Welcome to Quadratic and Simultaneous Equations!
Welcome to one of the most exciting and powerful toolkits in algebra! In this chapter, you will learn how to unlock two major mathematical puzzles: Quadratic Equations (equations with an \(x^2\) term) and Simultaneous Equations (pairs of equations that work together at the same time).
Why is this important? Real life is rarely a straight line. Whether engineers are calculating the trajectory of a rocket, architects are designing curved bridges, or businesses are finding the exact point where profit is maximised, quadratics and simultaneous equations are the secret maths making it all work.
Don't worry if algebra has felt confusing in the past. We will take everything step-by-step, breaking each concept down into small, bite-sized pieces with plenty of worked examples.
Part 1: Understanding Quadratic Equations
What is a Quadratic Equation?
An equation is called quadratic if the highest power of the variable (usually \(x\)) is \(2\). This means it contains an \(x^2\) term, but no higher power like \(x^3\) or \(x^4\).
The standard form of a quadratic equation looks like this:
\(ax^2 + bx + c = 0\)
Here, \(a\), \(b\), and \(c\) are just normal numbers (called coefficients), and \(a\) cannot be \(0\):
• \(a\) is the number in front of \(x^2\).
• \(b\) is the number in front of \(x\).
• \(c\) is the constant number on its own.
• The equation must equal \(0\) before we can solve it!
Did you know? The word "quadratic" comes from the Latin word quadratus, meaning "square", because the variable is raised to the power of two (squared)!
Method 1: Solving Quadratics by Factorising
Factorising means putting an expression back into brackets. When two brackets multiply together to give zero, at least one of those brackets must equal zero. This is called the Zero Product Property.
If \((A) \times (B) = 0\), then either \(A = 0\) or \(B = 0\).
Type A: Simple Trinomials where \(a = 1\) (e.g., \(x^2 + bx + c = 0\))
To factorise \(x^2 + bx + c = 0\), we look for two numbers that:
1. Multiply to give \(c\)
2. Add to give \(b\)
Example 1: Solve \(x^2 + 7x + 12 = 0\)
• Step 1: Find two numbers that multiply to \(+12\) and add to \(+7\).
Let's check factor pairs of \(12\): \(1 \times 12\) (sum = \(13\)), \(2 \times 6\) (sum = \(8\)), \(3 \times 4\) (sum = \(7\)). The numbers are \(+3\) and \(+4\).
• Step 2: Put them into two brackets: \((x + 3)(x + 4) = 0\)
• Step 3: Set each bracket to zero and solve:
\(x + 3 = 0 \implies x = -3\)
\(x + 4 = 0 \implies x = -4\)
Answer: \(x = -3\) or \(x = -4\)
Type B: The Difference of Two Squares (DOTS)
Whenever you have two squared terms separated by a minus sign, it follows the pattern:
\(a^2 - b^2 = (a + b)(a - b)\)
Example 2: Solve \(x^2 - 25 = 0\)
• Notice that \(25 = 5^2\).
• Factorise: \((x + 5)(x - 5) = 0\)
• Solve: \(x = -5\) or \(x = 5\) (often written as \(x = \pm 5\)).
Type C: Quadratics with No Constant Term (\(c = 0\))
Example 3: Solve \(x^2 - 6x = 0\)
• Factorise out the common factor \(x\): \(x(x - 6) = 0\)
• Solve: \(x = 0\) or \(x - 6 = 0 \implies x = 6\)
Answer: \(x = 0\) or \(x = 6\)
Common Mistake to Avoid: Never divide both sides by \(x\) to get \(x = 6\)! If you do that, you lose the solution \(x = 0\).
Type D: Harder Quadratics where \(a > 1\)
Example 4: Solve \(2x^2 + 7x + 3 = 0\)
• Step 1: Multiply \(a\) and \(c\): \(2 \times 3 = 6\).
• Step 2: Find two numbers that multiply to \(6\) and add to \(7\). These are \(6\) and \(1\).
• Step 3: Split the middle term: \(2x^2 + 6x + 1x + 3 = 0\)
• Step 4: Factorise in pairs:
\(2x(x + 3) + 1(x + 3) = 0\)
• Step 5: Group the brackets: \((2x + 1)(x + 3) = 0\)
• Step 6: Solve each bracket:
\(2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}\)
\(x + 3 = 0 \implies x = -3\)
Answer: \(x = -\frac{1}{2}\) or \(x = -3\)
Key Takeaway for Factorising: Always rearrange your equation to equal zero first. Factorise into brackets, set each bracket equal to zero, and solve for \(x\).
Part 2: The Quadratic Formula
What happens when you cannot find neat whole numbers to factorise? Don't worry! There is a master formula that solves any quadratic equation, every single time.
The Quadratic Formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Memory Aid: You can sing the formula to the tune of "Pop Goes the Weasel":
"\(x\) is equal to negative \(b\), plus or minus the square root, of \(b\) squared minus \(4ac\), all over \(2a\)!"
