Algebraic Expressions and Index Laws

Ever wondered how mathematicians simplify complex expressions? It all starts with understanding index laws. These powerful rules help you manipulate expressions containing powers.

The key index laws you need to remember are:

• When multiplying powers with the same base, add the indices: a^m × a^n = a^$m+n$
• When dividing, subtract the indices: a^m ÷ a^n = a^$m-n$
• For powers of powers, multiply the indices: $a^m$^n = a^(mn)
• For products raised to a power: (ab)^n = a^n b^n
• For negative indices: a^$-m$ = 1/a^m
• Any number to power zero equals 1: a^0 = 1

When expanding expressions with brackets, distribute each term. For example: -3x$7x - 4$ = -21x^2 - $-12x$ = -21x^2 + 12x

Pro Tip: When factorising, look for the highest common factor (HCF) first. For expressions like 3x + 9, pull out the common factor 3 to get 3$x + 3$.

For the difference of two squares, remember this pattern: a^2 - b^2 = $a+b$$a-b$. This transforms expressions like 4x^2 - 9y^2 into $2x+3y$$2x-3y$.

Working with Indices and Surds

Negative and fractional indices might look scary, but they follow simple rules that you can master. These skills are essential for handling complex algebraic problems.

With fractional indices, remember that a^$m/n$ means "the nth root of a^m". For example, x^(1/2) means √x and x^(1/3) means ∛x. When you see x^(-3), this equals 1/x^3, following our negative index rule.

Surds are irrational numbers expressed using root symbols. Key rules include:

• √(ab) = √a × √b
• √a/√b = √$a/b$

When simplifying surds, look for perfect square factors. For example, √12 = √(4 × 3) = √4 × √3 = 2√3.

Remember: When multiplying expressions with surds, treat them like algebraic terms. For instance, √2(5-√3) = 5√2 - √6.

Rationalising denominators is a technique to remove surds from the denominator of a fraction. For √3 in the denominator, multiply both numerator and denominator by √3 to get (√3)/3. For expressions like 1/(√5+√2), multiply by (√5-√2)/(√5-√2) to eliminate the surd in the denominator.

Solving Quadratic Equations

Quadratic equations appear everywhere in maths and science. Being able to solve them quickly gives you a major advantage on exams.

The three main methods for solving quadratic equations are:

1. Factorising: For equations like x² - 2x - 15 = 0, find factors of -15 that add up to -2 $which are -5 and 3$. This gives us $x-5$$x+3$ = 0, so x = 5 or x = -3.

2. Using the quadratic formula: For ax² + bx + c = 0, the solution is: x = $-b ± √(b² - 4ac)$/2a This works for any quadratic, even those that don't factorise nicely.

3. Completing the square: Rewrite the quadratic in the form $x+p$² + q. For example, x² + 8x can be rewritten as $x+4$² - 16.

Exam tip: When the question asks for the "roots of the function," it means to find the values of x where f(x) = 0.

Functions are mathematical relationships that map inputs to outputs. For a function f(x), the notation f(5) means "the value of the function when x = 5". When you see f(x) = g(x), you're looking for values where two different functions have the same output.

Understanding Quadratic Graphs

Quadratic graphs are parabolas that help us visualise solutions to quadratic equations. They're absolutely essential for understanding function behaviour.

The standard form of a quadratic function is f(x) = ax² + bx + c. The shape of the graph depends on a:

• If a > 0, the parabola opens upward (∪)
• If a < 0, the parabola opens downward (∩)

Key points on a quadratic graph include:

• The y-intercept $where the graph crosses the y-axis$: (0, c)
• The x-intercepts $where the graph crosses the x-axis$: solutions to ax² + bx + c = 0
• The turning point (minimum or maximum point of the graph)

Quick trick: Complete the square to find the turning point easily! For y = x² - 5x + 4, rewrite as y = $x - 5/2$² - 9/4, so the turning point is at (5/2, -9/4).

When analysing a quadratic graph, identify whether it has a minimum or maximum value. For y = 4x - 2x² - 3, the coefficient of x² is negative, so it's a downward-facing parabola with a maximum point. Completing the square gives y = -2$x - 1$² - 1, so the maximum point is at (1, -1).

The Discriminant and Quadratic Modelling

The discriminant is a powerful tool that quickly tells you the nature of a quadratic equation's solutions without having to solve it completely.

