Algebra Revision Notes

What is a Quadratic Equation?

A quadratic equation is any equation that can be rearranged into the standard form ax² + bx + c = 0, where 'x' is the variable, 'a', 'b', and 'c' are constant numbers, and 'a' cannot be 0. The defining feature is the x² term. If 'a' were 0, it would be a linear equation. Examples include x² + 3x + 2 = 0 or 2x² - 5 = 0. The solutions to a quadratic equation are called its roots.

Solving by Factorisation

One common method for solving quadratic equations on the foundation tier is factorisation. If an equation is in the form x² + bx + c = 0, you need to find two numbers that multiply to 'c' and add to 'b'. For example, to solve x² + 7x + 10 = 0: find two numbers that multiply to 10 and add to 7 (which are 2 and 5). This factors into (x + 2)(x + 5) = 0. For the product of two terms to be zero, at least one term must be zero. So, x + 2 = 0 or x + 5 = 0. This gives the solutions x = -2 or x = -5. This method is effective when the quadratic can be easily factored.

Solving by Rearrangement and Square Roots

Some quadratic equations, especially those without a 'bx' term (i.e., in the form ax² + c = 0 or simply x² = k), can be solved by rearranging and taking square roots. For example, to solve x² - 16 = 0: Add 16 to both sides to get x² = 16. Then, take the square root of both sides. Remember that a number has both a positive and a negative square root. So, x = √16 or x = -√16, which means x = 4 or x = -4. Always consider both positive and negative roots when taking a square root during solving.

Checking Your Solutions

After finding potential solutions (roots) for a quadratic equation, it's good practice to substitute each solution back into the original equation to verify your answer. If both sides of the equation balance (e.g., both equal 0), then your solution is correct. This helps catch any algebraic errors. For example, if you solved x² + 5x + 6 = 0 and found x = -2, substitute: (-2)² + 5(-2) + 6 = 4 - 10 + 6 = 0. This confirms x = -2 is a correct root.

Completing the Square

Completing the square is a technique used to rewrite a quadratic expression of the form $ax^2 + bx + c$ into the vertex form $a(x+p)^2 + q$. To solve $ax^2 + bx + c = 0$ by completing the square, first divide by $a$ (if $a \neq 1$) to get $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$. Then, move the constant term to the right side: $x^2 + \frac{b}{a}x = -\frac{c}{a}$. Take half of the coefficient of $x$, square it, and add it to both sides: $x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 = -\frac{c}{a} + (\frac{b}{2a})^2$. This creates a perfect square trinomial on the left: $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$. Finally, take the square root of both sides and solve for $x$ by rearranging: $x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$, leading to $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$.

The Quadratic Formula

The quadratic formula provides a direct way to find the roots of any quadratic equation $ax^2 + bx + c = 0$. The formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. It is derived directly from the completing the square method. It's essential to correctly identify the values of $a$, $b$, and $c$ from the equation when it's in the standard form. Remember that $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$, and $c$ is the constant term. Pay close attention to signs. For example, in $3x^2 - 5x + 2 = 0$, $a=3$, $b=-5$, $c=2$. If the equation is not in standard form, rearrange it first, e.g., $x^2 = 4x - 1$ becomes $x^2 - 4x + 1 = 0$ before applying the formula.

The Discriminant

The discriminant, denoted by $\Delta$ (delta) or simply $b^2 - 4ac$, is a critical part of the quadratic formula. It determines the nature of the roots of a quadratic equation without actually solving for them. There are three cases:

1. If $b^2 - 4ac > 0$, the equation has two distinct real roots. The parabola intersects the x-axis at two different points.
2. If $b^2 - 4ac = 0$, the equation has exactly one real root (a repeated root). The parabola touches the x-axis at exactly one point, which is its vertex.
3. If $b^2 - 4ac < 0$, the equation has no real roots. In this case, the roots are two complex conjugates. The parabola does not intersect the x-axis.

The Elimination Method

The elimination method involves manipulating the equations so that when they are added or subtracted, one of the variables is 'eliminated'. This usually means making the coefficients (the numbers in front of the variables) of one variable identical or additive inverses within both equations.

Steps:

1. Multiply one or both equations by a constant so that the coefficients of one variable are the same (or opposite in sign).
2. If the coefficients are the same, subtract one equation from the other. If they are opposite, add the equations.
3. Solve the resulting single-variable equation.
4. Substitute the value found back into one of the original equations to find the value of the other variable.
5. Check your solution by substituting both values into the *other* original equation.

The Substitution Method

The substitution method involves expressing one variable in terms of the other from one equation, and then substituting this expression into the second equation. This reduces the number of variables in the second equation to one, making it solvable.

Steps:

1. Rearrange one of the equations to make one variable the subject (e.g., isolate 'y' in terms of 'x', or 'x' in terms of 'y'). This is often easiest when a variable has a coefficient of 1 or -1.
2. Substitute this expression for the variable into the *other* equation.
3. Solve the resulting single-variable equation.
4. Substitute the value found back into the rearranged equation from step 1 (or any original equation) to find the value of the second variable.
5. Check your solution by substituting both values into the *other* original equation.

Checking Your Solution

After finding values for both variables (e.g., x and y), it's essential to check if they satisfy *both* original equations. If they satisfy only one, your solution is incorrect.

Example: Equations:

1. x + y = 7
2. x - y = 3

Solution found: x=5, y=2

Check: For equation 1: 5 + 2 = 7 (True) For equation 2: 5 - 2 = 3 (True)

Since both equations are satisfied, the solution (x=5, y=2) is correct. This step helps catch arithmetic errors.

Understanding Linear and Non-Linear Systems

When solving simultaneous equations involving one linear equation and one non-linear equation, the goal is to find the points where their graphs intersect. A linear equation represents a straight line. A non-linear equation can represent various shapes, such as a parabola ($y=x^2$), a circle ($x^2+y^2=r^2$), an ellipse, or a hyperbola. Unlike two linear equations, which usually have a single solution, a linear and non-linear system can have two solutions, one solution (if the line is tangent to the curve), or no real solutions (if they do not intersect). Visualising the shapes can help anticipate the number of possible solutions.

The Substitution Method: Step-by-Step

The most effective and common method for these types of simultaneous equations is substitution. Follow these steps meticulously:

1. Isolate a variable in the *linear* equation. For example, from $x + y = 5$, you could get $y = 5 - x$ or $x = 5 - y$. Choose the easiest one to avoid fractions.
2. Substitute this expression into the *non-linear* equation. This will eliminate one variable, leaving you with a single equation in a single variable, usually a quadratic one.
3. Solve the resulting quadratic equation. This might require factoring, using the quadratic formula ($x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$), or completing the square. Remember to consider both positive and negative roots if taking square roots.
4. Find corresponding values for the other variable. For each solution found in step 3, substitute it back into the *rearranged linear equation* from step 1 to find the corresponding value of the other variable. Always use the linear equation for this step to simplify calculations and avoid extraneous solutions.

Presenting and Verifying Solutions

Your solutions should be presented as coordinate pairs $(x, y)$, as they represent points of intersection on a graph. For example, if you find $x=1$ and $y=4$, one solution is $(1,4)$. If you find $x=4$ and $y=1$, another solution is $(4,1)$. It's crucial to substitute *both* $x$ and $y$ values from each solution pair into *both original equations* to verify that they satisfy both. This helps catch any algebraic errors and ensures the solution is valid for the complete system. If a solution only works in one equation, it is incorrect.