Step-by-Step Guide to Using the Formula
Example: Solve \(2x^2 + 5x - 4 = 0\). Give your answers to 2 decimal places.
• Step 1: Identify \(a\), \(b\), and \(c\).
Here, \(a = 2\), \(b = 5\), and \(c = -4\) (always keep the negative sign with the number!).
• Step 2: Substitute carefully into the formula using brackets around negative values.
\(x = \frac{-(5) \pm \sqrt{(5)^2 - 4(2)(-4)}}{2(2)}\)
• Step 3: Simplify the terms inside the square root (the discriminant).
\((5)^2 = 25\)
\(-4 \times 2 \times (-4) = +32\)
So, \(b^2 - 4ac = 25 + 32 = 57\).
• Step 4: Write out the simplified form.
\(x = \frac{-5 \pm \sqrt{57}}{4}\)
• Step 5: Calculate both answers (one with \(+\), one with \(-\)).
Using the plus sign: \(x = \frac{-5 + \sqrt{57}}{4} = \frac{-5 + 7.5498...}{4} = 0.637... \approx 0.64\)
Using the minus sign: \(x = \frac{-5 - \sqrt{57}}{4} = \frac{-5 - 7.5498...}{4} = -3.137... \approx -3.14\)
Answer: \(x = 0.64\) or \(x = -3.14\) (to \(2\) d.p.)
Top Tip: Look out for questions that say "Give your answers to 2 decimal places" or "Give your answers to 3 significant figures". This is an immediate clue that factorising won't work easily, and you should use the Quadratic Formula!
Key Takeaway for the Formula: Be extremely cautious with negative signs, especially when calculating \(-4ac\) and \((-b)\). Putting brackets around numbers on your calculator prevents simple sign errors.
Part 3: Completing the Square
Completing the square is another way of rewriting a quadratic equation. It changes \(x^2 + bx + c\) into the form \((x + p)^2 + q\).
This method is super useful because:
1. It gives us another way to solve equations.
2. It reveals the exact turning point (minimum or maximum point) of a quadratic curve!
How to Complete the Square for \(x^2 + bx + c\)
The general rule is:
\(x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2\)
Simple 3-Step Recipe:
1. Halve the coefficient of \(x\) (this is \(\frac{b}{2}\)) and put it inside the bracket: \(\left(x + \frac{b}{2}\right)^2\).
2. Subtract the square of that number: \(-\left(\frac{b}{2}\right)^2\).
3. Add/subtract the original constant number \(c\).
Example: Solving by Completing the Square
Solve \(x^2 + 6x - 7 = 0\) by completing the square.
• Step 1: Look at \(b = 6\). Half of \(6\) is \(3\).
• Step 2: Write \((x + 3)^2 - 3^2 - 7 = 0\)
• Step 3: Simplify the numbers: \((x + 3)^2 - 9 - 7 = 0 \implies (x + 3)^2 - 16 = 0\)
• Step 4: Rearrange and take the square root of both sides:
\((x + 3)^2 = 16\)
\(x + 3 = \pm \sqrt{16}\)
\(x + 3 = \pm 4\)
• Step 5: Split into two equations:
\(x = -3 + 4 \implies x = 1\)
\(x = -3 - 4 \implies x = -7\)
Answer: \(x = 1\) or \(x = -7\)
Finding Turning Points
For a quadratic written in the form \(y = (x + p)^2 + q\):
• The minimum turning point is at the coordinates \((-p, q)\).
Notice that the sign inside the bracket flips, but the number outside stays the same!
Example: The turning point of \(y = (x - 4)^2 + 5\) is \((4, 5)\).
Key Takeaway for Completing the Square: Always halve the middle coefficient, subtract its square, and remember to flip the sign inside the bracket when finding the coordinates of the turning point.
Part 4: Linear Simultaneous Equations
Simultaneous equations are two (or more) equations containing two unknown letters (usually \(x\) and \(y\)) that must be true at the same time. Graphically, the solution is the point where the two lines intersect!
Method 1: Elimination Method
The goal is to eliminate one of the variables by adding or subtracting the two equations.
Memory Trick: SSS vs DSA
• Same Signs \(\implies\) Subtract the equations.
• Different Signs \(\implies\) Add the equations.
Example: Solve simultaneously:
(1) \(3x + 2y = 19\)
(2) \(x + 2y = 9\)
• Step 1: Check coefficients. The \(y\) terms both have \(+2y\) (Same Signs).
• Step 2: Subtract Equation (2) from Equation (1):
\((3x - x) + (2y - 2y) = 19 - 9\)
\(2x = 10 \implies x = 5\)
• Step 3: Substitute \(x = 5\) back into Equation (2) to find \(y\):
\(5 + 2y = 9\)
\(2y = 9 - 5\)
\(2y = 4 \implies y = 2\)
• Step 4: Check in Equation (1): \(3(5) + 2(2) = 15 + 4 = 19\) (Correct!).
