Worksheet for solving quadratic equations by completing the square, designed for GCSE Higher level algebra practice.

Let's solve each section of this worksheet step by step, focusing on solving quadratic equations by completing the square.

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Section A: Solve the following equations

These are already in a form that is easy to solve directly.

#### 1) $(x + 7)^2 - 9 = 0$

$$
(x + 7)^2 = 9 \\
x + 7 = \pm 3 \\
x = -7 \pm 3 \\
x = -4 \quad \text{or} \quad x = -10
$$

✔ Answer: $x = -4$ or $x = -10$

---

#### 2) $(x - 4)^2 - 25 = 0$

$$
(x - 4)^2 = 25 \\
x - 4 = \pm 5 \\
x = 4 \pm 5 \\
x = 9 \quad \text{or} \quad x = -1
$$

✔ Answer: $x = 9$ or $x = -1$

---

#### 3) $2(x + 5)^2 - 32 = 0$

$$
2(x + 5)^2 = 32 \\
(x + 5)^2 = 16 \\
x + 5 = \pm 4 \\
x = -5 \pm 4 \\
x = -1 \quad \text{or} \quad x = -9
$$

✔ Answer: $x = -1$ or $x = -9$

---

#### 4) $5(x + 4)^2 - 180 = 0$

$$
5(x + 4)^2 = 180 \\
(x + 4)^2 = 36 \\
x + 4 = \pm 6 \\
x = -4 \pm 6 \\
x = 2 \quad \text{or} \quad x = -10
$$

✔ Answer: $x = 2$ or $x = -10$

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Section B: Solve by completing the square (leave in surd form)

We'll use the method of completing the square.

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#### 1) $x^2 + 4x - 30 = 0$

Move constant:
$$
x^2 + 4x = 30
$$
Complete the square:
Take half of 4 → 2, square it → 4.
Add to both sides:
$$
x^2 + 4x + 4 = 30 + 4 = 34 \\
(x + 2)^2 = 34 \\
x + 2 = \pm \sqrt{34} \\
x = -2 \pm \sqrt{34}
$$

✔ Answer: $x = -2 \pm \sqrt{34}$

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#### 2) $x^2 + 18x - 21 = 0$

$$
x^2 + 18x = 21 \\
\text{Half of 18 is 9, } 9^2 = 81 \\
x^2 + 18x + 81 = 21 + 81 = 102 \\
(x + 9)^2 = 102 \\
x + 9 = \pm \sqrt{102} \\
x = -9 \pm \sqrt{102}
$$

✔ Answer: $x = -9 \pm \sqrt{102}$

---

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

$$
x^2 - 6x = 3 \\
\text{Half of -6 is -3, } (-3)^2 = 9 \\
x^2 - 6x + 9 = 3 + 9 = 12 \\
(x - 3)^2 = 12 \\
x - 3 = \pm \sqrt{12} = \pm 2\sqrt{3} \\
x = 3 \pm 2\sqrt{3}
$$

✔ Answer: $x = 3 \pm 2\sqrt{3}$

---

#### 4) $x^2 - x - 7 = 0$

$$
x^2 - x = 7 \\
\text{Half of -1 is } -0.5, \quad (-0.5)^2 = 0.25 \\
x^2 - x + 0.25 = 7 + 0.25 = 7.25 = \frac{29}{4} \\
(x - 0.5)^2 = \frac{29}{4} \\
x - \frac{1}{2} = \pm \frac{\sqrt{29}}{2} \\
x = \frac{1}{2} \pm \frac{\sqrt{29}}{2} = \frac{1 \pm \sqrt{29}}{2}
$$

✔ Answer: $x = \frac{1 \pm \sqrt{29}}{2}$

---

#### 5) $x^2 + 12x - 5 = 23$

Bring all terms to one side:
$$
x^2 + 12x - 28 = 0 \\
x^2 + 12x = 28 \\
\text{Half of 12 is 6, } 6^2 = 36 \\
x^2 + 12x + 36 = 28 + 36 = 64 \\
(x + 6)^2 = 64 \\
x + 6 = \pm 8 \\
x = -6 \pm 8 \\
x = 2 \quad \text{or} \quad x = -14
$$

But wait — the question says "leave in surd form", but here we get rational answers.

