Worksheet featuring algebraic equations including quadratic and rational expressions for solving.

Problem Analysis

The task involves solving a variety of equations, including quadratic equations, rational equations, and some algebraic manipulations. Let's solve each equation step by step.

---

Part (a): \( 2x^2 - 11x + 14 = 0 \)

This is a quadratic equation. We can solve it using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = -11 \), and \( c = 14 \).

1. Calculate the discriminant:
\[
b^2 - 4ac = (-11)^2 - 4(2)(14) = 121 - 112 = 9
\]

2. Apply the quadratic formula:
\[
x = \frac{-(-11) \pm \sqrt{9}}{2(2)} = \frac{11 \pm 3}{4}
\]

3. Solve for the two roots:
\[
x = \frac{11 + 3}{4} = \frac{14}{4} = \frac{7}{2}
\]
\[
x = \frac{11 - 3}{4} = \frac{8}{4} = 2
\]

Solution:
\[
x = 2 \quad \text{or} \quad x = \frac{7}{2}
\]

---

Part (b): \( x^2 + 2x - 63 = 0 \)

This is another quadratic equation. We can factorize it directly:
\[
x^2 + 2x - 63 = (x + 9)(x - 7) = 0
\]

1. Set each factor to zero:
\[
x + 9 = 0 \quad \Rightarrow \quad x = -9
\]
\[
x - 7 = 0 \quad \Rightarrow \quad x = 7
\]

Solution:
\[
x = -9 \quad \text{or} \quad x = 7
\]

---

Part (c): \( x^2 - 16x + 48 = 0 \)

Factorize the quadratic equation:
\[
x^2 - 16x + 48 = (x - 12)(x - 4) = 0
\]

1. Set each factor to zero:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]

Solution:
\[
x = 4 \quad \text{or} \quad x = 12
\]

---

Part (d): \( x^2 + 7x - 1 = 0 \)

This quadratic equation does not factorize easily, so we use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 7 \), and \( c = -1 \).

1. Calculate the discriminant:
\[
b^2 - 4ac = 7^2 - 4(1)(-1) = 49 + 4 = 53
\]

2. Apply the quadratic formula:
\[
x = \frac{-7 \pm \sqrt{53}}{2}
\]

Solution:
\[
x = \frac{-7 + \sqrt{53}}{2} \quad \text{or} \quad x = \frac{-7 - \sqrt{53}}{2}
\]

---

Part (e): \( 2x^2 + 7x - 9 = 0 \)

Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = 7 \), and \( c = -9 \).

1. Calculate the discriminant:
\[
b^2 - 4ac = 7^2 - 4(2)(-9) = 49 + 72 = 121
\]

2. Apply the quadratic formula:
\[
x = \frac{-7 \pm \sqrt{121}}{2(2)} = \frac{-7 \pm 11}{4}
\]

3. Solve for the two roots:
\[
x = \frac{-7 + 11}{4} = \frac{4}{4} = 1
\]
\[
x = \frac{-7 - 11}{4} = \frac{-18}{4} = -\frac{9}{2}
\]

Solution:
\[
x = 1 \quad \text{or} \quad x = -\frac{9}{2}
\]

---

Part (f): \( 7 - 5x^2 + 2x = 0 \)

Rearrange the equation:
\[
-5x^2 + 2x + 7 = 0
\]
Multiply through by \(-1\) to simplify:
\[
5x^2 - 2x - 7 = 0
\]

Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 5 \), \( b = -2 \), and \( c = -7 \).

1. Calculate the discriminant:
\[
b^2 - 4ac = (-2)^2 - 4(5)(-7) = 4 + 140 = 144
\]

2. Apply the quadratic formula:
\[
x = \frac{-(-2) \pm \sqrt{144}}{2(5)} = \frac{2 \pm 12}{10}
\]

3. Solve for the two roots:
\[
x = \frac{2 + 12}{10} = \frac{14}{10} = \frac{7}{5}
\]
\[
x = \frac{2 - 12}{10} = \frac{-10}{10} = -1
\]

Solution:
\[
x = \frac{7}{5} \quad \text{or} \quad x = -1
\]

---

Part (g): \( x^2 + 3x + 7 = 0 \)

Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 3 \), and \( c = 7 \).

