Quadratics: Solving using Completing the Square Textbook Exercise ... - Free Printable

Let's solve each of these quadratic equations using completing the square method. I'll go through Question 1, then Question 2, and finally Question 3, explaining each step clearly.

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🔷 General Method: Completing the Square

For a quadratic equation in the form:

$$
x^2 + bx + c = 0
$$

We rewrite it as:

$$
x^2 + bx = -c
$$

Then add $\left(\frac{b}{2}\right)^2$ to both sides to complete the square:

$$
x^2 + bx + \left(\frac{b}{2}\right)^2 = -c + \left(\frac{b}{2}\right)^2
$$

This becomes:

$$
\left(x + \frac{b}{2}\right)^2 = \text{RHS}
$$

Then take square roots and solve for $x$.

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## ✔ Question 1: Solve using completing the square

(a) $ x^2 + 6x + 8 = 0 $

Step 1: Move constant to right
$$
x^2 + 6x = -8
$$

Step 2: Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides
$$
x^2 + 6x + 9 = -8 + 9 = 1
$$

Step 3: Write as square
$$
(x + 3)^2 = 1
$$

Step 4: Take square root
$$
x + 3 = \pm 1 \Rightarrow x = -3 \pm 1
$$

So:
$$
x = -2 \quad \text{or} \quad x = -4
$$

✔ Answer: $ x = -2, -4 $

---

(b) $ x^2 + 10x + 24 = 0 $

Move constant:
$$
x^2 + 10x = -24
$$

Add $\left(\frac{10}{2}\right)^2 = 25$:
$$
x^2 + 10x + 25 = -24 + 25 = 1
$$

$$
(x + 5)^2 = 1
\Rightarrow x + 5 = \pm 1 \Rightarrow x = -5 \pm 1
$$

$$
x = -4 \quad \text{or} \quad x = -6
$$

✔ Answer: $ x = -4, -6 $

---

(c) $ x^2 + 14x + 40 = 0 $

$$
x^2 + 14x = -40
$$

Add $\left(\frac{14}{2}\right)^2 = 49$:
$$
x^2 + 14x + 49 = -40 + 49 = 9
$$

$$
(x + 7)^2 = 9 \Rightarrow x + 7 = \pm 3 \Rightarrow x = -7 \pm 3
$$

$$
x = -4 \quad \text{or} \quad x = -10
$$

✔ Answer: $ x = -4, -10 $

---

(d) $ x^2 - 4x - 45 = 0 $

$$
x^2 - 4x = 45
$$

Add $\left(\frac{-4}{2}\right)^2 = 4$:
$$
x^2 - 4x + 4 = 45 + 4 = 49
$$

$$
(x - 2)^2 = 49 \Rightarrow x - 2 = \pm 7 \Rightarrow x = 2 \pm 7
$$

$$
x = 9 \quad \text{or} \quad x = -5
$$

✔ Answer: $ x = 9, -5 $

---

(e) $ x^2 - 12x + 35 = 0 $

$$
x^2 - 12x = -35
$$

Add $\left(\frac{-12}{2}\right)^2 = 36$:
$$
x^2 - 12x + 36 = -35 + 36 = 1
$$

$$
(x - 6)^2 = 1 \Rightarrow x - 6 = \pm 1 \Rightarrow x = 6 \pm 1
$$

$$
x = 7 \quad \text{or} \quad x = 5
$$

✔ Answer: $ x = 7, 5 $

---

(f) $ x^2 - 2x - 3 = 0 $

$$
x^2 - 2x = 3
$$

Add $\left(\frac{-2}{2}\right)^2 = 1$:
$$
x^2 - 2x + 1 = 3 + 1 = 4
$$

$$
(x - 1)^2 = 4 \Rightarrow x - 1 = \pm 2 \Rightarrow x = 1 \pm 2
$$

$$
x = 3 \quad \text{or} \quad x = -1
$$

✔ Answer: $ x = 3, -1 $

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(g) $ x^2 + 14x - 51 = 0 $

$$
x^2 + 14x = 51
$$

Add $(7)^2 = 49$:
$$
x^2 + 14x + 49 = 51 + 49 = 100
$$

$$
(x + 7)^2 = 100 \Rightarrow x + 7 = \pm 10 \Rightarrow x = -7 \pm 10
$$

$$
x = 3 \quad \text{or} \quad x = -17
$$

✔ Answer: $ x = 3, -17 $

---

(h) $ x^2 - 6x - 16 = 0 $

$$
x^2 - 6x = 16
$$

Add $(3)^2 = 9$:
$$
x^2 - 6x + 9 = 16 + 9 = 25
$$

$$
(x - 3)^2 = 25 \Rightarrow x - 3 = \pm 5 \Rightarrow x = 3 \pm 5
$$

$$
x = 8 \quad \text{or} \quad x = -2
$$

✔ Answer: $ x = 8, -2 $

---

(i) $ x^2 - 2x + 1 = 0 $

This is already a perfect square!