So technically:
$$
x = -6 \pm 8 = -14 \text{ or } 2
$$

But since they’re integers, no surds, so just write:

✔ Answer: $x = -14$ or $x = 2$

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#### 6) $x^2 - 32x + 45 = -35$

Bring all to left:
$$
x^2 - 32x + 80 = 0 \\
x^2 - 32x = -80 \\
\text{Half of -32 is -16, } (-16)^2 = 256 \\
x^2 - 32x + 256 = -80 + 256 = 176 \\
(x - 16)^2 = 176 \\
x - 16 = \pm \sqrt{176} = \pm \sqrt{16 \cdot 11} = \pm 4\sqrt{11} \\
x = 16 \pm 4\sqrt{11}
$$

✔ Answer: $x = 16 \pm 4\sqrt{11}$

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#### 7) $x^2 + 3x - 24 = 1$

$$
x^2 + 3x - 25 = 0 \\
x^2 + 3x = 25 \\
\text{Half of 3 is } 1.5, \quad (1.5)^2 = 2.25 \\
x^2 + 3x + 2.25 = 25 + 2.25 = 27.25 = \frac{109}{4} \\
(x + 1.5)^2 = \frac{109}{4} \\
x + \frac{3}{2} = \pm \frac{\sqrt{109}}{2} \\
x = -\frac{3}{2} \pm \frac{\sqrt{109}}{2} = \frac{-3 \pm \sqrt{109}}{2}
$$

✔ Answer: $x = \frac{-3 \pm \sqrt{109}}{2}$

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#### 8) $x^2 - 5x - 4 = 5$

$$
x^2 - 5x - 9 = 0 \\
x^2 - 5x = 9 \\
\text{Half of -5 is } -2.5, \quad (-2.5)^2 = 6.25 \\
x^2 - 5x + 6.25 = 9 + 6.25 = 15.25 = \frac{61}{4} \\
(x - 2.5)^2 = \frac{61}{4} \\
x - \frac{5}{2} = \pm \frac{\sqrt{61}}{2} \\
x = \frac{5}{2} \pm \frac{\sqrt{61}}{2} = \frac{5 \pm \sqrt{61}}{2}
$$

✔ Answer: $x = \frac{5 \pm \sqrt{61}}{2}$

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Section C: Solve by completing the square, leave to 1 decimal place

We will complete the square and approximate.

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#### 1) $2x^2 + 4x - 18 = 0$

Divide by 2 first:
$$
x^2 + 2x - 9 = 0 \\
x^2 + 2x = 9 \\
\text{Half of 2 is 1, } 1^2 = 1 \\
x^2 + 2x + 1 = 9 + 1 = 10 \\
(x + 1)^2 = 10 \\
x + 1 = \pm \sqrt{10} \approx \pm 3.162 \\
x = -1 \pm 3.162 \\
x \approx 2.2 \quad \text{or} \quad x \approx -4.2
$$

✔ Answer: $x \approx 2.2$ or $x \approx -4.2$

---

#### 2) $2x^2 + 8x - 5 = 0$

Divide by 2:
$$
x^2 + 4x - 2.5 = 0 \\
x^2 + 4x = 2.5 \\
\text{Half of 4 is 2, } 2^2 = 4 \\
x^2 + 4x + 4 = 2.5 + 4 = 6.5 \\
(x + 2)^2 = 6.5 \\
x + 2 = \pm \sqrt{6.5} \approx \pm 2.55 \\
x = -2 \pm 2.55 \\
x \approx 0.6 \quad \text{or} \quad x \approx -4.6
$$

✔ Answer: $x \approx 0.6$ or $x \approx -4.6$

---

#### 3) $3x^2 - 12x + 4 = 0$

Divide by 3:
$$
x^2 - 4x + \frac{4}{3} = 0 \\
x^2 - 4x = -\frac{4}{3} \\
\text{Half of -4 is -2, } (-2)^2 = 4 \\
x^2 - 4x + 4 = -\frac{4}{3} + 4 = \frac{8}{3} \approx 2.6667 \\
(x - 2)^2 = \frac{8}{3} \\
x - 2 = \pm \sqrt{\frac{8}{3}} \approx \pm \sqrt{2.6667} \approx \pm 1.633 \\
x = 2 \pm 1.633 \\
x \approx 3.6 \quad \text{or} \quad x \approx 0.4
$$

✔ Answer: $x \approx 3.6$ or $x \approx 0.4$

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#### 4) $2x^2 + 5x - 10 = 0$

Divide by 2:
$$
x^2 + \frac{5}{2}x - 5 = 0 \\
x^2 + 2.5x = 5 \\
\text{Half of 2.5 is 1.25, } (1.25)^2 = 1.5625 \\
x^2 + 2.5x + 1.5625 = 5 + 1.5625 = 6.5625 \\
(x + 1.25)^2 = 6.5625 \\
x + 1.25 = \pm \sqrt{6.5625} \approx \pm 2.561 \\
x = -1.25 \pm 2.561 \\
x \approx 1.3 \quad \text{or} \quad x \approx -3.8
$$