1. Calculate the discriminant:
\[
b^2 - 4ac = 3^2 - 4(1)(7) = 9 - 28 = -19
\]

Since the discriminant is negative, there are no real solutions. The solutions are complex:
\[
x = \frac{-3 \pm \sqrt{-19}}{2} = \frac{-3 \pm i\sqrt{19}}{2}
\]

Solution:
\[
x = \frac{-3 + i\sqrt{19}}{2} \quad \text{or} \quad x = \frac{-3 - i\sqrt{19}}{2}
\]

---

Part (h): \( 4x^2 + 4x + 1 = 0 \)

Factorize the quadratic equation:
\[
4x^2 + 4x + 1 = (2x + 1)^2 = 0
\]

1. Set the factor to zero:
\[
2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}
\]

Solution:
\[
x = -\frac{1}{2}
\]

---

Part (i): \( x^2 - 7x - 30 = 0 \)

Factorize the quadratic equation:
\[
x^2 - 7x - 30 = (x - 10)(x + 3) = 0
\]

1. Set each factor to zero:
\[
x - 10 = 0 \quad \Rightarrow \quad x = 10
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]

Solution:
\[
x = 10 \quad \text{or} \quad x = -3
\]

---

Part (j): \( x^2 + 4x - 96 = 0 \)

Factorize the quadratic equation:
\[
x^2 + 4x - 96 = (x + 12)(x - 8) = 0
\]

1. Set each factor to zero:
\[
x + 12 = 0 \quad \Rightarrow \quad x = -12
\]
\[
x - 8 = 0 \quad \Rightarrow \quad x = 8
\]

Solution:
\[
x = 8 \quad \text{or} \quad x = -12
\]

---

Part (k): \( x^2 + 0.9x - 0.36 = 0 \)

Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 0.9 \), and \( c = -0.36 \).

1. Calculate the discriminant:
\[
b^2 - 4ac = (0.9)^2 - 4(1)(-0.36) = 0.81 + 1.44 = 2.25
\]

2. Apply the quadratic formula:
\[
x = \frac{-0.9 \pm \sqrt{2.25}}{2(1)} = \frac{-0.9 \pm 1.5}{2}
\]

3. Solve for the two roots:
\[
x = \frac{-0.9 + 1.5}{2} = \frac{0.6}{2} = 0.3
\]
\[
x = \frac{-0.9 - 1.5}{2} = \frac{-2.4}{2} = -1.2
\]

Solution:
\[
x = 0.3 \quad \text{or} \quad x = -1.2
\]

---

Part (l): \( \frac{6}{x-1} + \frac{5}{x+1} = \frac{6}{x-2} \)

1. Find a common denominator:
\[
\frac{6(x+1)(x-2) + 5(x-1)(x-2)}{(x-1)(x+1)(x-2)} = \frac{6(x-1)(x+1)}{(x-1)(x+1)(x-2)}
\]

2. Simplify the numerators:
\[
6(x+1)(x-2) + 5(x-1)(x-2) = 6(x^2 - x - 2) + 5(x^2 - 3x + 2)
\]
\[
= 6x^2 - 6x - 12 + 5x^2 - 15x + 10
\]
\[
= 11x^2 - 21x - 2
\]

3. Equate the numerators:
\[
11x^2 - 21x - 2 = 6(x^2 - 1)
\]
\[
11x^2 - 21x - 2 = 6x^2 - 6
\]

4. Simplify:
\[
11x^2 - 21x - 2 - 6x^2 + 6 = 0
\]
\[
5x^2 - 21x + 4 = 0
\]

5. Solve the quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 5 \), \( b = -21 \), and \( c = 4 \).

\[
b^2 - 4ac = (-21)^2 - 4(5)(4) = 441 - 80 = 361
\]
\[
x = \frac{21 \pm \sqrt{361}}{2(5)} = \frac{21 \pm 19}{10}
\]

