Comprehensive algebra worksheet guiding students through solving quadratic equations by completing the square, ranging from basic forms to proofs.

Let’s solve each section step by step. We’ll start with Section A, then B, then C, and finally the Extension.

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Section A: Solve the following equations.

These are already in completed square form — we just need to isolate x.

1) (x + 7)² – 9 = 0
→ Add 9 to both sides:
(x + 7)² = 9
→ Take square root of both sides:
x + 7 = ±3
→ Subtract 7:
x = -7 + 3 = -4 or x = -7 – 3 = -10

2) (x – 4)² – 25 = 0
→ Add 25:
(x – 4)² = 25
→ Square root:
x – 4 = ±5
→ Add 4:
x = 4 + 5 = 9 or x = 4 – 5 = -1

3) 2(x + 5)² – 32 = 0
→ Add 32:
2(x + 5)² = 32
→ Divide by 2:
(x + 5)² = 16
→ Square root:
x + 5 = ±4
→ Subtract 5:
x = -5 + 4 = -1 or x = -5 – 4 = -9

4) 5(x + 4)² – 180 = 0
→ Add 180:
5(x + 4)² = 180
→ Divide by 5:
(x + 4)² = 36
→ Square root:
x + 4 = ±6
→ Subtract 4:
x = -4 + 6 = 2 or x = -4 – 6 = -10

✔ Section A Answers:
1) x = -4, -10
2) x = 9, -1
3) x = -1, -9
4) x = 2, -10

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Section B: Solve by completing the square. Leave answers in surd form.

We’ll complete the square for each quadratic.

1) x² + 4x – 30 = 0
→ Move constant: x² + 4x = 30
→ Complete square: take half of 4 → 2, square it → 4
→ Add 4 to both sides: x² + 4x + 4 = 34
→ (x + 2)² = 34
→ x + 2 = ±√34
→ x = -2 ± √34

2) x² + 18x – 21 = 0
→ x² + 18x = 21
→ Half of 18 is 9, square is 81
→ x² + 18x + 81 = 21 + 81 = 102
→ (x + 9)² = 102
→ x = -9 ± √102

3) x² – 6x – 3 = 0
→ x² – 6x = 3
→ Half of -6 is -3, square is 9
→ x² – 6x + 9 = 3 + 9 = 12
→ (x – 3)² = 12
→ x – 3 = ±√12 = ±2√3
→ x = 3 ± 2√3

4) x² – x – 7 = 0
→ x² – x = 7
→ Half of -1 is -0.5, square is 0.25
→ x² – x + 0.25 = 7 + 0.25 = 7.25 = 29/4
→ (x – 0.5)² = 29/4
→ x – 0.5 = ±√(29)/2
→ x = 0.5 ± √29 / 2 = (1 ± √29)/2

5) x² + 12x – 5 = 23
→ First, move all to left: x² + 12x – 28 = 0
→ Or better: x² + 12x = 28
→ Half of 12 is 6, square is 36
→ x² + 12x + 36 = 28 + 36 = 64
→ (x + 6)² = 64
→ x + 6 = ±8
→ x = -6 + 8 = 2 or x = -6 – 8 = -14
Wait — but this gives integer answers. The problem says “leave in surd form” — but if it simplifies nicely, that’s fine. Let me check original:
Original: x² + 12x – 5 = 23 → x² + 12x = 28 → yes, leads to perfect square. So answer is integers. But maybe they want surd anyway? No — if it’s exact, we write exact. So:

x = 2 or x = -14

But let’s double-check:
(x+6)^2 = 64 → x+6=±8 → x=2 or -14 → correct.

However, the instruction says “leave your answers in surd form” — which usually means if there’s a radical, leave it. If not, just give exact value. So we can write as integers.

But wait — perhaps I misread. Let me recheck problem 5:
“x² + 12x – 5 = 23” → so x² + 12x = 28 → yes, completes to (x+6)^2 = 64 → no surd needed.

So answer: x = 2, -14

But let’s continue — maybe others have surds.

