PDF Year 11 Maths Worksheets | Fun and engaging worksheets - Free Printable

Let's solve the quadratic equations step by step using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

We'll go through Section A, Section B, and then the Extension.

---

✔ Section A

#### 1) $ x^2 + 3x - 1 = 0 $

- $ a = 1 $
- $ b = 3 $
- $ c = -1 $

Now apply the quadratic formula:

$$
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2}
$$

$$
\sqrt{13} \approx 3.6056
$$

So:
- $ x_1 = \frac{-3 + 3.6056}{2} = \frac{0.6056}{2} \approx 0.30 $
- $ x_2 = \frac{-3 - 3.6056}{2} = \frac{-6.6056}{2} \approx -3.30 $

✔ Answer: $ x \approx 0.30 $ or $ x \approx -3.30 $

---

#### 2) $ 6 - 3x - 2x^2 = 0 $

Rewrite in standard form:
$ -2x^2 - 3x + 6 = 0 $

- $ a = -2 $
- $ b = -3 $
- $ c = 6 $

$$
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-2)(6)}}{2(-2)} = \frac{3 \pm \sqrt{9 + 48}}{-4} = \frac{3 \pm \sqrt{57}}{-4}
$$

$$
\sqrt{57} \approx 7.55
$$

- $ x_1 = \frac{3 + 7.55}{-4} = \frac{10.55}{-4} \approx -2.64 $
- $ x_2 = \frac{3 - 7.55}{-4} = \frac{-4.55}{-4} \approx 1.14 $

✔ Answer: $ x \approx -2.64 $ or $ x \approx 1.14 $

---

#### 3) $ 8x^2 - 7 = 0 $

Standard form: $ 8x^2 + 0x - 7 = 0 $

- $ a = 8 $
- $ b = 0 $
- $ c = -7 $

$$
x = \frac{-0 \pm \sqrt{0^2 - 4(8)(-7)}}{2(8)} = \frac{\pm \sqrt{224}}{16}
$$

$$
\sqrt{224} = \sqrt{16 \times 14} = 4\sqrt{14} \approx 4 \times 3.7417 = 14.9668
$$

So:
- $ x_1 = \frac{14.9668}{16} \approx 0.935 \approx 0.94 $
- $ x_2 = \frac{-14.9668}{16} \approx -0.935 \approx -0.94 $

✔ Answer: $ x \approx 0.94 $ or $ x \approx -0.94 $

---

✔ Section B

We’ll solve each equation to 2 decimal places.

---

#### 1) $ 4x^2 + 9x + 1 = 0 $

- $ a = 4, b = 9, c = 1 $

$$
x = \frac{-9 \pm \sqrt{81 - 16}}{8} = \frac{-9 \pm \sqrt{65}}{8}
$$

$ \sqrt{65} \approx 8.06 $

- $ x_1 = \frac{-9 + 8.06}{8} = \frac{-0.94}{8} \approx -0.12 $
- $ x_2 = \frac{-9 - 8.06}{8} = \frac{-17.06}{8} \approx -2.13 $

✔ Answer: $ x \approx -0.12 $ or $ x \approx -2.13 $

---

#### 2) $ x^2 - 8x + 1 = 0 $

- $ a = 1, b = -8, c = 1 $

$$
x = \frac{8 \pm \sqrt{64 - 4}}{2} = \frac{8 \pm \sqrt{60}}{2}
$$

$ \sqrt{60} \approx 7.75 $

- $ x_1 = \frac{8 + 7.75}{2} = \frac{15.75}{2} = 7.875 \approx 7.88 $
- $ x_2 = \frac{8 - 7.75}{2} = \frac{0.25}{2} = 0.125 \approx 0.13 $

✔ Answer: $ x \approx 7.88 $ or $ x \approx 0.13 $

---

#### 3) $ 7x^2 + 3x - 2 = 0 $

- $ a = 7, b = 3, c = -2 $

$$
x = \frac{-3 \pm \sqrt{9 + 56}}{14} = \frac{-3 \pm \sqrt{65}}{14}
$$

$ \sqrt{65} \approx 8.06 $

- $ x_1 = \frac{-3 + 8.06}{14} = \frac{5.06}{14} \approx 0.36 $
- $ x_2 = \frac{-3 - 8.06}{14} = \frac{-11.06}{14} \approx -0.79 $