6. Solve for the two roots:
\[
x = \frac{21 + 19}{10} = \frac{40}{10} = 4
\]
\[
x = \frac{21 - 19}{10} = \frac{2}{10} = \frac{1}{5}
\]

Solution:
\[
x = 4 \quad \text{or} \quad x = \frac{1}{5}
\]

---

Part (m): \( \frac{x-1}{x-2} + \frac{x-2}{x-1} = \frac{5}{2} \)

1. Let \( y = \frac{x-1}{x-2} \). Then \( \frac{x-2}{x-1} = \frac{1}{y} \).

2. Substitute into the equation:
\[
y + \frac{1}{y} = \frac{5}{2}
\]

3. Multiply through by \( y \):
\[
y^2 + 1 = \frac{5}{2}y
\]

4. Rearrange into a quadratic equation:
\[
2y^2 - 5y + 2 = 0
\]

5. Solve the quadratic equation using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = -5 \), and \( c = 2 \).

\[
b^2 - 4ac = (-5)^2 - 4(2)(2) = 25 - 16 = 9
\]
\[
y = \frac{5 \pm \sqrt{9}}{2(2)} = \frac{5 \pm 3}{4}
\]

6. Solve for the two roots:
\[
y = \frac{5 + 3}{4} = 2
\]
\[
y = \frac{5 - 3}{4} = \frac{1}{2}
\]

7. Back-substitute \( y = \frac{x-1}{x-2} \):
- For \( y = 2 \):
\[
\frac{x-1}{x-2} = 2 \quad \Rightarrow \quad x - 1 = 2(x - 2) \quad \Rightarrow \quad x - 1 = 2x - 4 \quad \Rightarrow \quad x = 3
\]
- For \( y = \frac{1}{2} \):
\[
\frac{x-1}{x-2} = \frac{1}{2} \quad \Rightarrow \quad 2(x - 1) = x - 2 \quad \Rightarrow \quad 2x - 2 = x - 2 \quad \Rightarrow \quad x = 0
\]

Solution:
\[
x = 3 \quad \text{or} \quad x = 0
\]

---

Final Answer:

\[
\boxed{
\begin{aligned}
&\text{(a)} \quad x = 2 \quad \text{or} \quad x = \frac{7}{2} \\
&\text{(b)} \quad x = -9 \quad \text{or} \quad x = 7 \\
&\text{(c)} \quad x = 4 \quad \text{or} \quad x = 12 \\
&\text{(d)} \quad x = \frac{-7 + \sqrt{53}}{2} \quad \text{or} \quad x = \frac{-7 - \sqrt{53}}{2} \\
&\text{(e)} \quad x = 1 \quad \text{or} \quad x = -\frac{9}{2} \\
&\text{(f)} \quad x = \frac{7}{5} \quad \text{or} \quad x = -1 \\
&\text{(g)} \quad x = \frac{-3 + i\sqrt{19}}{2} \quad \text{or} \quad x = \frac{-3 - i\sqrt{19}}{2} \\
&\text{(h)} \quad x = -\frac{1}{2} \\
&\text{(i)} \quad x = 10 \quad \text{or} \quad x = -3 \\
&\text{(j)} \quad x = 8 \quad \text{or} \quad x = -12 \\
&\text{(k)} \quad x = 0.3 \quad \text{or} \quad x = -1.2 \\
&\text{(l)} \quad x = 4 \quad \text{or} \quad x = \frac{1}{5} \\
&\text{(m)} \quad x = 3 \quad \text{or} \quad x = 0 \\
\end{aligned}
}
\]