6) x² – 32x + 45 = -35
→ Bring all to left: x² – 32x + 80 = 0
Or: x² – 32x = -80
→ Half of -32 is -16, square is 256
→ x² – 32x + 256 = -80 + 256 = 176
→ (x – 16)² = 176
→ 176 = 16 * 11 → √176 = 4√11
→ x – 16 = ±4√11
→ x = 16 ± 4√11

7) x² + 3x – 24 = 1
→ x² + 3x = 25
→ Half of 3 is 1.5, square is 2.25 = 9/4
→ x² + 3x + 9/4 = 25 + 9/4 = 100/4 + 9/4 = 109/4
→ (x + 3/2)² = 109/4
→ x + 3/2 = ±√109 / 2
→ x = (-3 ± √109)/2

8) x² – 5x – 4 = 5
→ x² – 5x = 9
→ Half of -5 is -2.5, square is 6.25 = 25/4
→ x² – 5x + 25/4 = 9 + 25/4 = 36/4 + 25/4 = 61/4
→ (x – 5/2)² = 61/4
→ x – 5/2 = ±√61 / 2
→ x = (5 ± √61)/2

✔ Section B Answers:
1) x = -2 ± √34
2) x = -9 ± √102
3) x = 3 ± 2√3
4) x = (1 ± √29)/2
5) x = 2, -14
6) x = 16 ± 4√11
7) x = (-3 ± √109)/2
8) x = (5 ± √61)/2

Note: For #5, even though it says “surd form”, since the solution is rational, we write the exact values.

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Section C: Solve by completing the square. Leave answers to one decimal place.

We’ll complete the square and then approximate.

1) 2x² + 4x – 18 = 0
First, divide entire equation by 2:
x² + 2x – 9 = 0
→ x² + 2x = 9
→ Half of 2 is 1, square is 1
→ x² + 2x + 1 = 10
→ (x + 1)² = 10
→ x + 1 = ±√10 ≈ ±3.162
→ x = -1 + 3.162 ≈ 2.2 or x = -1 – 3.162 ≈ -4.2

2) 2x² + 8x – 5 = 0
Divide by 2: x² + 4x – 2.5 = 0
→ x² + 4x = 2.5
→ Half of 4 is 2, square is 4
→ x² + 4x + 4 = 2.5 + 4 = 6.5
→ (x + 2)² = 6.5
→ x + 2 = ±√6.5 ≈ ±2.550
→ x = -2 + 2.550 ≈ 0.6 or x = -2 – 2.550 ≈ -4.6

3) 3x² – 12x + 4 = 0
Divide by 3: x² – 4x + 4/3 = 0
→ x² – 4x = -4/3
→ Half of -4 is -2, square is 4
→ x² – 4x + 4 = -4/3 + 4 = -4/3 + 12/3 = 8/3
→ (x – 2)² = 8/3 ≈ 2.6667
→ x – 2 = ±√(8/3) = ±√(2.6667) ≈ ±1.633
→ x = 2 + 1.633 ≈ 3.6 or x = 2 – 1.633 ≈ 0.4

4) 2x² + 5x – 10 = 0
Divide by 2: x² + 2.5x – 5 = 0
→ x² + 2.5x = 5
→ Half of 2.5 is 1.25, square is 1.5625
→ x² + 2.5x + 1.5625 = 5 + 1.5625 = 6.5625
→ (x + 1.25)² = 6.5625
→ x + 1.25 = ±√6.5625 = ±2.5617 (since 2.5617² ≈ 6.5625)
Actually, 6.5625 = 105/16? Wait, 2.5625² = ? Better calculate numerically.

√6.5625 = ? Note: 2.56^2 = 6.5536, 2.562^2 = 6.563844, close. Actually, 6.5625 = 105/16? 105÷16=6.5625 yes. √(105/16)=√105 / 4 ≈ 10.24695 / 4 ≈ 2.5617

So x = -1.25 + 2.5617 ≈ 1.3 or x = -1.25 – 2.5617 ≈ -3.8

5) 5x² = 5x + 1
Bring all to left: 5x² – 5x – 1 = 0
Divide by 5: x² – x – 0.2 = 0
→ x² – x = 0.2
→ Half of -1 is -0.5, square is 0.25
→ x² – x + 0.25 = 0.2 + 0.25 = 0.45
→ (x – 0.5)² = 0.45
→ x – 0.5 = ±√0.45 ≈ ±0.6708
→ x = 0.5 + 0.6708 ≈ 1.2 or x = 0.5 – 0.6708 ≈ -0.2

