A quadratic equation contains a squared term, like x². Many can be solved by factorising into two brackets, then using the fact that if two things multiply to zero, at least one of them must be zero. For x² + bx + c, find two numbers that multiply to c and add to b — these become the numbers in your brackets.
Example
Solve x² + 5x + 6 = 0: find two numbers that multiply to 6 and add to 5 — that's 2 and 3. Factorised: (x + 2)(x + 3) = 0. So x = −2 or x = −3.
Key terms
Quadratic:
An equation containing a squared term (like x²).
Factorise:
Rewrite an expression as a product of brackets.
Null factor law:
If two factors multiply to zero, at least one of them must be zero.
Questions
1. A quadratic equation contains a term with:
A squared variable (x²)
No variables at all
Only fractions
Only negative numbers
2. Factorising an equation means:
Rewriting it as a product of brackets
Adding more terms to it
Removing all numbers
Making it longer
3. The null factor law states that if two factors multiply to zero:
At least one of them must be zero
Both must be positive
Neither can be zero
They must be equal
4. Solve: (x − 2)(x − 3) = 0
x = 2 or x = 3
x = −2 or x = −3
x = 5 only
x = 6 only
5. Solve: (x + 1)(x + 4) = 0
x = −1 or x = −4
x = 1 or x = 4
x = 5 only
x = −5 only
6. To factorise x² + bx + c, you look for two numbers that:
Multiply to c and add to b
Add to c and multiply to b
Are both equal to b
Are both equal to c
7. Solve: x(x − 5) = 0
x = 0 or x = 5
x = 5 only
x = 0 only
x has no solutions
8. Solve: x² + 5x + 6 = 0
x = −2 or x = −3
x = 2 or x = 3
x = 5 or x = 6
x = −5 or x = −6
9. Solve: x² − 7x + 12 = 0
x = 3 or x = 4
x = −3 or x = −4
x = 7 or x = 12
x = 12 only
10. Solve: x² + 3x − 10 = 0
x = 2 or x = −5
x = −2 or x = 5
x = 3 or x = −10
x = 10 only
11. Solve: x² − 9 = 0
x = 3 or x = −3
x = 9 only
x = 3 only
x = 81
12. Solve: x² − x − 6 = 0
x = 3 or x = −2
x = −3 or x = 2
x = 6 or x = 1
x = −6 only
13. Solve: x² + 8x + 15 = 0
x = −3 or x = −5
x = 3 or x = 5
x = 8 or x = 15
x = −8 only
14. Solve: x² − 4x = 0
x = 0 or x = 4
x = 4 only
x = 0 only
x = −4 only
15. A rectangle has length (x + 3) and width x, with an area of 40. Which equation represents this?
x² + 3x − 40 = 0
x² + 3x + 40 = 0
x² − 3x = 40
3x = 40
16. Solve: 2x² + 6x = 0
x = 0 or x = −3
x = 0 or x = 3
x = 6 only
x = −6 only
17. Solve: x² − 2x − 15 = 0
x = 5 or x = −3
x = −5 or x = 3
x = 2 or x = 15
x = 15 only
18. A ball's height is modelled by h = −(t² − 6t), where h = 0 at the start and end. At what times (t) is h = 0?
t = 0 or t = 6
t = 6 only
t = 0 only
t = 3 only
19. Solve: x² + 2x − 24 = 0
x = 4 or x = −6
x = −4 or x = 6
x = 2 or x = 24
x = 24 only
20. A garden bed's area (x + 5)(x − 2) = 0 represents a length and width. Which value of x gives a valid (positive) width?
x = 2, since the other solution gives a negative width
x = −5, since it is also a valid width
Both solutions give valid widths
Neither solution is valid
21. Solve: x² − 6x + 9 = 0
x = 3 (a repeated solution)
x = 3 or x = −3
x = 9 only
x = 6 only
Answer key (parent copy)
1. A squared variable (x²)
2. Rewriting it as a product of brackets
3. At least one of them must be zero
4. x = 2 or x = 3
5. x = −1 or x = −4
6. Multiply to c and add to b
7. x = 0 or x = 5
8. x = −2 or x = −3
9. x = 3 or x = 4
10. x = 2 or x = −5
11. x = 3 or x = −3
12. x = 3 or x = −2
13. x = −3 or x = −5
14. x = 0 or x = 4
15. x² + 3x − 40 = 0
16. x = 0 or x = −3
17. x = 5 or x = −3
18. t = 0 or t = 6
19. x = 4 or x = −6
20. x = 2, since the other solution gives a negative width