$$
(x - 1)^2 = 0 \Rightarrow x = 1
$$

✔ Answer: $ x = 1 $ (repeated root)

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## ✔ Question 2: Solve using completing the square

(a) $ x^2 + 5x + 4 = 0 $

$$
x^2 + 5x = -4
$$

Add $\left(\frac{5}{2}\right)^2 = \frac{25}{4}$:
$$
x^2 + 5x + \frac{25}{4} = -4 + \frac{25}{4} = \frac{-16 + 25}{4} = \frac{9}{4}
$$

$$
\left(x + \frac{5}{2}\right)^2 = \frac{9}{4}
\Rightarrow x + \frac{5}{2} = \pm \frac{3}{2}
\Rightarrow x = -\frac{5}{2} \pm \frac{3}{2}
$$

$$
x = -\frac{5}{2} + \frac{3}{2} = -1, \quad x = -\frac{5}{2} - \frac{3}{2} = -4
$$

✔ Answer: $ x = -1, -4 $

---

(b) $ x^2 - 3x - 18 = 0 $

$$
x^2 - 3x = 18
$$

Add $\left(\frac{-3}{2}\right)^2 = \frac{9}{4}$:
$$
x^2 - 3x + \frac{9}{4} = 18 + \frac{9}{4} = \frac{72 + 9}{4} = \frac{81}{4}
$$

$$
\left(x - \frac{3}{2}\right)^2 = \frac{81}{4}
\Rightarrow x - \frac{3}{2} = \pm \frac{9}{2}
\Rightarrow x = \frac{3}{2} \pm \frac{9}{2}
$$

$$
x = \frac{12}{2} = 6, \quad x = \frac{-6}{2} = -3
$$

✔ Answer: $ x = 6, -3 $

---

(c) $ x^2 + x - 12 = 0 $

$$
x^2 + x = 12
$$

Add $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$:
$$
x^2 + x + \frac{1}{4} = 12 + \frac{1}{4} = \frac{48 + 1}{4} = \frac{49}{4}
$$

$$
\left(x + \frac{1}{2}\right)^2 = \frac{49}{4}
\Rightarrow x + \frac{1}{2} = \pm \frac{7}{2}
\Rightarrow x = -\frac{1}{2} \pm \frac{7}{2}
$$

$$
x = 3, \quad x = -4
$$

✔ Answer: $ x = 3, -4 $

---

(d) $ x^2 - 7x + 12 = 0 $

$$
x^2 - 7x = -12
$$

Add $\left(\frac{-7}{2}\right)^2 = \frac{49}{4}$:
$$
x^2 - 7x + \frac{49}{4} = -12 + \frac{49}{4} = \frac{-48 + 49}{4} = \frac{1}{4}
$$

$$
\left(x - \frac{7}{2}\right)^2 = \frac{1}{4}
\Rightarrow x - \frac{7}{2} = \pm \frac{1}{2}
\Rightarrow x = \frac{7}{2} \pm \frac{1}{2}
$$

$$
x = 4, \quad x = 3
$$

✔ Answer: $ x = 4, 3 $

---

(e) $ x^2 - 11x + 24 = 0 $

$$
x^2 - 11x = -24
$$

Add $\left(\frac{-11}{2}\right)^2 = \frac{121}{4}$:
$$
x^2 - 11x + \frac{121}{4} = -24 + \frac{121}{4} = \frac{-96 + 121}{4} = \frac{25}{4}
$$

$$
\left(x - \frac{11}{2}\right)^2 = \frac{25}{4}
\Rightarrow x - \frac{11}{2} = \pm \frac{5}{2}
\Rightarrow x = \frac{11}{2} \pm \frac{5}{2}
$$

