Solve Equations Revision Notes for HSC SSCE Mathematics Advanced

Solve Quadratic Equations

A quadratic equation is a fundamental type of equation in algebra that involves a squared term. Learning to solve quadratic equations is an essential skill in mathematics, as these equations appear frequently in many real-world applications and advanced mathematical concepts. There are three main methods for solving quadratic equations, each with its own advantages depending on the specific equation you're working with.

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Quadratic equations are essential in physics (projectile motion), engineering (structural design), economics (profit optimization), and computer graphics. Mastering these solution methods will serve you well across many fields of study.

What is a quadratic equation?

A quadratic equation is an equation that can be written in the standard form $ax^2 + bx + c = 0$ , where $a \neq 0$ , and $b$ and $c$ are constants. The term "quadratic" comes from the Latin word for "square," referring to the squared term $x^2$ .

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Definition: An equation of the form $ax^2 + bx + c = 0$ , where $a \neq 0$ , $b$ and $c$ are constants.

The condition $a \neq 0$ is crucial—if $a = 0$ , the equation would no longer have a squared term and would become linear, not quadratic.

Understanding the terminology associated with quadratic equations will help you communicate clearly about them:

Monic quadratic: A quadratic equation where the coefficient of $x^2$ equals 1. For example, $x^2 + 5x + 6 = 0$ is monic.
Non-monic quadratic: A quadratic equation where the coefficient of $x^2$ does not equal 1 (and is not 0). For example, $2x^2 + 7x - 4 = 0$ is non-monic.
Trinomial: An expression with exactly three non-zero terms. Most standard quadratic equations are trinomials.
Constant term: A fixed number with no variable attached. In the standard form $ax^2 + bx + c = 0$ , the value $c$ is the constant term.

Solve by factorisation

Factorisation is often the quickest method for solving quadratic equations when the equation can be easily factored. This method relies on an important mathematical principle called the null factor law.

The null factor law

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The Null Factor Law

If the product of two or more factors equals zero, then at least one of those factors must equal zero. Mathematically, if $ab = 0$ , then either $a = 0$ or $b = 0$ (or both).

This is the fundamental principle that makes factorisation work as a solution method.

This property is the key to solving quadratic equations by factorisation. When we write a quadratic equation in factored form as $a(x - \alpha)(x - \beta) = 0$ , we can apply the null factor law to determine that either $(x - \alpha) = 0$ or $(x - \beta) = 0$ . This means the solutions (also called roots) are $x = \alpha$ and $x = \beta$ .

Steps for solving by factorisation

To solve a quadratic equation by factorisation:

Ensure the equation is written in standard form with one side equal to 0.
Factorise the quadratic expression into the form $(x - \alpha)(x - \beta) = 0$ for monic quadratics, or $a(x - \alpha)(x - \beta) = 0$ for non-monic quadratics.
Apply the null factor law by setting each factor equal to 0.
Solve each resulting linear equation to find the values of $x$ .

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Not all quadratic equations can be easily factored. When factorisation is difficult or impossible, you'll need to use completing the square or the quadratic formula instead.

Worked example: Factorising a monic quadratic

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Worked Example: Solving a Monic Quadratic by Factorisation

Let's solve $x^2 + 6x - 55 = 0$ by factorisation.

This is a monic quadratic (the coefficient of $x^2$ is 1). To factorise it, we need to find two integers that multiply to give the constant term $c = -55$ and add to give the coefficient of $x$ , which is $b = 6$ . These integers are 11 and $-5$ because $11 \times (-5) = -55$ and $11 + (-5) = 6$ .

Step-by-step solution:

$x^2 + 6x - 55 = 0$

$(x + 11)(x - 5) = 0$

$x + 11 = 0, \quad x - 5 = 0$

For $x + 11 = 0$ :

$x = -11$

For $x - 5 = 0$ :

$x = 5$

Answer: The solutions are $x = -11$ and $x = 5$ .

Worked example: Factorising a non-monic quadratic

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Worked Example: Solving a Non-Monic Quadratic by Factorisation

Now let's solve $2x^2 + 7x - 4 = 0$ by factorisation.