✔ Answer: $ x \approx 0.36 $ or $ x \approx -0.79 $

---

#### 4) $ 3x^2 - 4x - 5 = 0 $

- $ a = 3, b = -4, c = -5 $

$$
x = \frac{4 \pm \sqrt{16 + 60}}{6} = \frac{4 \pm \sqrt{76}}{6}
$$

$ \sqrt{76} \approx 8.72 $

- $ x_1 = \frac{4 + 8.72}{6} = \frac{12.72}{6} \approx 2.12 $
- $ x_2 = \frac{4 - 8.72}{6} = \frac{-4.72}{6} \approx -0.79 $

✔ Answer: $ x \approx 2.12 $ or $ x \approx -0.79 $

---

#### 5) $ 5x - 1 - x^2 = 0 $

Rewriting: $ -x^2 + 5x - 1 = 0 $ → Multiply by -1: $ x^2 - 5x + 1 = 0 $

- $ a = 1, b = -5, c = 1 $

$$
x = \frac{5 \pm \sqrt{25 - 4}}{2} = \frac{5 \pm \sqrt{21}}{2}
$$

$ \sqrt{21} \approx 4.58 $

- $ x_1 = \frac{5 + 4.58}{2} = \frac{9.58}{2} = 4.79 $
- $ x_2 = \frac{5 - 4.58}{2} = \frac{0.42}{2} = 0.21 $

✔ Answer: $ x \approx 4.79 $ or $ x \approx 0.21 $

---

#### 6) $ 4 - 3x - 2x^2 = 0 $

Rewriting: $ -2x^2 - 3x + 4 = 0 $

- $ a = -2, b = -3, c = 4 $

$$
x = \frac{3 \pm \sqrt{9 + 32}}{-4} = \frac{3 \pm \sqrt{41}}{-4}
$$

$ \sqrt{41} \approx 6.40 $

- $ x_1 = \frac{3 + 6.40}{-4} = \frac{9.40}{-4} = -2.35 $
- $ x_2 = \frac{3 - 6.40}{-4} = \frac{-3.40}{-4} = 0.85 $

✔ Answer: $ x \approx -2.35 $ or $ x \approx 0.85 $

---

#### 7) $ 2x^2 - 2x = 7x $

Bring all terms to one side:

$ 2x^2 - 2x - 7x = 0 \Rightarrow 2x^2 - 9x = 0 $

Factor: $ x(2x - 9) = 0 $

Solutions: $ x = 0 $ or $ x = \frac{9}{2} = 4.5 $

✔ Answer: $ x = 0 $ or $ x = 4.50 $

(Note: This is factorable, but we could also use the formula.)

---

#### 8) $ x(2x + 5) = 10 $

Expand: $ 2x^2 + 5x = 10 $

$ 2x^2 + 5x - 10 = 0 $

- $ a = 2, b = 5, c = -10 $

$$
x = \frac{-5 \pm \sqrt{25 + 80}}{4} = \frac{-5 \pm \sqrt{105}}{4}
$$

$ \sqrt{105} \approx 10.25 $

- $ x_1 = \frac{-5 + 10.25}{4} = \frac{5.25}{4} = 1.3125 \approx 1.31 $
- $ x_2 = \frac{-5 - 10.25}{4} = \frac{-15.25}{4} = -3.8125 \approx -3.81 $

✔ Answer: $ x \approx 1.31 $ or $ x \approx -3.81 $

---

#### 9) $ (x - 1)^2 = 17 $

Expand: $ x^2 - 2x + 1 = 17 $

$ x^2 - 2x - 16 = 0 $

- $ a = 1, b = -2, c = -16 $

$$
x = \frac{2 \pm \sqrt{4 + 64}}{2} = \frac{2 \pm \sqrt{68}}{2}
$$

$ \sqrt{68} \approx 8.25 $

- $ x_1 = \frac{2 + 8.25}{2} = \frac{10.25}{2} = 5.125 \approx 5.13 $
- $ x_2 = \frac{2 - 8.25}{2} = \frac{-6.25}{2} = -3.125 \approx -3.13 $