6) 2x² – 3x = 3
Bring to standard: 2x² – 3x – 3 = 0
Divide by 2: x² – 1.5x – 1.5 = 0
→ x² – 1.5x = 1.5
→ Half of -1.5 is -0.75, square is 0.5625
→ x² – 1.5x + 0.5625 = 1.5 + 0.5625 = 2.0625
→ (x – 0.75)² = 2.0625
→ x – 0.75 = ±√2.0625
√2.0625 = ? 1.436^2 = 2.062096, close. Actually, 2.0625 = 33/16? 33÷16=2.0625 yes. √(33/16)=√33 / 4 ≈ 5.7446 / 4 ≈ 1.43615

So x = 0.75 + 1.436 ≈ 2.2 or x = 0.75 – 1.436 ≈ -0.7

✔ Section C Answers (to 1 decimal place):
1) x ≈ 2.2, -4.2
2) x ≈ 0.6, -4.6
3) x ≈ 3.6, 0.4
4) x ≈ 1.3, -3.8
5) x ≈ 1.2, -0.2
6) x ≈ 2.2, -0.7

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Extension:

A. The solutions to 2x² + ax + b = 0 are x = 2 + √(11/2)

Since it's a quadratic, there should be two solutions. Given only one, but typically if one solution is given like this, the other is with minus: x = 2 ± √(11/2)

So assume solutions are x = 2 + √(11/2) and x = 2 – √(11/2)

For a quadratic equation with roots r1 and r2, we can write:

x² – (r1+r2)x + r1*r2 = 0

But here the leading coefficient is 2, so:

2x² + ax + b = 2(x – r1)(x – r2)

Sum of roots: r1 + r2 = [2 + √(11/2)] + [2 – √(11/2)] = 4

Product: r1 * r2 = (2 + √(11/2))(2 – √(11/2)) = 2² – (√(11/2))² = 4 – 11/2 = 8/2 – 11/2 = -3/2

So the monic quadratic would be: x² – (sum)x + product = x² – 4x – 3/2

Multiply by 2: 2x² – 8x – 3

Compare to 2x² + ax + b → so a = -8, b = -3

Check: if 2x² -8x -3 =0, use quadratic formula:

x = [8 ± √(64 + 24)] / 4 = [8 ± √88]/4 = [8 ± 2√22]/4 = [4 ± √22]/2 = 2 ± √22 / 2

But √22 / 2 = √(22/4) = √(11/2)? Let's see: (√22 / 2)^2 = 22/4 = 11/2 → yes!

So x = 2 ± √(11/2) → matches.

Thus, a = -8, b = -3

B. Prove quadratic formula from ax² + bx + c = 0

Start: ax² + bx + c = 0

Divide by a (a ≠ 0): x² + (b/a)x + c/a = 0

Move constant: x² + (b/a)x = -c/a

Complete the square: take half of b/a → b/(2a), square it → b²/(4a²)

Add to both sides:

x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²)

Left side: (x + b/(2a))²

Right side: (-4ac + b²)/(4a²) = (b² – 4ac)/(4a²)

So:

(x + b/(2a))² = (b² – 4ac)/(4a²)

Take square root:

x + b/(2a) = ± √(b² – 4ac) / (2a)

Subtract b/(2a):

x = -b/(2a) ± √(b² – 4ac)/(2a) = [ -b ± √(b² – 4ac) ] / (2a)

Proved.

✔ Extension Answers:
A. a = -8, b = -3
B. Proof shown above.

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Final Answer:

Section A:
1) x = -4, -10
2) x = 9, -1
3) x = -1, -9
4) x = 2, -10

Section B:
1) x = -2 ± √34
2) x = -9 ± √102
3) x = 3 ± 2√3
4) x = (1 ± √29)/2
5) x = 2, -14
6) x = 16 ± 4√11
7) x = (-3 ± √109)/2
8) x = (5 ± √61)/2

Section C:
1) x ≈ 2.2, -4.2
2) x ≈ 0.6, -4.6
3) x ≈ 3.6, 0.4
4) x ≈ 1.3, -3.8
5) x ≈ 1.2, -0.2
6) x ≈ 2.2, -0.7

Extension:
A. a = -8, b = -3
B. Proof completed as shown.