This is a non-monic quadratic because the coefficient of $x^2$ is 2. To factorise this equation, we use the grouping method. We need to find two numbers that multiply to $a \times c = 2 \times (-4) = -8$ and add to the coefficient of $x$ , which is $b = 7$ . These numbers are 8 and $-1$ .

Step-by-step solution:

We split the middle term using these numbers, then factor by grouping:

$2x^2 + 7x - 4 = 0$

$2x^2 + 8x - x - 4 = 0$

$(2x^2 + 8x) - (x + 4) = 0$

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

$(2x - 1)(x + 4) = 0$

$2x - 1 = 0, \quad x + 4 = 0$

For $2x - 1 = 0$ :

$2x = 1$

$x = \frac{1}{2}$

For $x + 4 = 0$ :

$x = -4$

Answer: The solutions are $x = \frac{1}{2}$ and $x = -4$ .

Solve by completing the square

Completing the square is a powerful algebraic technique that can be used to solve any quadratic equation, whether it can be easily factored or not. This method involves converting the quadratic equation into a perfect square trinomial form, which can then be solved by taking square roots.

Understanding perfect square trinomials

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What is a Perfect Square Trinomial?

A perfect square trinomial is an expression that can be written as $(x + k)^2$ or $(x - k)^2$ .

For example:

$(x + 3)^2 = x^2 + 6x + 9$
$(x - 5)^2 = x^2 - 10x + 25$

The completing the square method transforms any quadratic equation $ax^2 + bx + c = 0$ into the form $(x + k)^2 = d$ , where $d$ is a constant. Once in this form, we can solve by taking the square root of both sides, remembering to include both the positive and negative square roots.

Steps for completing the square

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Steps for Completing the Square:

Move the constant term $c$ to the right-hand side of the equation.
If the equation is non-monic (that is, $a \neq 1$ ), divide all terms by $a$ to make it monic.
Take the coefficient of the $x$ term, divide it by 2, and square the result. Add this value to both sides of the equation.
Factorise the left-hand side as a perfect square and simplify the right-hand side.
Take the square root of both sides, remembering to include the $\pm$ symbol, and then isolate $x$ .

Worked example: Completing the square for a monic quadratic

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Worked Example: Completing the Square for a Monic Quadratic

Let's solve $x^2 + 6x - 7 = 0$ by completing the square.

Since this is already monic, we can begin by moving the constant to the right side. The coefficient of $x$ is 6, so we calculate $\left(\frac{6}{2}\right)^2 = 9$ and add this to both sides.

Step-by-step solution:

$x^2 + 6x - 7 = 0$

$x^2 + 6x = 7$

$x^2 + 6x + \left(\frac{6}{2}\right)^2 = 7 + \left(\frac{6}{2}\right)^2$

$x^2 + 6x + 9 = 7 + 9$

$(x + 3)^2 = 16$

$x + 3 = \pm\sqrt{16}$

$x + 3 = \pm 4$

$x = -3 \pm 4$

This gives us two solutions: $x = -3 + 4 = 1$ and $x = -3 - 4 = -7$ .

Answer: The solutions are $x = 1$ and $x = -7$ .

Worked example: Completing the square for a non-monic quadratic

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Worked Example: Completing the Square for a Non-Monic Quadratic

Let's solve $2x^2 - 8x + 6 = 0$ by completing the square.

This is non-monic, so we first divide all terms by the leading coefficient 2 to make it monic.

Step-by-step solution:

$2x^2 - 8x + 6 = 0$

$x^2 - 4x + 3 = 0$

$x^2 - 4x = -3$

Now we add $\left(\frac{-4}{2}\right)^2 = 4$ to both sides:

$x^2 - 4x + 4 = -3 + 4$

$(x - 2)^2 = 1$

$x - 2 = \pm\sqrt{1}$

$x - 2 = \pm 1$

$x = 2 \pm 1$

Answer: The two solutions are $x = 2 + 1 = 3$ and $x = 2 - 1 = 1$ .

Solve by quadratic formula

The quadratic formula provides a direct method for finding the solutions to any quadratic equation in standard form. This formula is derived by applying the method of completing the square to the general form $ax^2 + bx + c = 0$ .

