question 2: Solve the simultaneous equations
x2 + y2 = 20
y = x + 3
Question 3: Solve the simultaneous equations
2x2 + y2 = 10
2x - y = 5
Question 4: Solve the simultaneous equations
y = x2 + x - 7
4x + 2y + 1 = 0
Question 5: Solve the simultaneous equations
y = x - 2
2x2 - xy = 11