Quadratic Equations · Answer guide

Lesson 1 · Build and recognise a quadratic equation

Convert a product situation into ax² + bx + c = 0.

1. A rectangle has width x and length x + 2. Its area is 3. Which equation models it?

   Answer: x² + 2x − 3 = 0. Area is width × length: x(x + 2) = 3. Expanding and moving the area to the left gives x² + 2x − 3 = 0.

2. A rectangle has width x and length x + 2. Its area is 8. Which equation models it?

   Answer: x² + 2x − 8 = 0. Area is width × length: x(x + 2) = 8. Expanding and moving the area to the left gives x² + 2x − 8 = 0.

3. A rectangle has width x and length x + 2. Its area is 15. Which equation models it?

   Answer: x² + 2x − 15 = 0. Area is width × length: x(x + 2) = 15. Expanding and moving the area to the left gives x² + 2x − 15 = 0.

Discuss: Model two consecutive positive integers whose product is 56, then identify the positive pair.

Teaching note: A quadratic requires a nonzero x² coefficient after simplification. A negative algebraic root is not automatically a valid physical length. Open responses need discussion; they are not automatically marked.

Lesson 2 · Solve by factorisation

Use the zero-product rule to retain both roots.

1. Solve x² − 7x + 10 = 0.

   Answer: x = 2 or 5. Factor as (x − 2)(x − 5) = 0. Either factor can be zero, giving x = 2 or 5.

2. Solve x² − 7x + 12 = 0.

   Answer: x = 3 or 4. Factor as (x − 3)(x − 4) = 0. Either factor can be zero, giving x = 3 or 4.

3. Solve x² − 7x + 6 = 0.

   Answer: x = 1 or 6. Factor as (x − 1)(x − 6) = 0. Either factor can be zero, giving x = 1 or 6.

Discuss: Make a quadratic whose roots are 0 and −4. Explain why dividing by x would lose the zero root.

Teaching note: Do not divide by x when x might be zero; doing so can lose a root. For x(x − 5) = 0, both 0 and 5 are roots. Open responses need discussion; they are not automatically marked.

Lesson 3 · Use the discriminant and formula

Decide the real-root type and calculate roots when D is nonnegative.

1. What is the nature of the roots of x² + (-6)x + (5) = 0?

   Answer: Two distinct real roots. D = b² − 4ac = 36 − 20 = 16. Its sign gives the stated root type.

2. What is the nature of the roots of x² + (-4)x + (4) = 0?

   Answer: Two equal real roots. D = b² − 4ac = 16 − 16 = 0. Its sign gives the stated root type.

3. What is the nature of the roots of x² + (2)x + (5) = 0?

   Answer: No real roots. D = b² − 4ac = 4 − 20 = -16. Its sign gives the stated root type.

Discuss: Compare x² − 4x + c = 0 for c = 3, 4 and 5. Use D to explain the three different real-root cases.

Teaching note: The whole numerator −b ± √D is divided by 2a. Squaring a negative b produces a positive number. Open responses need discussion; they are not automatically marked.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. A rectangle has width x and length x + 2. Its area is 35. Which equation models it?

   x² + 2x − 35 = 0. Area is width × length: x(x + 2) = 35. Expanding and moving the area to the left gives x² + 2x − 35 = 0.

2. A rectangle has width x and length x + 2. Its area is 48. Which equation models it?

   x² + 2x − 48 = 0. Area is width × length: x(x + 2) = 48. Expanding and moving the area to the left gives x² + 2x − 48 = 0.

3. Solve x² − 10x + 16 = 0.

   x = 2 or 8. Factor as (x − 2)(x − 8) = 0. Either factor can be zero, giving x = 2 or 8.

4. Solve x² − 14x + 45 = 0.

   x = 5 or 9. Factor as (x − 5)(x − 9) = 0. Either factor can be zero, giving x = 5 or 9.

5. What is the nature of the roots of x² + (6)x + (9) = 0?

   Two equal real roots. D = b² − 4ac = 36 − 36 = 0. Its sign gives the stated root type.

6. What is the nature of the roots of x² + (4)x + (7) = 0?

   No real roots. D = b² − 4ac = 16 − 28 = -12. Its sign gives the stated root type.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. A rectangle has width x and length x + 2. Its area is 63. Which equation models it?

   x² + 2x − 63 = 0. Area is width × length: x(x + 2) = 63. Expanding and moving the area to the left gives x² + 2x − 63 = 0.

2. A rectangle has width x and length x + 2. Its area is 80. Which equation models it?

   x² + 2x − 80 = 0. Area is width × length: x(x + 2) = 80. Expanding and moving the area to the left gives x² + 2x − 80 = 0.

3. Solve x² − 11x + 24 = 0.

   x = 3 or 8. Factor as (x − 3)(x − 8) = 0. Either factor can be zero, giving x = 3 or 8.

4. Solve x² − 13x + 42 = 0.

   x = 6 or 7. Factor as (x − 6)(x − 7) = 0. Either factor can be zero, giving x = 6 or 7.

5. What is the nature of the roots of x² + (-10)x + (21) = 0?

   Two distinct real roots. D = b² − 4ac = 100 − 84 = 16. Its sign gives the stated root type.

6. What is the nature of the roots of x² + (2)x + (1) = 0?

   Two equal real roots. D = b² − 4ac = 4 − 4 = 0. Its sign gives the stated root type.