The Discriminant Revision Notes for VCE SSCE Mathematical Methods

The Discriminant

Introduction

When we solve quadratic equations using the quadratic formula, we encounter a special expression that tells us important information about the solutions. This expression is called the discriminant.

The quadratic formula for solving $ax^2 + bx + c = 0$ is:

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

The expression under the square root sign, $b^2 - 4ac$ , is the discriminant. We use the Greek letter delta (∆) to represent it:

\Delta = b^2 - 4ac

infoNote

The discriminant is a powerful analytical tool that tells us how many solutions a quadratic equation has and what type of solutions they are, without actually solving the equation. This makes it invaluable for quickly analyzing quadratic functions and their graphs.

The number of x-axis intercepts

A parabola can intersect the x-axis in three different ways:

Zero intercepts – the graph sits entirely above or entirely below the x-axis
One intercept – the graph touches the x-axis at exactly one point (the turning point)
Two intercepts – the graph crosses the x-axis at two distinct points

The discriminant determines which of these situations occurs for any quadratic function $y = ax^2 + bx + c$ .

Case 1: No x-axis intercepts (∆ < 0)

When the discriminant is negative ( $b^2 - 4ac < 0$ ), the equation $ax^2 + bx + c = 0$ has no real solutions. This means the parabola does not intersect the x-axis at all.

infoNote

Why does this happen? Because you cannot take the square root of a negative number in the real number system. The parabola either sits completely above the x-axis (if $a > 0$ ) or completely below it (if $a < 0$ ).

Case 2: One x-axis intercept (∆ = 0)

When the discriminant is zero ( $b^2 - 4ac = 0$ ), the equation $ax^2 + bx + c = 0$ has exactly one solution. The parabola touches the x-axis at one point, which is the turning point (vertex) of the parabola.

We sometimes say the equation has two coincident solutions – two identical solutions that coincide at the same point.

Case 3: Two x-axis intercepts (∆ > 0)

When the discriminant is positive ( $b^2 - 4ac > 0$ ), the equation $ax^2 + bx + c = 0$ has two distinct real solutions. The parabola crosses the x-axis at two different points.

Here are the three cases depicted graphically:

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Quick Reference for Number of Solutions:

$\Delta < 0$ → No real solutions (no x-axis intercepts)
$\Delta = 0$ → One solution (one x-axis intercept)
$\Delta > 0$ → Two distinct solutions (two x-axis intercepts)

Worked example: Finding discriminants

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Worked Example: Finding Discriminants and Determining Intercepts

Question: Find the discriminant of each of the following quadratics and state whether the graph crosses the x-axis, touches the x-axis, or does not intersect the x-axis.

a) $y = x^2 - 6x + 8$

b) $y = x^2 - 8x + 16$

c) $y = 2x^2 - 3x + 4$

Solution

a) For $y = x^2 - 6x + 8$ :

Here, $a = 1$ , $b = -6$ , and $c = 8$ .

\Delta = b^2 - 4ac

\Delta = (-6)^2 - (4 \times 1 \times 8)

\Delta = 36 - 32

\Delta = 4

Since ∆ > 0, the graph intersects the x-axis at two distinct points.

This means there are two distinct solutions to the equation $x^2 - 6x + 8 = 0$ .

b) For $y = x^2 - 8x + 16$ :

Here, $a = 1$ , $b = -8$ , and $c = 16$ .

\Delta = b^2 - 4ac

\Delta = (-8)^2 - (4 \times 1 \times 16)

\Delta = 64 - 64

\Delta = 0

Since ∆ = 0, the graph touches the x-axis at exactly one point.

This means there is one solution to the equation $x^2 - 8x + 16 = 0$ .

c) For $y = 2x^2 - 3x + 4$ :

Here, $a = 2$ , $b = -3$ , and $c = 4$ .

\Delta = b^2 - 4ac

\Delta = (-3)^2 - (4 \times 2 \times 4)

\Delta = 9 - 32

\Delta = -23

Since ∆ < 0, the graph does not intersect the x-axis.

This means there are no real solutions to the equation $2x^2 - 3x + 4 = 0$ .

Worked example: Finding parameter values

Sometimes we need to find values of a parameter (like $m$ ) that give us a specific number of solutions.

