Solve Quadratic Systems Revision Notes for HSC SSCE Mathematics Advanced

Solve Quadratic Systems

Introduction

A quadratic system involves finding where two functions intersect. This could be where a quadratic function meets a linear function, or where two quadratic functions meet. Understanding how to solve these systems is essential for analysing the relationship between different mathematical functions.

In this topic, you will learn to:

Find intersection points of a quadratic and a linear function using both algebraic and graphical methods
Find intersection points of two quadratic functions
Use the discriminant to determine how many intersection points exist
Interpret solutions as coordinates where graphs meet

Intersections of quadratics and linear graphs

When we want to find where a quadratic function $f(x) = ax^2 + bx + c$ and a linear function $g(x) = mx + d$ meet, we are looking for their intersection points. These are the points where the two graphs cross each other.

Finding intersection points

The intersection points occur where f(x) = g(x). We can find these points using two different approaches:

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Algebraic method:

To solve algebraically, follow these steps:

Set the two functions equal: $ax^2 + bx + c = mx + d$
Rearrange to form a standard quadratic equation: $ax^2 + (b - m)x + (c - d) = 0$
Solve the quadratic equation to find the $x$ -values of the intersection points
Substitute each $x$ -value back into either original function to find the corresponding $y$ -coordinates

Using the linear equation to find the $y$ -values is usually quicker than using the quadratic equation.

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Graphical method:

To solve graphically:

Plot both functions on the same coordinate plane
Identify where the graphs intersect
Read the coordinates of the intersection points from the graph

This method works well when the intersection points have integer or simple decimal coordinates. However, it may be less accurate for complex values.

Using the discriminant

The discriminant ( $\Delta$ ) provides a quick way to determine how many intersection points exist without solving the entire equation. After rearranging the system into the form $ax^2 + bx + c = 0$ , the discriminant is:

$\Delta = b^2 - 4ac$

chatImportant

The value of the discriminant tells us:

$\Delta > 0$ : Two intersection points (the line cuts through the parabola at two places)
$\Delta = 0$ : One intersection point (the line is tangent to the parabola, touching it at exactly one point)
$\Delta < 0$ : No intersection points (the line and parabola don't meet)

Exam tip: Always check the discriminant first if you're asked about the number of intersections. It's much faster than solving the full equation!

Worked example: Quadratic and linear intersection

Let's find the intersection points of $f(x) = x^2 - 4x + 4$ and $g(x) = 2x - 1$ .

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Worked Example: Finding Intersection Points Algebraically

Step 1: Set the functions equal to each other.

$f(x) = g(x)$

$x^2 - 4x + 4 = 2x - 1$

Step 2: Rearrange to form a quadratic equation.

$x^2 - 4x - 2x + 4 + 1 = 0$

$x^2 - 6x + 5 = 0$

Step 3: Factorise the quadratic.

$(x - 5)(x - 1) = 0$

Step 4: Solve for $x$ .

$x = 5 \text{ or } x = 1$

Step 5: Find the corresponding $y$ -coordinates by substituting into $g(x) = 2x - 1$ .

When $x = 5$ :

$g(5) = 2 \times 5 - 1 = 9$

When $x = 1$ :

$g(1) = 2 \times 1 - 1 = 1$

Solution: The intersection points are (5, 9) and (1, 1).

Graphical solution

To solve graphically, we plot both functions on a coordinate plane and identify where they meet.

The parabola $f(x) = x^2 - 4x + 4$ and the line $g(x) = 2x - 1$ intersect at (5, 9) and (1, 1), confirming our algebraic solution.

Intersections of two quadratics

Finding where two quadratic functions intersect follows a similar process to quadratic-linear systems. If we have $f(x) = a_1x^2 + b_1x + c_1$ and $g(x) = a_2x^2 + b_2x + c_2$ , their intersection points occur where f(x) = g(x).

