Introduction to Quadratic and Simultaneous Equations

Welcome to one of the most powerful chapters in Algebra! In this section, we move beyond simple "find \(x\)" problems and start looking at equations where \(x\) is squared (\(x^2\)) or where we have two different equations working at the same time. These skills are like the "Swiss Army Knife" of mathematics—they help us solve everything from the path of a ball thrown in the air to finding where two paths cross on a map.

Quick Review: Before we start, remember that an expression is just a collection of terms (like \(3x + 2\)), while an equation has an equals sign (like \(3x + 2 = 11\)). In this chapter, we are always looking for the value of the unknown letters that make the equation true.

1. Solving Quadratic Equations by Factorising

A quadratic equation usually looks like this: \(ax^2 + bx + c = 0\). The "squared" term (\(x^2\)) is what makes it a quadratic. For Foundation students, you will mostly see equations where \(a = 1\), such as \(x^2 + 5x + 6 = 0\).

How to Solve by Factorising (The "Multiply/Add" Method):

To solve \(x^2 + bx + c = 0\), we want to put it into two brackets: \((x + \text{something})(x + \text{something}) = 0\).

  1. Find two numbers that multiply to give the end number (\(c\)).
  2. Those same two numbers must add together to give the middle number (\(b\)).
  3. Put these numbers into the brackets.
  4. Set each bracket to zero to find your two solutions.

Example: Solve \(x^2 + 7x + 10 = 0\)
1. We need two numbers that multiply to give \(10\) and add to give \(7\).
2. Those numbers are \(2\) and \(5\) (because \(2 \times 5 = 10\) and \(2 + 5 = 7\)).
3. Put them in brackets: \((x + 2)(x + 5) = 0\).
4. For the answer to be zero, either \(x + 2 = 0\) (so \(x = -2\)) or \(x + 5 = 0\) (so \(x = -5\)).

Common Mistake: Don't forget to flip the sign at the end! If your bracket is \((x + 5)\), your solution for \(x\) is \(-5\).

Higher Tier Note (H): You may also need to factorise quadratics where the number in front of \(x^2\) is greater than 1, like \(2x^2 + 7x + 3 = 0\). You must also be able to recognize the Difference of Two Squares, such as \(x^2 - 16 = 0\), which factorises to \((x - 4)(x + 4) = 0\).

Key Takeaway: Factorising turns a difficult addition problem into a simple multiplication problem where one part must equal zero.

2. Solving Quadratics Graphically

You can find the solutions (roots) of a quadratic equation by looking at its graph. The solutions are simply the x-coordinates where the curve crosses the x-axis (where \(y = 0\)).

  • If the graph touches the x-axis at one point, there is one "repeated" solution.
  • If the graph crosses at two points, there are two solutions.
  • If it doesn't touch the x-axis at all, there are no real solutions.

3. Advanced Solving Methods (Higher Tier Only)

(H) Sometimes, a quadratic equation cannot be factorised easily because the answers are decimals. In these cases, we use two other methods:

A. The Quadratic Formula

This formula is provided on your Exam Aid (formula sheet), so you don't need to memorise it, but you must know how to use it! For \(ax^2 + bx + c = 0\):

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

Step-by-step:
1. Write down the values of \(a\), \(b\), and \(c\). (Include the minus signs if they are there!)
2. Plug them into the formula.
3. Use the \(\pm\) symbol to get two different answers—one where you add the square root and one where you subtract it.

B. Completing the Square

This involves rewriting the equation in the form \((x + p)^2 + q = 0\).
1. Take half of the middle number (\(b\)) and put it inside the bracket: \((x + \frac{b}{2})^2\).
2. Subtract the square of that same number: \(-( \frac{b}{2} )^2\).
3. Add your constant \(c\) and solve for \(x\).

Key Takeaway: The Formula and Completing the Square work for any quadratic, even if it doesn't factorise.

4. Linear Simultaneous Equations

Simultaneous equations are a pair of equations with two unknowns (usually \(x\) and \(y\)). To solve them, we need to find the values that work for both equations at the same time.

The Elimination Method:

  1. Make the number in front of one of the letters (either \(x\) or \(y\)) the same in both equations by multiplying the whole equation.
  2. Add or Subtract the equations to "eliminate" that letter.
    Rule: Same Signs Subtract (SSS), Different Signs Add.
  3. Solve the simple equation that is left.
  4. Substitute your answer back into one of the original equations to find the second letter.

Example:
\(2x + y = 7\)
\(3x - y = 8\)
Since the \(y\) terms have different signs (\(+y\) and \(-y\)), we Add the equations:
\((2x + 3x) + (y - y) = (7 + 8)\)
\(5x = 15 \implies x = 3\)
Now, put \(x = 3\) back into the first equation: \(2(3) + y = 7 \implies 6 + y = 7 \implies y = 1\).

Did you know? On a graph, the solution to simultaneous equations is the exact point where the two lines cross!

5. Linear and Quadratic Simultaneous Equations (Higher Tier Only)

(H) Sometimes you will have one linear equation (like \(y = x + 2\)) and one quadratic equation (like \(x^2 + y^2 = 10\)).

The best way to solve these is the Substitution Method:

  1. Rearrange the linear equation to get one letter on its own (e.g., \(y = ...\)).
  2. Replace that letter in the quadratic equation with your new expression. (Use brackets!)
  3. Expand the brackets and rearrange the equation to equal zero.
  4. Solve the resulting quadratic (by factorising or using the formula).
  5. Substitute your two \(x\) values back into the linear equation to find two matching \(y\) values.

Common Mistake: Students often find the two values for \(x\) but forget to find the two values for \(y\). You need pairs of coordinates as your final answer!

Summary & Quick Review

  • Quadratic Equations: Look for \(x^2\). Solve by factorising (Find factors of \(c\) that add to \(b\)).
  • Graphical Solutions: Look at where the curve crosses the x-axis.
  • (H) Quadratic Formula: Use \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) when you can't factorise.
  • Simultaneous Equations: Find where two lines meet. Use "Elimination" for two linear equations.
  • (H) Substitution: Use this when one equation is quadratic and the other is linear.
  • Problem Solving: Always read the context. If a question asks for the area of a rectangle in terms of \(x\), you might be setting up a quadratic equation!

Don't worry if this seems tricky at first! Algebra is a language; the more you "speak" it by practicing these methods, the more natural it will feel.