Solved SOLVING QUADRATICS WORKSHEET Solve using the Square | Chegg.com - Free Printable

Solving Quadratics Worksheet

We will solve each problem step by step using the specified methods.

---

#### 1) Solve using the Square Root Method:

##### Problem 1: $ -5x^2 - 4 = -504 $

1. Start by isolating the $ x^2 $ term:
\[
-5x^2 - 4 = -504
\]
Add 4 to both sides:
\[
-5x^2 = -500
\]
Divide by $-5$:
\[
x^2 = 100
\]

2. Take the square root of both sides:
\[
x = \pm \sqrt{100}
\]
\[
x = \pm 10
\]

Solution: $ x = 10 $ or $ x = -10 $

---

##### Problem 2: $ x^2 = 81 $

1. Take the square root of both sides:
\[
x = \pm \sqrt{81}
\]
\[
x = \pm 9
\]

Solution: $ x = 9 $ or $ x = -9 $

---

#### 2) Solve by Factoring:

##### Problem 3: $ 5x^2 - 16x + 3 = 0 $

1. Factor the quadratic expression:
We need two numbers that multiply to $ 5 \cdot 3 = 15 $ and add to $-16$. These numbers are $-15$ and $-1$.
\[
5x^2 - 16x + 3 = (5x - 1)(x - 3)
\]

2. Set each factor equal to zero:
\[
5x - 1 = 0 \quad \text{or} \quad x - 3 = 0
\]
Solve for $ x $:
\[
5x - 1 = 0 \implies x = \frac{1}{5}
\]
\[
x - 3 = 0 \implies x = 3
\]

Solution: $ x = \frac{1}{5} $ or $ x = 3 $

---

##### Problem 4: $ x^2 + 5x + 6 = 0 $

1. Factor the quadratic expression:
We need two numbers that multiply to $ 6 $ and add to $ 5 $. These numbers are $ 2 $ and $ 3 $.
\[
x^2 + 5x + 6 = (x + 2)(x + 3)
\]

2. Set each factor equal to zero:
\[
x + 2 = 0 \quad \text{or} \quad x + 3 = 0
\]
Solve for $ x $:
\[
x + 2 = 0 \implies x = -2
\]
\[
x + 3 = 0 \implies x = -3
\]

Solution: $ x = -2 $ or $ x = -3 $

---

##### Problem 5: $ x^2 - 16 = 0 $

1. Recognize this as a difference of squares:
\[
x^2 - 16 = (x - 4)(x + 4)
\]

2. Set each factor equal to zero:
\[
x - 4 = 0 \quad \text{or} \quad x + 4 = 0
\]
Solve for $ x $:
\[
x - 4 = 0 \implies x = 4
\]
\[
x + 4 = 0 \implies x = -4
\]

Solution: $ x = 4 $ or $ x = -4 $

---

##### Problem 6: $ x^2 - 2x = -1 $

1. Rewrite the equation in standard form:
\[
x^2 - 2x + 1 = 0
\]

2. Factor the quadratic expression:
\[
x^2 - 2x + 1 = (x - 1)^2
\]

3. Set the factor equal to zero:
\[
(x - 1)^2 = 0
\]
\[
x - 1 = 0 \implies x = 1
\]

Solution: $ x = 1 $

---

#### 3) Solve by Completing the Square:

##### Problem 7: $ 3x^2 + 18x + 10 = 0 $

1. Divide the entire equation by 3 to simplify:
\[
x^2 + 6x + \frac{10}{3} = 0
\]

2. Move the constant term to the right side:
\[
x^2 + 6x = -\frac{10}{3}
\]

3. Complete the square:
- Take half of the coefficient of $ x $, which is $ 6 $, divide by 2 to get $ 3 $, and square it to get $ 9 $.
- Add $ 9 $ to both sides:
\[
x^2 + 6x + 9 = -\frac{10}{3} + 9
\]
- Simplify the right side:
\[
x^2 + 6x + 9 = -\frac{10}{3} + \frac{27}{3} = \frac{17}{3}
\]

