Solving Quadratic Equations
From the curve of a bridge to the path of a football, quadratic equations describe the world around us. This guide will show you how to solve these important equations by factorising and by interpreting their graphs.
The Core Concepts
Solving by Factorising
To solve a quadratic equation like $x^2 + bx + c = 0$, we first factorise it into two brackets: $(x+p)(x+q)=0$.
We then use the Zero Product Property, which states that if two things multiply to make zero, one of them must be zero.
then either $x+p = 0$ or $x+q = 0$.
Solving from a Graph
The graph of a quadratic equation is a U-shaped curve called a parabola. The solutions to the equation $ax^2+bx+c=0$ are the points where the graph crosses the x-axis. These points are called the roots or x-intercepts.
Worked Examples
Example 1: Solving with Factorising
Solve $x^2 + 7x + 10 = 0$.
- Find two numbers that multiply to 10 and add to 7. The numbers are 2 and 5.
- Write in factorised form:
$(x+2)(x+5) = 0$. - Apply the Zero Product Property:
Either $x+2=0$ or $x+5=0$.
Solutions: $x = -2$ and $x = -5$
Example 2: With Negative Numbers
Solve $x^2 – 3x – 18 = 0$.
- Find two numbers that multiply to -18 and add to -3. The numbers are 3 and -6.
- Write in factorised form:
$(x+3)(x-6) = 0$. - Apply the Zero Product Property:
Either $x+3=0$ or $x-6=0$.
Solutions: $x = -3$ and $x = 6$
Tutor Insights
🤔 Common Misunderstandings
- Forgetting the “= 0” part: Factorising an expression is only the first step. You must set each factor to zero to find the solutions.
- Sign errors: Getting mixed up with positive and negative numbers when finding the factor pairs.
📝 Common Exam Mistakes
- Leaving the answer in factorised form instead of giving the final values for $x$.
- Only finding one solution when there are often two.
- Algebraic errors when rearranging an equation to make it equal to zero before starting.
Practice Questions
Solve the following quadratic equations by factorising.
- $x^2 + 9x + 14 = 0$
- $x^2 – 3x – 10 = 0$
- $x^2 – 11x + 28 = 0$
- $x^2 – 49 = 0$ (Hint: Difference of two squares)
- $x^2 + 5x = -4$ (Hint: Rearrange first!)
Show Answers
- Working: $(x+2)(x+7)=0$.
Solutions: $x=-2, x=-7$. - Working: $(x+2)(x-5)=0$.
Solutions: $x=-2, x=5$. - Working: $(x-4)(x-7)=0$.
Solutions: $x=4, x=7$. - Working: $(x-7)(x+7)=0$.
Solutions: $x=7, x=-7$. - Working: Rearrange to $x^2+5x+4=0$, which is $(x+1)(x+4)=0$.
Solutions: $x=-1, x=-4$.
FAQs
Can a quadratic equation have only one solution?
Yes! This happens if the quadratic is a perfect square, like $(x-3)^2 = 0$. The only solution is $x=3$. On a graph, this means the parabola just touches the x-axis at one point.
What if I can’t factorise the quadratic?
Not all quadratics factorise neatly. If you’re stuck, you’ll need to use other methods like the Quadratic Formula or Completing the Square, which are key skills for the Higher Tier paper.