$$
x = 8, \quad x = 3
$$

✔ Answer: $ x = 8, 3 $

---

(f) $ x^2 - 7x - 30 = 0 $

$$
x^2 - 7x = 30
$$

Add $\left(\frac{-7}{2}\right)^2 = \frac{49}{4}$:
$$
x^2 - 7x + \frac{49}{4} = 30 + \frac{49}{4} = \frac{120 + 49}{4} = \frac{169}{4}
$$

$$
\left(x - \frac{7}{2}\right)^2 = \frac{169}{4}
\Rightarrow x - \frac{7}{2} = \pm \frac{13}{2}
\Rightarrow x = \frac{7}{2} \pm \frac{13}{2}
$$

$$
x = 10, \quad x = -3
$$

✔ Answer: $ x = 10, -3 $

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## ✔ Question 3: Solve using completing the square — write in simplified surd form

We now deal with equations that may not factor nicely — answers will involve surds.

(a) $ x^2 + 4x - 3 = 0 $

$$
x^2 + 4x = 3
$$

Add $\left(\frac{4}{2}\right)^2 = 4$:
$$
x^2 + 4x + 4 = 3 + 4 = 7
\Rightarrow (x + 2)^2 = 7
\Rightarrow x + 2 = \pm \sqrt{7}
\Rightarrow x = -2 \pm \sqrt{7}
$$

✔ Answer: $ x = -2 \pm \sqrt{7} $

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(b) $ x^2 + 6x - 10 = 0 $

$$
x^2 + 6x = 10
$$

Add $9$:
$$
x^2 + 6x + 9 = 10 + 9 = 19
\Rightarrow (x + 3)^2 = 19
\Rightarrow x = -3 \pm \sqrt{19}
$$

✔ Answer: $ x = -3 \pm \sqrt{19} $

---

(c) $ x^2 - 2x - 5 = 0 $

$$
x^2 - 2x = 5
$$

Add $1$:
$$
x^2 - 2x + 1 = 5 + 1 = 6
\Rightarrow (x - 1)^2 = 6
\Rightarrow x = 1 \pm \sqrt{6}
$$

✔ Answer: $ x = 1 \pm \sqrt{6} $

---

(d) $ x^2 - 10x + 1 = 0 $

$$
x^2 - 10x = -1
$$

Add $25$:
$$
x^2 - 10x + 25 = -1 + 25 = 24
\Rightarrow (x - 5)^2 = 24
\Rightarrow x = 5 \pm \sqrt{24}
$$

Simplify $ \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6} $

✔ Answer: $ x = 5 \pm 2\sqrt{6} $

---

(e) $ x^2 + 8x + 3 = 0 $

$$
x^2 + 8x = -3
$$

Add $16$:
$$
x^2 + 8x + 16 = -3 + 16 = 13
\Rightarrow (x + 4)^2 = 13
\Rightarrow x = -4 \pm \sqrt{13}
$$

✔ Answer: $ x = -4 \pm \sqrt{13} $

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(f) $ x^2 - 8x - 22 = 0 $

$$
x^2 - 8x = 22
$$

Add $16$:
$$
x^2 - 8x + 16 = 22 + 16 = 38
\Rightarrow (x - 4)^2 = 38
\Rightarrow x = 4 \pm \sqrt{38}
$$

✔ Answer: $ x = 4 \pm \sqrt{38} $

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## 📌 Summary of Answers

Question 1:

- (a) $ x = -2, -4 $
- (b) $ x = -4, -6 $
- (c) $ x = -4, -10 $
- (d) $ x = 9, -5 $
- (e) $ x = 7, 5 $
- (f) $ x = 3, -1 $
- (g) $ x = 3, -17 $
- (h) $ x = 8, -2 $
- (i) $ x = 1 $

Question 2:

- (a) $ x = -1, -4 $
- (b) $ x = 6, -3 $
- (c) $ x = 3, -4 $
- (d) $ x = 4, 3 $
- (e) $ x = 8, 3 $
- (f) $ x = 10, -3 $

Question 3 (surd form):

- (a) $ x = -2 \pm \sqrt{7} $
- (b) $ x = -3 \pm \sqrt{19} $
- (c) $ x = 1 \pm \sqrt{6} $
- (d) $ x = 5 \pm 2\sqrt{6} $
- (e) $ x = -4 \pm \sqrt{13} $
- (f) $ x = 4 \pm \sqrt{38} $

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