Solving Quadratic Equations (C) - By Quadratic Formula Worksheet ... - Free Printable

Let’s solve each problem step by step using the quadratic formula:

> Quadratic Formula:
If \( ax^2 + bx + c = 0 \), then
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

We’ll round all answers to 2 decimal places.

---

## ✔ Section A

Problem 1: \( x^2 + 3x - 1 = 0 \)

- \( a = 1 \)
- \( b = 3 \)
- \( c = -1 \)

Discriminant:
\( b^2 - 4ac = 3^2 - 4(1)(-1) = 9 + 4 = 13 \)

Solutions:
\[
x = \frac{-3 \pm \sqrt{13}}{2} = \frac{-3 \pm 3.6056}{2}
\]

→ \( x_1 = \frac{-3 + 3.6056}{2} = \frac{0.6056}{2} ≈ 0.30 \)
→ \( x_2 = \frac{-3 - 3.6056}{2} = \frac{-6.6056}{2} ≈ -3.30 \)

✔ Answer: 0.30, -3.30

---

Problem 2: \( 6 - 3x - 2x^2 = 0 \)

First, rewrite in standard form:
\( -2x^2 - 3x + 6 = 0 \) → or multiply by -1 to make it easier:
\( 2x^2 + 3x - 6 = 0 \) (same solutions)

So:
- \( a = 2 \)
- \( b = 3 \)
- \( c = -6 \)

Discriminant:
\( 3^2 - 4(2)(-6) = 9 + 48 = 57 \)

Solutions:
\[
x = \frac{-3 \pm \sqrt{57}}{4} = \frac{-3 \pm 7.5498}{4}
\]

→ \( x_1 = \frac{-3 + 7.5498}{4} = \frac{4.5498}{4} ≈ 1.14 \)
→ \( x_2 = \frac{-3 - 7.5498}{4} = \frac{-10.5498}{4} ≈ -2.64 \)

✔ Answer: 1.14, -2.64

*(Note: If you used original coefficients without flipping signs, you’d get same answer — just be consistent with signs!)*

---

Problem 3: \( 8x^2 - 7 = 0 \)

This is missing the “bx” term → so \( b = 0 \)

- \( a = 8 \)
- \( b = 0 \)
- \( c = -7 \)

Discriminant:
\( 0^2 - 4(8)(-7) = 0 + 224 = 224 \)

Solutions:
\[
x = \frac{0 \pm \sqrt{224}}{16} = \frac{\pm 14.9666}{16}
\]

→ \( x_1 = \frac{14.9666}{16} ≈ 0.94 \)
→ \( x_2 = \frac{-14.9666}{16} ≈ -0.94 \)

✔ Answer: 0.94, -0.94

---

## ✔ Section B

We’ll go one by one.

---

1) \( 4x^2 + 9x + 1 = 0 \)

- \( a=4, b=9, c=1 \)
- Discriminant: \( 81 - 16 = 65 \)
- \( x = \frac{-9 \pm \sqrt{65}}{8} = \frac{-9 \pm 8.0623}{8} \)

→ \( x_1 = \frac{-0.9377}{8} ≈ -0.12 \)
→ \( x_2 = \frac{-17.0623}{8} ≈ -2.13 \)

✔ Answer: -0.12, -2.13

---

2) \( x^2 - 8x + 1 = 0 \)

- \( a=1, b=-8, c=1 \)
- Discriminant: \( 64 - 4 = 60 \)
- \( x = \frac{8 \pm \sqrt{60}}{2} = \frac{8 \pm 7.746}{2} \)

→ \( x_1 = \frac{15.746}{2} ≈ 7.87 \)
→ \( x_2 = \frac{0.254}{2} ≈ 0.13 \)

✔ Answer: 7.87, 0.13

---

3) \( 7x^2 + 3x - 2 = 0 \)

- \( a=7, b=3, c=-2 \)
- Discriminant: \( 9 + 56 = 65 \)
- \( x = \frac{-3 \pm \sqrt{65}}{14} = \frac{-3 \pm 8.0623}{14} \)

→ \( x_1 = \frac{5.0623}{14} ≈ 0.36 \)
→ \( x_2 = \frac{-11.0623}{14} ≈ -0.79 \)

✔ Answer: 0.36, -0.79

---

4) \( 3x^2 - 4x - 5 = 0 \)

- \( a=3, b=-4, c=-5 \)
- Discriminant: \( 16 + 60 = 76 \)
- \( x = \frac{4 \pm \sqrt{76}}{6} = \frac{4 \pm 8.7178}{6} \)

→ \( x_1 = \frac{12.7178}{6} ≈ 2.12 \)
→ \( x_2 = \frac{-4.7178}{6} ≈ -0.79 \)

✔ Answer: 2.12, -0.79

---

5) \( 5x - 1 - x^2 = 0 \)

Rewrite: \( -x^2 + 5x - 1 = 0 \) → or multiply by -1: \( x^2 - 5x + 1 = 0 \)

Use \( a=1, b=-5, c=1 \)

Discriminant: \( 25 - 4 = 21 \)

\( x = \frac{5 \pm \sqrt{21}}{2} = \frac{5 \pm 4.5826}{2} \)

→ \( x_1 = \frac{9.5826}{2} ≈ 4.79 \)
→ \( x_2 = \frac{0.4174}{2} ≈ 0.21 \)

✔ Answer: 4.79, 0.21

---

6) \( 4 - 3x - 2x^2 = 0 \)

Rewrite: \( -2x^2 - 3x + 4 = 0 \) → or use as is.

