Edexcel Mathematics (4XMAF) -Unit 1 - 2.7 Quadratic Equations- Study Notes- New Syllabus
Edexcel Mathematics (4XMAF) -Unit 1 – 2.7 Quadratic Equations- Study Notes- New syllabus
Edexcel Mathematics (4XMAF) -Unit 1 – 2.7 Quadratic Equations- Study Notes -Edexcel iGCSE Mathematics – per latest Syllabus.
Key Concepts:
A solve quadratic equations by factorisation (limited to x² + bx + c = 0)
Solve:
x² + x − 30 = 0
Edexcel iGCSE Mathematics -Concise Summary Notes- All Topics
Solving Quadratic Equations by Factorisation
A quadratic equation contains \( x^2 \) as the highest power.
General form: \( x^2 + bx + c = 0 \)
We solve it by factorising and then finding the values of \( x \).
Key Idea
If \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
So after factorising, we set each bracket equal to zero.
Method
1. Find two numbers that multiply to \( c \).
2. Their sum must equal \( b \).
3. Write the expression as two brackets.
4. Set each bracket equal to zero.
Example Idea
Solve \( x^2 + x – 30 = 0 \)
We need two numbers that multiply to \( -30 \) and add to \( 1 \).
\( 6 \) and \( -5 \)
\( (x+6)(x-5)=0 \)
\( x+6=0 \Rightarrow x=-6 \)
\( x-5=0 \Rightarrow x=5 \)
Example 1:
Solve \( x^2 + 7x + 12 = 0 \).
▶️ Answer/Explanation
Numbers multiply to \( 12 \) and add to \( 7 \): \( 3,\;4 \)
\( (x+3)(x+4)=0 \)
\( x=-3,\;x=-4 \)
Conclusion: \( x=-3,\,-4 \).
Example 2:
Solve \( x^2 – 2x – 15 = 0 \).
▶️ Answer/Explanation
Multiply to \( -15 \), add to \( -2 \): \( -5,\;3 \)
\( (x-5)(x+3)=0 \)
\( x=5,\;x=-3 \)
Conclusion: \( x=5,\,-3 \).
Example 3:
Solve \( x^2 + x – 30 = 0 \).
▶️ Answer/Explanation
Multiply to \( -30 \), add to \( 1 \): \( 6,\;-5 \)
\( (x+6)(x-5)=0 \)
\( x=-6 \) or \( x=5 \)
Conclusion: \( x=-6 \) or \( x=5 \).