✔ Answer: $ x \approx 5.13 $ or $ x \approx -3.13 $

---

#### 10) $ 2(3 - x) = (4x + 3)^2 + 6 $

First expand both sides:

Left: $ 6 - 2x $

Right: $ (4x + 3)^2 + 6 = 16x^2 + 24x + 9 + 6 = 16x^2 + 24x + 15 $

Equation:
$$
6 - 2x = 16x^2 + 24x + 15
$$

Bring all terms to one side:
$$
0 = 16x^2 + 24x + 15 - 6 + 2x = 16x^2 + 26x + 9
$$

So: $ 16x^2 + 26x + 9 = 0 $

- $ a = 16, b = 26, c = 9 $

$$
x = \frac{-26 \pm \sqrt{676 - 576}}{32} = \frac{-26 \pm \sqrt{100}}{32}
$$

$ \sqrt{100} = 10 $

- $ x_1 = \frac{-26 + 10}{32} = \frac{-16}{32} = -0.50 $
- $ x_2 = \frac{-26 - 10}{32} = \frac{-36}{32} = -1.125 \approx -1.13 $

✔ Answer: $ x \approx -0.50 $ or $ x \approx -1.13 $

---

✔ Extension

Given: $ 3x^2 - x + 7 = 0 $

#### A. What happens when you try to solve using the quadratic formula?

We compute discriminant:

$$
b^2 - 4ac = (-1)^2 - 4(3)(7) = 1 - 84 = -83
$$

Since the discriminant is negative, the square root of a negative number is not real.

So, no real solutions exist.

✔ Answer: The equation has no real solutions because the discriminant is negative.

---

#### B. How does $ b^2 - 4ac $ explain this?

The discriminant $ D = b^2 - 4ac $ determines the nature of the roots:

- If $ D > 0 $: two distinct real solutions
- If $ D = 0 $: one repeated real solution
- If $ D < 0 $: no real solutions (complex solutions)

Here, $ D = -83 < 0 $, so no real solutions.

✔ Answer: Since $ b^2 - 4ac = -83 < 0 $, the square root is not real, so there are no real solutions.

---

#### C. Conditions for number of solutions:

For $ ax^2 + bx + c = 0 $:

- No solutions: $ b^2 - 4ac < 0 $
- Two solutions: $ b^2 - 4ac > 0 $
- One solution: $ b^2 - 4ac = 0 $

✔ Answer:
- No solutions: $ b^2 - 4ac < 0 $
- Two solutions: $ b^2 - 4ac > 0 $
- One solution: $ b^2 - 4ac = 0 $

---

✔ Final Summary

#### Section A Answers:
1. $ a=1, b=3, c=-1 $ → $ x \approx 0.30, -3.30 $
2. $ a=-2, b=-3, c=6 $ → $ x \approx -2.64, 1.14 $
3. $ a=8, b=0, c=-7 $ → $ x \approx 0.94, -0.94 $

#### Section B Answers:
1. $ x \approx -0.12, -2.13 $
2. $ x \approx 7.88, 0.13 $
3. $ x \approx 0.36, -0.79 $
4. $ x \approx 2.12, -0.79 $
5. $ x \approx 4.79, 0.21 $
6. $ x \approx -2.35, 0.85 $
7. $ x = 0, 4.50 $
8. $ x \approx 1.31, -3.81 $
9. $ x \approx 5.13, -3.13 $
10. $ x \approx -0.50, -1.13 $

#### Extension:
A. No real solutions (discriminant negative)
B. $ b^2 - 4ac = -83 < 0 $ → complex roots
C.
- No solutions: $ b^2 - 4ac < 0 $
- Two solutions: $ b^2 - 4ac > 0 $
- One solution: $ b^2 - 4ac = 0 $

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