Better: Multiply by -1 → \( 2x^2 + 3x - 4 = 0 \)

- \( a=2, b=3, c=-4 \)
- Discriminant: \( 9 + 32 = 41 \)
- \( x = \frac{-3 \pm \sqrt{41}}{4} = \frac{-3 \pm 6.4031}{4} \)

→ \( x_1 = \frac{3.4031}{4} ≈ 0.85 \)
→ \( x_2 = \frac{-9.4031}{4} ≈ -2.35 \)

✔ Answer: 0.85, -2.35

---

7) \( 2x^2 - 2x = 7x \)

Bring all terms to left:
\( 2x^2 - 2x - 7x = 0 \) → \( 2x^2 - 9x = 0 \)

Wait — this factors easily: \( x(2x - 9) = 0 \) → but let’s use formula anyway.

Actually, we can write as: \( 2x^2 - 9x + 0 = 0 \)

- \( a=2, b=-9, c=0 \)
- Discriminant: \( 81 - 0 = 81 \)
- \( x = \frac{9 \pm 9}{4} \)

→ \( x_1 = \frac{18}{4} = 4.50 \)
→ \( x_2 = \frac{0}{4} = 0.00 \)

✔ Answer: 4.50, 0.00

---

8) \( x(2x + 5) = 10 \)

Expand: \( 2x^2 + 5x = 10 \) → \( 2x^2 + 5x - 10 = 0 \)

- \( a=2, b=5, c=-10 \)
- Discriminant: \( 25 + 80 = 105 \)
- \( x = \frac{-5 \pm \sqrt{105}}{4} = \frac{-5 \pm 10.24695}{4} \)

→ \( x_1 = \frac{5.24695}{4} ≈ 1.31 \)
→ \( x_2 = \frac{-15.24695}{4} ≈ -3.81 \)

✔ Answer: 1.31, -3.81

---

9) \( (x - 1)^2 = 17 \)

Take square root both sides:
\( x - 1 = \pm \sqrt{17} \) → \( x = 1 \pm \sqrt{17} \)

√17 ≈ 4.123

→ \( x_1 = 1 + 4.123 = 5.12 \)
→ \( x_2 = 1 - 4.123 = -3.12 \)

✔ Answer: 5.12, -3.12

*(Alternatively, expand and use formula — same result.)*

---

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

Left side: \( 6 - 2x \)

Right side: Expand \( (4x+3)^2 = 16x^2 + 24x + 9 \), then add 6 → \( 16x^2 + 24x + 15 \)

Equation:
\( 6 - 2x = 16x^2 + 24x + 15 \)

Bring all to right:
\( 0 = 16x^2 + 24x + 15 - 6 + 2x \)
→ \( 16x^2 + 26x + 9 = 0 \)

Check:
Left: 6 - 2x
Right: 16x² + 24x + 15
Subtract left from both sides:
0 = 16x² + 24x + 15 - 6 + 2x → yes, 16x² + 26x + 9 = 0

Now solve:

- \( a=16, b=26, c=9 \)
- Discriminant: \( 26^2 - 4*16*9 = 676 - 576 = 100 \)
- \( x = \frac{-26 \pm \sqrt{100}}{32} = \frac{-26 \pm 10}{32} \)

→ \( x_1 = \frac{-16}{32} = -0.50 \)
→ \( x_2 = \frac{-36}{32} = -1.125 ≈ -1.13 \)

✔ Answer: -0.50, -1.13

---

## ✔ Extension: \( 3x^2 - x + 7 = 0 \)

A. What happens when you try to solve?

Plug into formula:

- \( a=3, b=-1, c=7 \)
- Discriminant: \( (-1)^2 - 4*3*7 = 1 - 84 = -83 \)

Square root of negative number → not real!

So, no real solutions.

B. How does \( b^2 - 4ac \) explain?

The discriminant is negative → means no real roots. The parabola never touches the x-axis.

C. Conditions for number of solutions:

For \( ax^2 + bx + c = 0 \):

- No solutions → if \( b^2 - 4ac < 0 \) (discriminant negative)
- Two solutions → if \( b^2 - 4ac > 0 \) (positive)
- One solution → if \( b^2 - 4ac = 0 \) (zero)

---

## Final Answer:

Section A:

1. a=1, b=3, c=-1 → 0.30, -3.30
2. a=-2, b=-3, c=6 (or a=2,b=3,c=-6) → 1.14, -2.64
3. a=8, b=0, c=-7 → 0.94, -0.94

Section B:

1) -0.12, -2.13
2) 7.87, 0.13
3) 0.36, -0.79
4) 2.12, -0.79
5) 4.79, 0.21
6) 0.85, -2.35
7) 4.50, 0.00
8) 1.31, -3.81
9) 5.12, -3.12
10) -0.50, -1.13

Extension:

A. You get a square root of a negative number → no real solutions.
B. Because discriminant = -83 < 0 → no real roots.
C.
- No solutions: \( b^2 - 4ac < 0 \)
- Two solutions: \( b^2 - 4ac > 0 \)
- One solution: \( b^2 - 4ac = 0 \)