Answer: \(x = 5, y = 2\)
What if the numbers don't match?
If neither the \(x\) nor \(y\) coefficients match, multiply one (or both) equations to create matching numbers.
Example:
(1) \(2x + 3y = 13\)
(2) \(5x - 2y = 4\)
• Multiply Equation (1) by \(2\): \(4x + 6y = 26\)
• Multiply Equation (2) by \(3\): \(15x - 6y = 12\)
• The \(y\) terms now have Different Signs (\(+6y\) and \(-6y\)), so we Add the equations:
\((4x + 15x) + (6y - 6y) = 26 + 12\)
\(19x = 38 \implies x = 2\)
• Substitute \(x = 2\) into Equation (1):
\(2(2) + 3y = 13 \implies 4 + 3y = 13 \implies 3y = 9 \implies y = 3\)
Answer: \(x = 2, y = 3\)
Method 2: Substitution Method
This is especially handy when one equation already has \(y =\) or \(x =\) on its own.
Example: Solve \(y = 2x - 1\) and \(3x + 2y = 12\).
• Substitute \((2x - 1)\) in place of \(y\) in the second equation:
\(3x + 2(2x - 1) = 12\)
\(3x + 4x - 2 = 12\)
\(7x - 2 = 12 \implies 7x = 14 \implies x = 2\)
• Now find \(y\): \(y = 2(2) - 1 = 3\)
Answer: \(x = 2, y = 3\)
Key Takeaway for Linear Simultaneous: Always align terms clearly. Decide whether to add or subtract to eliminate one letter first, solve for the remaining letter, then substitute back to find the second value.
Part 5: Simultaneous Equations (One Linear, One Quadratic)
In higher tier questions, you will often need to solve a system where one equation is linear (a straight line) and one equation is quadratic (a curve or circle). A line can cross a curve at up to two points, which means you will usually get two pairs of answers!
Standard Procedure: Always Use Substitution
1. Rearrange the linear equation to make either \(y\) or \(x\) the subject.
2. Substitute this expression into the quadratic equation.
3. Expand and simplify to form a single quadratic equation equal to \(0\).
4. Solve the quadratic (by factorising or formula) to get two values (e.g., \(x_1\) and \(x_2\)).
5. Substitute each value back into the linear equation to find the corresponding pair (e.g., \(y_1\) and \(y_2\)).
Full Worked Example
Solve the simultaneous equations:
(1) \(y - x = 3\)
(2) \(x^2 + y^2 = 29\)
• Step 1: Rearrange the linear equation.
From (1), add \(x\) to both sides: \(y = x + 3\)
• Step 2: Substitute into the quadratic equation.
Replace \(y\) with \((x + 3)\) in Equation (2):
\(x^2 + (x + 3)^2 = 29\)
• Step 3: Expand brackets and simplify.
Remember: \((x + 3)^2 = (x + 3)(x + 3) = x^2 + 6x + 9\)
\(x^2 + x^2 + 6x + 9 = 29\)
\(2x^2 + 6x + 9 - 29 = 0\)
\(2x^2 + 6x - 20 = 0\)
• Step 4: Solve the quadratic.
Divide the entire equation by \(2\) to make it simpler:
\(x^2 + 3x - 10 = 0\)
Factorise (two numbers multiplying to \(-10\) and adding to \(+3\) are \(+5\) and \(-2\)):
\((x + 5)(x - 2) = 0\)
So, \(x = -5\) or \(x = 2\).
• Step 5: Find the corresponding \(y\) values using \(y = x + 3\).
When \(x = -5\): \(y = -5 + 3 = -2\)
When \(x = 2\): \(y = 2 + 3 = 5\)
• Step 6: Write down the matched solution pairs clearly:
Pair 1: \(x = -5, y = -2\)
Pair 2: \(x = 2, y = 5\)
Common Mistake to Avoid: Do not mix and match the \(x\) and \(y\) values! The solution \(x = -5\) belongs exclusively with \(y = -2\). Writing them clearly as matched pairs or coordinates (e.g. \((-5, -2)\) and \((2, 5)\)) prevents any loss of marks.
Key Takeaway for Linear/Quadratic Systems: Substitute the linear equation into the quadratic. Solve for the two values of the first variable, then find both matching values for the second variable.
Quick Summary & Exam Checklist
Before entering the exam, make sure you can:
• Factorise standard quadratics (\(x^2 + bx + c\)), difference of two squares (\(x^2 - a^2\)), and quadratics with coefficients (\(ax^2 + bx + c\)).
• Quote and accurately calculate with the Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Complete the square into \((x + p)^2 + q\) and state the turning point \((-p, q)\).
• Solve linear simultaneous equations using Elimination (matching coefficients) or Substitution.
• Solve combined linear and quadratic simultaneous equations, ensuring you present two distinct pairs of solutions.