The quadratic formula

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The Quadratic Formula

For a quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$ , the roots are given by:

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

In this formula:

$x$ represents the solutions (roots) of the equation
$a$ , $b$ , and $c$ are the coefficients from the quadratic equation $ax^2 + bx + c = 0$ , where $a \neq 0$

The quadratic formula is particularly useful when factorisation is difficult or impossible, or when you need to find exact solutions quickly. The $\pm$ symbol indicates that there are typically two solutions: one using addition and one using subtraction.

Using the quadratic formula

To solve a quadratic equation using the formula:

Ensure the equation is in standard form $ax^2 + bx + c = 0$ .
Identify the coefficients $a$ , $b$ , and $c$ .
Substitute these values into the quadratic formula.
Simplify the expression, working carefully with the square root and arithmetic operations.
Write out both solutions separately, one using the positive value and one using the negative value.

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Common Mistakes to Avoid:

Forgetting to include the negative sign in front of $b$
Missing the $\pm$ symbol, which gives two solutions
Making sign errors when $b$ or $c$ are negative
Forgetting to multiply by 4 in the term $4ac$

Worked example: Solving with the quadratic formula

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Worked Example: Using the Quadratic Formula

Let's solve $-x^2 + 6x - 8 = 0$ using the quadratic formula.

First, we identify the coefficients: $a = -1$ , $b = 6$ , and $c = -8$ .

Step-by-step solution:

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

$x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-8)}}{2(-1)}$

$x = \frac{-6 \pm \sqrt{36 - 32}}{-2}$

$x = \frac{-6 \pm \sqrt{4}}{-2}$

$x = \frac{-6 \pm 2}{-2}$

For the positive value:

$x = \frac{-6 + 2}{-2} = \frac{-4}{-2} = 2$

For the negative value:

$x = \frac{-6 - 2}{-2} = \frac{-8}{-2} = 4$

Answer: The solutions are $x = 2$ and $x = 4$ .

Another worked example with surds

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Worked Example: Quadratic Formula with Surd Solutions

Let's solve $5x^2 = 8x + 1$ using the quadratic formula.

First, we rewrite in standard form: $5x^2 - 8x - 1 = 0$

The coefficients are $a = 5$ , $b = -8$ , and $c = -1$ .

Step-by-step solution:

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

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(-1)}}{2(5)}$

$x = \frac{8 \pm \sqrt{64 + 20}}{10}$

$x = \frac{8 \pm \sqrt{84}}{10}$

$x = \frac{8 \pm \sqrt{4 \times 21}}{10}$

$x = \frac{8 \pm 2\sqrt{21}}{10}$

We can factorise the numerator:

$x = \frac{2(4 \pm \sqrt{21})}{10}$

$x = \frac{4 \pm \sqrt{21}}{5}$

Answer: The solutions are $x = \frac{4 + \sqrt{21}}{5}$ and $x = \frac{4 - \sqrt{21}}{5}$ .

Remember!

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Key Points to Remember:

A quadratic equation has the form $ax^2 + bx + c = 0$ where $a \neq 0$ . The three main methods for solving are factorisation, completing the square, and using the quadratic formula.
When using factorisation, write the equation in the form $(x - \alpha)(x - \beta) = 0$ and apply the null factor law: if a product equals zero, at least one factor must equal zero.
Completing the square transforms a quadratic into the form $(x + k)^2 = d$ . For non-monic quadratics, divide through by $a$ first. Remember to add the square of half the $x$ -coefficient to both sides.
The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ works for any quadratic equation and is especially useful when factorisation is difficult.
Always check your solutions by substituting them back into the original equation to verify they satisfy the equation.

Solve Equations (HSC SSCE Mathematics Advanced): Revision Notes

Solve Quadratic Equations

What is a quadratic equation?

Solve by factorisation

The null factor law

Steps for solving by factorisation

Worked example: Factorising a monic quadratic

Worked example: Factorising a non-monic quadratic

Solve by completing the square

Understanding perfect square trinomials

Steps for completing the square

Worked example: Completing the square for a monic quadratic

Worked example: Completing the square for a non-monic quadratic

Solve by quadratic formula

The quadratic formula

Using the quadratic formula

Worked example: Solving with the quadratic formula

Another worked example with surds

Remember!

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