For a quadratic equation ax² + bx + c = 0, the discriminant is b² - 4ac:

• If b² - 4ac > 0: Two distinct real roots $the parabola crosses the x-axis twice$
• If b² - 4ac = 0: One repeated root $the parabola touches the x-axis at exactly one point$
• If b² - 4ac < 0: No real roots $the parabola doesn't cross the x-axis at all$

For example, to find values of k where x² + kx + 9 = 0 has exactly one solution, we set the discriminant equal to zero: k² - 36 = 0, giving k = ±6.

Application alert: Quadratics are brilliant for modelling real-world situations like projectile motion!

In modelling problems, completing the square helps identify maximum height and flight time. For a function like h(t) = 12.25 + 14.7t - 4.9t², rewrite it as h(t) = 23.275 - 4.9$t - 1.5$², which tells us the maximum height is 23.275 units, occurring at t = 1.5 seconds. To find when the object hits the ground, solve h(t) = 0 using the quadratic formula.

Linear and Quadratic Simultaneous Equations

Simultaneous equations help us find values that satisfy multiple conditions at once. They're incredibly useful in everything from physics to economics.

For linear simultaneous equations like: 2x + 3y = 8 3x - y = 23

The elimination method works brilliantly. Multiply the second equation by 3 to get 9x - 3y = 69, then add this to the first equation to eliminate y: 2x + 3y = 8 9x - 3y = 69 11x = 77 → x = 7

Substitute back to find y = -2.

Quadratic simultaneous equations involve at least one quadratic equation. The key strategy is to substitute from the linear equation into the quadratic one.

For example, with: x + 2y = 3 x² + 3xy = 10

Rearrange the first equation to get x = 3 - 2y, then substitute this into the second: $3 - 2y$² + 3y$3 - 2y$ = 10

Problem-solving tip: Always check your solutions by substituting back into both original equations to verify they work!

Expanding and simplifying gives 2y² + 3y + 1 = 0, which factorises to $2y + 1$$y + 1$ = 0, giving y = -1/2 or y = -1, and corresponding x-values of 4 and 5.

Graphical Solutions to Simultaneous Equations

Graphs give us visual insights into solutions that algebraic methods sometimes hide. They're especially valuable for understanding the relationship between equations.

When solving simultaneous equations graphically:

• The solution points are where the graphs intersect
• Linear and quadratic equation pairs can have 0, 1, or 2 intersection points
• The number of solutions relates directly to the discriminant of the resulting quadratic

For example, when a line y = 2x + 1 intersects with a quadratic curve kx² + 2y + $k-2$ = 0, we can determine the number of solutions by analysing the discriminant of the resulting equation kx² + 4x + k = 0.

Visual insight: The discriminant b² - 4ac determines not just the number of solutions algebraically, but also how the graphs intersect visually!

For this particular example, setting 16 - 4k² = 0 gives k = ±2. When k = 2, the line is tangent to the quadratic curve, giving exactly one solution. For other values of k, there will be either two solutions (the line cuts the curve twice) or no solutions (the line doesn't intersect the curve).

Inequalities - Linear and Quadratic

Inequalities describe ranges of values rather than exact solutions. They're crucial for understanding boundaries and constraints in real-world problems.

For linear inequalities like 5x + 9 > x + 20:

1. Rearrange to get all terms with x on one side: 4x > 11
2. Divide both sides by the coefficient of x (being careful about the direction of the inequality if dividing by a negative): x > 2.75
3. Represent the solution on a number line using open or closed circles (○ for < or >, ● for ≤ or ≥)

When working with multiple inequalities, find the intersection or union of the individual solution sets.

Direction matters: When multiplying or dividing by a negative number, the inequality sign flips direction (> becomes < and vice versa)!

Quadratic inequalities like 3 - 5x - 2x² < 0 require a slightly different approach:

1. Rearrange to standard form: 2x² + 5x - 3 > 0
2. Find the roots of the corresponding equation: $2x - 1$$x + 3$ = 0 gives x = 1/2 and x = -3
3. Test regions between and outside these values to determine where the inequality is satisfied
4. In this case, the solution is x < -3 or x > 1/2

The sign of the coefficient of x² determines whether the parabola opens upward or downward, which affects the solution regions.