Finding intersection points for two quadratics

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Algebraic method:

Set $f(x) = g(x)$
Rearrange to form: $(a_1 - a_2)x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0$
Solve using factorisation, completing the square, or the quadratic formula
Substitute the $x$ -values into either original function to find corresponding $y$ -coordinates

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Graphical method:

Plot both parabolas on the same coordinate plane
Identify where the curves intersect
Read the coordinates of intersection points

Note that reading exact values from a graph can be challenging when the intersection points involve surds or complex decimals.

Using the discriminant for two quadratics

After rearranging into standard form, the discriminant still tells us about the number of intersections:

chatImportant

$\Delta > 0$ : Two intersection points (the parabolas cross at two places)
$\Delta = 0$ : One intersection point (the parabolas are tangent, touching at exactly one point)
$\Delta < 0$ : No intersection points (the parabolas don't meet)

Exam tip: When solving systems of two quadratics, the quadratic formula is often the most reliable method, especially when the equation doesn't factorise easily.

Worked example: Two quadratics intersection

Find the intersection points of $f(x) = x^2 - 2x - 3$ and $g(x) = -x^2 + 3x + 1$ both graphically and algebraically.

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Worked Example: Finding Where Two Parabolas Meet

Step 1: Set the functions equal.

$f(x) = g(x)$

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

Step 2: Rearrange to standard form.

$x^2 - 2x - 3 + x^2 - 3x - 1 = 0$

$2x^2 - 5x - 4 = 0$

Step 3: Use the quadratic formula since this doesn't factorise easily.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute $a = 2$ , $b = -5$ , and $c = -4$ :

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 2 \times (-4)}}{2 \times 2}$

$x = \frac{5 \pm \sqrt{25 + 32}}{4}$

$x = \frac{5 \pm \sqrt{57}}{4}$

Step 4: Find the corresponding $y$ -coordinates by substituting into $f(x) = x^2 - 2x - 3$ .

For $x = \frac{5 + \sqrt{57}}{4}$ :

$f\left(\frac{5 + \sqrt{57}}{4}\right) = \left(\frac{5 + \sqrt{57}}{4}\right)^2 - 2 \times \frac{5 + \sqrt{57}}{4} - 3$

$= \frac{\sqrt{57} - 3}{8}$

For $x = \frac{5 - \sqrt{57}}{4}$ :

$f\left(\frac{5 - \sqrt{57}}{4}\right) = \left(\frac{5 - \sqrt{57}}{4}\right)^2 - 2 \times \frac{5 - \sqrt{57}}{4} - 3$

$= \frac{-3 - \sqrt{57}}{8}$

Solution: The intersection points are approximately $\left(\frac{5 + \sqrt{57}}{4}, \frac{\sqrt{57} - 3}{8}\right)$ and $\left(\frac{5 - \sqrt{57}}{4}, \frac{-3 - \sqrt{57}}{8}\right)$ .

In decimal form, these are approximately (3.1, 0.6) and (-0.6, -1.3).

Graphical solution

We can verify our solution by plotting both parabolas.

The parabolas $f(x) = x^2 - 2x - 3$ and $g(x) = -x^2 + 3x + 1$ intersect at approximately (3.1, 0.6) and (-0.6, -1.3), which matches our algebraic solution.

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Key Points to Remember:

Intersection points occur where two functions are equal: solve by setting f(x) = g(x)
For algebraic solutions, rearrange into a quadratic equation, solve for $x$ -values, then substitute back to find $y$ -coordinates
For graphical solutions, plot both functions and identify where they cross
The discriminant ( $\Delta = b^2 - 4ac$ $Δ = b^{2} - 4 a c$ ) quickly determines the number of intersections:
- $\Delta > 0$ : two intersections
- $\Delta = 0$ : one intersection (tangent)
- $\Delta < 0$ : no intersections
When working with two quadratics, the quadratic formula is often the most reliable solving method
Always verify your solutions by checking that the coordinates satisfy both original equations

Solve Quadratic Systems (HSC SSCE Mathematics Advanced): Revision Notes