4. Write the left side as a perfect square:
\[
(x + 3)^2 = \frac{17}{3}
\]

5. Take the square root of both sides:
\[
x + 3 = \pm \sqrt{\frac{17}{3}}
\]
\[
x + 3 = \pm \frac{\sqrt{51}}{3}
\]

6. Solve for $ x $:
\[
x = -3 \pm \frac{\sqrt{51}}{3}
\]

Solution: $ x = -3 + \frac{\sqrt{51}}{3} $ or $ x = -3 - \frac{\sqrt{51}}{3} $

---

##### Problem 8: $ x^2 - 2x - 15 = 0 $

1. Move the constant term to the right side:
\[
x^2 - 2x = 15
\]

2. Complete the square:
- Take half of the coefficient of $ x $, which is $-2$, divide by 2 to get $-1$, and square it to get $ 1 $.
- Add $ 1 $ to both sides:
\[
x^2 - 2x + 1 = 15 + 1
\]
- Simplify the right side:
\[
x^2 - 2x + 1 = 16
\]

3. Write the left side as a perfect square:
\[
(x - 1)^2 = 16
\]

4. Take the square root of both sides:
\[
x - 1 = \pm \sqrt{16}
\]
\[
x - 1 = \pm 4
\]

5. Solve for $ x $:
\[
x - 1 = 4 \implies x = 5
\]
\[
x - 1 = -4 \implies x = -3
\]

Solution: $ x = 5 $ or $ x = -3 $

---

#### 4) Solve using the Quadratic Formula:

The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

##### Problem 9: $ 4x^2 - x - 5 = 0 $

1. Identify $ a = 4 $, $ b = -1 $, and $ c = -5 $.
2. Substitute into the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-5)}}{2(4)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 80}}{8}
\]
\[
x = \frac{1 \pm \sqrt{81}}{8}
\]
\[
x = \frac{1 \pm 9}{8}
\]

3. Solve for the two solutions:
\[
x = \frac{1 + 9}{8} = \frac{10}{8} = \frac{5}{4}
\]
\[
x = \frac{1 - 9}{8} = \frac{-8}{8} = -1
\]

Solution: $ x = \frac{5}{4} $ or $ x = -1 $

---

##### Problem 10: $ 2x^2 + 3x - 3 = 0 $

1. Identify $ a = 2 $, $ b = 3 $, and $ c = -3 $.
2. Substitute into the quadratic formula:
\[
x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-3)}}{2(2)}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 24}}{4}
\]
\[
x = \frac{-3 \pm \sqrt{33}}{4}
\]

Solution: $ x = \frac{-3 + \sqrt{33}}{4} $ or $ x = \frac{-3 - \sqrt{33}}{4} $

---

Final Answers:

1. $ x = 10 $ or $ x = -10 $
2. $ x = 9 $ or $ x = -9 $
3. $ x = \frac{1}{5} $ or $ x = 3 $
4. $ x = -2 $ or $ x = -3 $
5. $ x = 4 $ or $ x = -4 $
6. $ x = 1 $
7. $ x = -3 + \frac{\sqrt{51}}{3} $ or $ x = -3 - \frac{\sqrt{51}}{3} $
8. $ x = 5 $ or $ x = -3 $
9. $ x = \frac{5}{4} $ or $ x = -1 $
10. $ x = \frac{-3 + \sqrt{33}}{4} $ or $ x = \frac{-3 - \sqrt{33}}{4} $

\boxed{
\begin{aligned}
&1. \, x = 10 \, \text{or} \, x = -10 \\
&2. \, x = 9 \, \text{or} \, x = -9 \\
&3. \, x = \frac{1}{5} \, \text{or} \, x = 3 \\
&4. \, x = -2 \, \text{or} \, x = -3 \\
&5. \, x = 4 \, \text{or} \, x = -4 \\
&6. \, x = 1 \\
&7. \, x = -3 + \frac{\sqrt{51}}{3} \, \text{or} \, x = -3 - \frac{\sqrt{51}}{3} \\
&8. \, x = 5 \, \text{or} \, x = -3 \\
&9. \, x = \frac{5}{4} \, \text{or} \, x = -1 \\
&10. \, x = \frac{-3 + \sqrt{33}}{4} \, \text{or} \, x = \frac{-3 - \sqrt{33}}{4}
\end{aligned}
} $