The General Quadratic Formula Revision Notes for VCE SSCE Mathematical Methods

The General Quadratic Formula

Introduction

Some quadratic equations are difficult or impossible to solve using factorization by inspection. When this happens, we need an alternative method to find the solutions. The quadratic formula provides a reliable way to solve any quadratic equation in standard form, regardless of whether it can be factorized easily.

The quadratic formula is particularly useful for finding the $x$ -axis intercepts of a quadratic function when these values are not obvious through factorization.

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The quadratic formula is your go-to method when:

A quadratic doesn't factor nicely by inspection
You need exact values for the solutions
You're working with quadratics involving parameters or surds

The quadratic formula

For any quadratic equation written in the form $ax^2 + bx + c = 0$ where $a \neq 0$ , the solutions are given by the quadratic formula:

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

This formula works for all quadratic equations and will give you the exact solutions (if they exist).

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The quadratic formula provides an alternative to completing the square. However, completing the square is often more useful for curve sketching because it gives you the turning point coordinates directly. The quadratic formula is primarily used for finding solutions or $x$ -intercepts.

How the formula is derived

The quadratic formula comes from completing the square for the general quadratic equation. Here's a brief overview of the derivation:

Starting with the general quadratic in completed square form:

y = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c

We set this equal to zero to solve for $x$ :

a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c = 0

After rearranging:

a\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a} - c

Dividing both sides by $a$ :

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

Taking the square root of both sides:

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

Therefore:

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

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This derivation shows that the quadratic formula is really just a rearrangement of the completed square form. Understanding this connection helps you see why both methods work!

The axis of symmetry

An important relationship can be seen from the quadratic formula. The axis of symmetry of any quadratic is located exactly halfway between the two solutions (when they exist). This is given by:

x = -\frac{b}{2a}

Notice that this is the part of the quadratic formula before the $\pm$ symbol. This makes sense because the axis of symmetry is the average of the two solutions.

The discriminant and number of solutions

The expression under the square root sign in the quadratic formula, $b^2 - 4ac$ , is called the discriminant. It is often represented by the Greek letter delta: $\Delta = b^2 - 4ac$ .

The discriminant tells us important information about the number of solutions to a quadratic equation:

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Understanding the Discriminant:

If $b^2 - 4ac > 0$ : The discriminant is positive, meaning we can take its square root. This gives us two distinct real solutions. The parabola crosses the $x$ -axis at two points.

If $b^2 - 4ac = 0$ : The discriminant equals zero, so the $\pm$ part of the formula becomes zero. This gives us exactly one solution (a repeated root). The parabola just touches the $x$ -axis at its vertex.

If $b^2 - 4ac < 0$ : The discriminant is negative, meaning we cannot take its square root (in the real number system). There are no real solutions. The parabola does not intersect the $x$ -axis at all.

Understanding the discriminant helps you predict the nature of the solutions before calculating them.

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Exam tip: You can use a CAS calculator to solve quadratic equations quickly. On the TI-Nspire, use menu > Algebra > Solve. On the Casio ClassPad, select "solve" from the keyboard. However, you should still understand the quadratic formula for exam questions that require you to show working.

Worked example: solving simple quadratics

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Worked Example: Solving a Simple Quadratic

Solve $x^2 - x - 1 = 0$ using the quadratic formula.

First, identify the coefficients:

$a = 1$

$b = -1$

$c = -1$

Substitute these values into the quadratic formula:

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

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-1)}}{2 \times 1}

x = \frac{1 \pm \sqrt{1 + 4}}{2}

x = \frac{1 \pm \sqrt{5}}{2}

This gives us two solutions: $x = \frac{1 + \sqrt{5}}{2}$ or $x = \frac{1 - \sqrt{5}}{2}$ .

Worked example: quadratics with parameters

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Worked Example: Solving a Quadratic with Parameters

Solve $x^2 - 2kx - 3 = 0$ where $k$ is a constant.

Here we have a parameter $k$ instead of a specific number. The method is the same.

Identify the coefficients:

$a = 1$

$b = -2k$

$c = -3$

Apply the quadratic formula:

x = \frac{-(-2k) \pm \sqrt{(-2k)^2 - 4 \times 1 \times (-3)}}{2 \times 1}

x = \frac{2k \pm \sqrt{4k^2 + 12}}{2}

We can simplify this further. Factor out 4 from under the square root:

x = \frac{2k \pm \sqrt{4(k^2 + 3)}}{2}

x = \frac{2k \pm 2\sqrt{k^2 + 3}}{2}

x = k \pm \sqrt{k^2 + 3}

Note that since $k^2 \geq 0$ for all values of $k$ , we know that $k^2 + 3 \geq 3 > 0$ . This means the discriminant is always positive, so this equation always has two distinct real solutions regardless of the value of $k$ .

Worked example: sketching using the quadratic formula

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Worked Example: Sketching a Graph Using the Quadratic Formula

Sketch the graph of $y = -3x^2 - 12x - 7$ , using the quadratic formula to find the $x$ -axis intercepts.

First, find the $y$ -intercept. When $x = 0$ :

$y = -7$

So the $y$ -intercept is $-7$ .

Next, find the axis of symmetry:

x = -\frac{b}{2a} = -\frac{-12}{2 \times (-3)} = -\frac{-12}{-6} = -2

Now find the turning point. When $x = -2$ :

$y = -3(-2)^2 - 12(-2) - 7$

$y = -3(4) + 24 - 7$

$y = -12 + 24 - 7$

$y = 5$

The turning point is at $(-2, 5)$ .

To find the $x$ -axis intercepts, solve $-3x^2 - 12x - 7 = 0$ using the quadratic formula.

Here: $a = -3$ , $b = -12$ , $c = -7$

x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(-3)(-7)}}{2(-3)}

x = \frac{12 \pm \sqrt{144 - 84}}{-6}

x = \frac{12 \pm \sqrt{60}}{-6}

Simplify $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$ :

x = \frac{12 \pm 2\sqrt{15}}{-6}

x = \frac{2(6 \pm \sqrt{15})}{-6}

x = -\frac{6 \pm \sqrt{15}}{3}

x = -2 \mp \frac{\sqrt{15}}{3}

So the $x$ -intercepts are $x = -2 - \frac{\sqrt{15}}{3}$ and $x = -2 + \frac{\sqrt{15}}{3}$ .

As decimal approximations: $x \approx -3.29$ or $x \approx -0.71$ (to two decimal places).

The graph shows a downward-opening parabola (since $a = -3 < 0$ ) with vertex at $(-2, 5)$ and $x$ -intercepts at approximately $-3.29$ and $-0.71$ .

Remember!

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Key Points to Remember:

The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ can solve any quadratic equation in the form $ax^2 + bx + c = 0$ .
The axis of symmetry is given by $x = -\frac{b}{2a}$ , which is the $x$ -coordinate of the turning point.
The discriminant $\Delta = b^2 - 4ac$ determines the number of solutions: if positive, two solutions; if zero, one solution; if negative, no real solutions.
When using the quadratic formula, always identify $a$ , $b$ , and $c$ carefully, paying attention to negative signs.
The quadratic formula is derived from completing the square, showing the deep connection between these two methods.

The General Quadratic Formula (VCE SSCE Mathematical Methods): Revision Notes