Edexcel Mathematics (4XMAH) -Unit 1 - 2.7 Quadratic Equations- Study Notes- New Syllabus
Edexcel Mathematics (4XMAH) -Unit 1 – 2.7 Quadratic Equations- Study Notes- New syllabus
Edexcel Mathematics (4XMAH) -Unit 1 – 2.7 Quadratic Equations- Study Notes -Edexcel iGCSE Mathematics – per latest Syllabus.
Key Concepts:
A solve quadratic equations by factorisation
2x² − 3x + 1 = 0
x(3x − 2) = 5
B solve quadratic equations by using the quadratic formula or completing the square
C form and solve quadratic equations from data given in a context
Edexcel iGCSE Mathematics -Concise Summary Notes- All Topics
Solving Quadratic Equations by Factorisation
A quadratic equation is an equation that contains \( x^2 \) as the highest power.
\( ax^2+bx+c=0 \)
To solve it by factorisation, we write the quadratic as two brackets and use the zero product rule.
If \( AB=0 \), then \( A=0 \) or \( B=0 \).
Steps
1. Rearrange the equation so one side equals 0.
2. Factorise the quadratic.
3. Set each bracket equal to 0.
4. Solve for \( x \).
Example 1:
Solve \( 2x^2-3x+1=0 \).
▶️ Answer/Explanation
Factorise:
\( (2x-1)(x-1)=0 \)
Set each bracket to zero:
\( 2x-1=0 \Rightarrow x=\dfrac{1}{2} \)
\( x-1=0 \Rightarrow x=1 \)
Conclusion: \( x=\dfrac{1}{2},\;1 \).
Example 2:
Solve \( x(3x-2)=5 \).
▶️ Answer/Explanation
First rearrange:
\( 3x^2-2x-5=0 \)
Factorise:
\( (3x-5)(x+1)=0 \)
Solve:
\( 3x-5=0 \Rightarrow x=\dfrac{5}{3} \)
\( x+1=0 \Rightarrow x=-1 \)
Conclusion: \( x=\dfrac{5}{3},\;-1 \).
Example 3:
Solve \( x^2-9=0 \).
▶️ Answer/Explanation
Difference of squares:
\( (x-3)(x+3)=0 \)
\( x=3 \) or \( x=-3 \)
Conclusion: \( x=3,\;-3 \).
Solving Quadratic Equations Using the Quadratic Formula and Completing the Square
Some quadratic equations cannot be factorised easily. In these cases we solve them using either the quadratic formula or completing the square.
The Quadratic Formula
For the equation
\( ax^2+bx+c=0 \)
the solutions are
\( x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \)
The value inside the square root \( b^2-4ac \) is called the discriminant.
Completing the Square
We rewrite the quadratic into a squared bracket and then solve.
\( (x+b)^2=k \Rightarrow x=-b\pm\sqrt{k} \)
Example 1:
Solve \( x^2-5x+6=0 \).
▶️ Answer/Explanation
Here \( a=1,\;b=-5,\;c=6 \).
\( x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(6)}}{2(1)} \)
\( x=\dfrac{5\pm\sqrt{25-24}}{2} \)
\( x=\dfrac{5\pm1}{2} \)
\( x=3 \) or \( x=2 \)
Conclusion: \( x=2,\;3 \).
Example 2 :
Solve \( 2x^2+3x-2=0 \).
▶️ Answer/Explanation
\( a=2,\;b=3,\;c=-2 \)
\( x=\dfrac{-3\pm\sqrt{3^2-4(2)(-2)}}{4} \)
\( x=\dfrac{-3\pm\sqrt{9+16}}{4} \)
\( x=\dfrac{-3\pm5}{4} \)
\( x=\dfrac{1}{2} \) or \( x=-2 \)
Conclusion: \( x=\dfrac{1}{2},\;-2 \).
Example 3:
Solve \( x^2+4x+1=0 \) by completing the square.
▶️ Answer/Explanation
Complete the square:
\( x^2+4x+1=0 \)
\( (x+2)^2-4+1=0 \)
\( (x+2)^2-3=0 \)
\( (x+2)^2=3 \)
\( x+2=\pm\sqrt{3} \)
\( x=-2\pm\sqrt{3} \)
Conclusion: \( x=-2\pm\sqrt{3} \).
Forming and Solving Quadratic Equations from Context
In many problems, the quadratic equation is not given directly. We must first form an equation from the information and then solve it.
Typical Situations
- Area problems
- Number problems
- Geometry problems
Steps
1. Let the unknown be \( x \).
2. Translate the words into an algebraic equation.
3. Rearrange to form a quadratic equation.
4. Solve the equation.
5. Check the answer makes sense in context.
Example 1:
The area of a rectangle is \( 48 \text{ cm}^2 \). Its length is \( x+4 \) and width is \( x \). Find \( x \).
▶️ Answer/Explanation
Area \( =\text{length} \times \text{width} \)
\( x(x+4)=48 \)
\( x^2+4x-48=0 \)
\( (x+8)(x-6)=0 \)
\( x=-8 \) or \( x=6 \)
Length cannot be negative.
Conclusion: \( x=6 \).
Example 2:
The product of two consecutive integers is 72. Find the integers.
▶️ Answer/Explanation
Let first integer \( =x \)
Next integer \( =x+1 \)
\( x(x+1)=72 \)
\( x^2+x-72=0 \)
\( (x+9)(x-8)=0 \)
\( x=-9 \) or \( x=8 \)
So the integers are \( 8 \) and \( 9 \), or \( -9 \) and \( -8 \).
Conclusion: \( (8,9) \) or \( (-9,-8) \).
Example 3:
A square has area \( 121\text{ cm}^2 \). Find the side length.
▶️ Answer/Explanation
Let side \( =x \)
Area of square \( =x^2 \)
\( x^2=121 \)
\( x=11 \) (length must be positive)
Conclusion: \( 11\text{ cm} \).