Solving Simultaneous Linear and Quadratic Equations Revision Notes for VCE SSCE Mathematical Methods

Solving Simultaneous Linear and Quadratic Equations

Understanding the concept

When we solve simultaneous equations, we're finding where the graphs of these equations meet or intersect. Just as solving two simultaneous linear equations gives us the point where two straight lines cross, we can also find where a straight line meets a parabola by solving a linear and quadratic equation simultaneously.

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This concept extends what you already know about linear equations. Instead of finding where two straight lines cross, we're now finding where a straight line crosses a curved parabola. The mathematical process is similar, but the results can be different!

Possible intersection scenarios

A straight line can interact with a parabola in three different ways, giving us different numbers of intersection points:

Two points of intersection: The line crosses through the parabola at two separate points
One point of intersection: The line just touches the parabola at a single point
No points of intersection: The line and parabola don't meet at all

What is a tangent?

The term "tangent" has a specific meaning when we're talking about lines and parabolas:

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When there is exactly one point of intersection between the parabola and the straight line, we say the line is a tangent to the parabola. A tangent just touches the curve at one point without crossing through it.

The substitution method

To find the points of intersection between a linear equation and a quadratic equation, we use the substitution method. Here's how it works:

Step 1: Write both equations with $y$ as the subject.

For example:

Linear equation: $y = -2x + 4$
Quadratic equation: $y = x^2 - 8x + 12$

Step 2: Since both expressions equal $y$ , set them equal to each other and solve for $x$ .

Step 3: Substitute your $x$ -values back into either equation to find the corresponding $y$ -values.

Step 4: Write your answer as coordinate points.

Worked example: Two points of intersection

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Worked Example: Finding Two Intersection Points

Find the points of intersection of the line with equation $y = -2x + 4$ and the parabola with equation $y = x^2 - 8x + 12$ .

Solution:

At the point of intersection, both equations have the same $x$ and $y$ values, so:

x^2 - 8x + 12 = -2x + 4

Rearrange to get everything on one side:

x^2 - 6x + 8 = 0

Factorise:

(x - 2)(x - 4) = 0

Solve for $x$ :

x = 2 \text{ or } x = 4

Find the corresponding $y$ -values by substituting into the linear equation:

When $x = 2$ : $y = -2(2) + 4 = 0$

When $x = 4$ : $y = -2(4) + 4 = -4$

Therefore, the points of intersection are (2, 0) and (4, -4).

We can verify this graphically:

The graph confirms our two intersection points: one at $(2, 0)$ on the $x$ -axis, and another at $(4, -4)$ which is the vertex of the parabola.

Using technology

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Calculator Methods

These calculator methods can help you verify your solutions or solve more complex simultaneous equations quickly.

Using the TI-Nspire

To solve simultaneous equations on the TI-Nspire calculator:

Select menu > Algebra > Solve System of Equations > Solve System of Equations
Press enter to accept the default settings of two equations with variables $x$ and $y$
Complete the template by entering your equations
The calculator will display the solution(s)

Using the Casio ClassPad

To solve simultaneous equations on the Casio ClassPad:

In the Main screen, turn on the keyboard
Select the simultaneous equations icon from Math1
Enter your equations into the two lines
Enter $x, y$ as the variables
Tap EXE to solve

Worked example: Proving a line is tangent

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Worked Example: Proving a Line is a Tangent

Prove that the straight line with equation $y = 1 - x$ meets the parabola with equation $y = x^2 - 3x + 2$ once only.

Solution:

At the point of intersection:

x^2 - 3x + 2 = 1 - x

Rearrange:

x^2 - 2x + 1 = 0

Factorise (this is a perfect square):

(x - 1)^2 = 0

Solve for $x$ :

x = 1

Find the $y$ -value:

y = 1 - 1 = 0

Since we get only one solution for $x$ , the straight line just touches the parabola at the single point (1, 0). This proves the line is a tangent to the parabola.

We can see this graphically:

The graph shows that the blue line touches the red parabola at exactly one point, confirming it's a tangent.

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Key Indicator: When factorising gives you a perfect square like $(x - 1)^2 = 0$ , this tells you there's only one solution, which means the line is a tangent to the parabola.

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Key Points to Remember:

Solving simultaneous linear and quadratic equations finds where a straight line and parabola intersect
There can be 0, 1, or 2 points of intersection depending on how the line and parabola are positioned
When there's exactly one point of intersection, the line is a tangent to the parabola
Use the substitution method: set the two expressions for $y$ equal to each other and solve for $x$
A perfect square result like $(x - 1)^2 = 0$ indicates a tangent (one solution)

Solving Simultaneous Linear and Quadratic Equations (VCE SSCE Mathematical Methods): Revision Notes