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Worked Example: Finding Parameter Values for Different Numbers of Solutions

Question: Find the values of $m$ for which the equation $3x^2 - 2mx + 3 = 0$ has:

a) one solution

b) no solution

c) two distinct solutions

Solution

For the quadratic $3x^2 - 2mx + 3$ , we have $a = 3$ , $b = -2m$ , and $c = 3$ .

The discriminant is:

\Delta = b^2 - 4ac

\Delta = (-2m)^2 - 4 \times 3 \times 3

\Delta = 4m^2 - 36

a) For one solution, we need $\Delta = 0$ :

4m^2 - 36 = 0

4m^2 = 36

m^2 = 9

m = \pm 3

Therefore, the equation has one solution when m = 3 or m = -3.

b) For no solution, we need $\Delta < 0$ :

4m^2 - 36 < 0

4m^2 < 36

m^2 < 9

From the graph below, we can see this occurs when -3 < m < 3.

c) For two distinct solutions, we need $\Delta > 0$ :

4m^2 - 36 > 0

4m^2 > 36

m^2 > 9

From the graph, this occurs when m > 3 or m < -3.

The graph shows $\Delta = 4m^2 - 36$ plotted against $m$ . Notice that the parabola has its vertex at $(0, -36)$ and crosses the m-axis at $m = -3$ and $m = 3$ .

The nature of solutions

The discriminant not only tells us how many solutions a quadratic equation has, but also what type of solutions they are.

For a quadratic equation $ax^2 + bx + c = 0$ where $a$ , $b$ , and $c$ are rational numbers:

Rational solutions

Two rational solutions: If $\Delta = b^2 - 4ac$ is a perfect square and $\Delta \neq 0$ , then the equation has two rational solutions.

infoNote

A perfect square is a number like 4, 9, 16, 25, etc. – numbers whose square root is a whole number or fraction. When the discriminant is a perfect square, we can simplify $\sqrt{\Delta}$ to a rational number, which makes the entire quadratic formula result in rational solutions.

One rational solution: If $\Delta = 0$ , then the equation has one rational solution (or two coincident rational solutions).

Irrational solutions

Two irrational solutions: If $\Delta$ is not a perfect square and $\Delta > 0$ , then the equation has two irrational solutions.

For example, if $\Delta = 5$ , then $\sqrt{5}$ is irrational, so the solutions will be irrational.

Worked example: Proving rationality of solutions

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Worked Example: Proving Solutions are Rational

Question: Show that the solutions of the equation $3x^2 + (m - 3)x - m = 0$ are rational for all rational values of $m$ .

Solution

For this equation, $a = 3$ , $b = (m - 3)$ , and $c = -m$ .

Calculate the discriminant:

\Delta = b^2 - 4ac

\Delta = (m - 3)^2 - 4 \times 3 \times (-m)

\Delta = (m - 3)^2 + 12m

Expand $(m - 3)^2$ :

\Delta = m^2 - 6m + 9 + 12m

\Delta = m^2 + 6m + 9

Factor the expression:

\Delta = (m + 3)^2

Since $\Delta = (m + 3)^2$ , we know that:

$(m + 3)^2 \geq 0$ for all values of $m$

Furthermore, ∆ is a perfect square for all rational values of $m$ , because $(m + 3)^2$ is always a perfect square when $m$ is rational.

Therefore, the solutions are rational for all rational values of m.

chatImportant

Key Insight: When the discriminant can be expressed as a perfect square (like $(m + 3)^2$ in this example), we can immediately conclude that the solutions will be rational, provided the coefficients $a$ , $b$ , and $c$ are rational.

Remember!

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Key Points to Remember:

The discriminant of a quadratic $ax^2 + bx + c$ is $\Delta = b^2 - 4ac$
If $\Delta < 0$ : no real solutions and the parabola does not intersect the x-axis
If $\Delta = 0$ : one solution (two coincident solutions) and the parabola touches the x-axis at one point
If $\Delta > 0$ : two distinct solutions and the parabola crosses the x-axis at two points
For rational coefficients: if $\Delta$ is a perfect square (and $\Delta \neq 0$ ), solutions are rational; if $\Delta$ is not a perfect square (and $\Delta > 0$ ), solutions are irrational

The Discriminant (VCE SSCE Mathematical Methods): Revision Notes