✔ Answer: $x \approx 1.3$ or $x \approx -3.8$

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#### 5) $5x^2 = 5x + 1$

Bring all to one side:
$$
5x^2 - 5x - 1 = 0 \\
\text{Divide by 5: } x^2 - x - 0.2 = 0 \\
x^2 - x = 0.2 \\
\text{Half of -1 is -0.5, } (-0.5)^2 = 0.25 \\
x^2 - x + 0.25 = 0.2 + 0.25 = 0.45 \\
(x - 0.5)^2 = 0.45 \\
x - 0.5 = \pm \sqrt{0.45} \approx \pm 0.671 \\
x = 0.5 \pm 0.671 \\
x \approx 1.2 \quad \text{or} \quad x \approx -0.2
$$

✔ Answer: $x \approx 1.2$ or $x \approx -0.2$

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

Bring to one side:
$$
2x^2 - 3x - 3 = 0 \\
\text{Divide by 2: } x^2 - 1.5x - 1.5 = 0 \\
x^2 - 1.5x = 1.5 \\
\text{Half of -1.5 is -0.75, } (-0.75)^2 = 0.5625 \\
x^2 - 1.5x + 0.5625 = 1.5 + 0.5625 = 2.0625 \\
(x - 0.75)^2 = 2.0625 \\
x - 0.75 = \pm \sqrt{2.0625} \approx \pm 1.436 \\
x = 0.75 \pm 1.436 \\
x \approx 2.2 \quad \text{or} \quad x \approx -0.7
$$

✔ Answer: $x \approx 2.2$ or $x \approx -0.7$

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Extension

#### A. The solutions to $2x^2 + ax + b = 0$ are $x = 2 \pm \sqrt{\frac{11}{2}}$. Find $a$ and $b$.

Given roots:
$$
x = 2 \pm \sqrt{\frac{11}{2}}
$$

This means the equation can be written as:
$$
(x - (2 + \sqrt{11/2}))(x - (2 - \sqrt{11/2})) = 0
$$

Use identity:
$$
(x - r)(x - s) = x^2 - (r+s)x + rs
$$

Sum of roots: $r + s = 2 + \sqrt{11/2} + 2 - \sqrt{11/2} = 4$

Product: $rs = (2)^2 - (\sqrt{11/2})^2 = 4 - \frac{11}{2} = \frac{8 - 11}{2} = -\frac{3}{2}$

So the monic quadratic is:
$$
x^2 - 4x - \frac{3}{2} = 0
$$

But original equation is $2x^2 + ax + b = 0$, so multiply entire equation by 2:
$$
2x^2 - 8x - 3 = 0
$$

Thus:
- $a = -8$
- $b = -3$

✔ Answer: $a = -8$, $b = -3$

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#### B. Prove the quadratic formula starting from $ax^2 + bx + c = 0$

We start with:
$$
ax^2 + bx + c = 0
$$

Step 1: Divide both sides by $a$ (assuming $a \ne 0$):
$$
x^2 + \frac{b}{a}x + \frac{c}{a} = 0
$$

Step 2: Move constant:
$$
x^2 + \frac{b}{a}x = -\frac{c}{a}
$$

Step 3: Complete the square. Take half of $\frac{b}{a}$: $\frac{b}{2a}$, square it: $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$

Add to both sides:
$$
x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}
$$

Left side is a perfect square:
$$
\left(x + \frac{b}{2a}\right)^2 = \frac{-4ac + b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}
$$

Take square root:
$$
x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}
$$

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

✔ Quadratic Formula Proved!

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✔ Final Answers Summary

#### Section A:
1. $x = -4, -10$
2. $x = 9, -1$
3. $x = -1, -9$
4. $x = 2, -10$

#### Section B:
1. $x = -2 \pm \sqrt{34}$
2. $x = -9 \pm \sqrt{102}$
3. $x = 3 \pm 2\sqrt{3}$
4. $x = \frac{1 \pm \sqrt{29}}{2}$
5. $x = -14, 2$
6. $x = 16 \pm 4\sqrt{11}$
7. $x = \frac{-3 \pm \sqrt{109}}{2}$
8. $x = \frac{5 \pm \sqrt{61}}{2}$

#### Section C:
1. $x \approx 2.2, -4.2$
2. $x \approx 0.6, -4.6$
3. $x \approx 3.6, 0.4$
4. $x \approx 1.3, -3.8$
5. $x \approx 1.2, -0.2$
6. $x \approx 2.2, -0.7$

#### Extension:
A. $a = -8$, $b = -3$
B. Proof shown above.

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