Question 1:( 3 marks each)
a)	Solve the equations:
i)	(x^2- 169)/x^2 =0

ii)	2x^2-7x+6=0 (use the quadratic formula)

iii)	x^2-12x+5 =7 (Apply the square root property)
b)	Solve the following inequalities
ii)    5 |(2x+3)/3|  ≤ 10

iii)    |2x+1|≥ 13

Let's solve each of the equations and inequalities step by step.

### Part a) Solve the equations:

#### i) \(\frac{x^2 - 169}{x^2} = 0\)

To solve this equation:
1. Multiply both sides by \(x^2\) to eliminate the denominator:
   \[
   x^2 - 169 = 0
   \]
2. Add 169 to both sides:
   \[
   x^2 = 169
   \]
3. Take the square root of both sides:
   \[
   x = \pm \sqrt{169}
   \]
   \[
   x = \pm 13
   \]
So, the solutions are \(x = 13\) and \(x = -13\).

#### ii) \(2x^2 - 7x + 6 = 0\) (Use the quadratic formula)

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 2\), \(b = -7\), and \(c = 6\).

1. Calculate the discriminant:
   \[
   \Delta = b^2 - 4ac = (-7)^2 - 4(2)(6) = 49 - 48 = 1
   \]
2. Substitute the values into the quadratic formula:
   \[
   x = \frac{-(-7) \pm \sqrt{1}}{4} = \frac{7 \pm 1}{4}
   \]
3. Solve for the two possible values of \(x\):
   \[
   x_1 = \frac{7 + 1}{4} = \frac{8}{4} = 2
   \]
   \[
   x_2 = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2}
   \]
So, the solutions are \(x = 2\) and \(x = \frac{3}{2}\).

#### iii) \(x^2 - 12x + 5 = 7\) (Apply the square root property)

1. Subtract 7 from both sides to set the equation to zero:
   \[
   x^2 - 12x - 2 = 0
   \]
2. Complete the square:
   \[
   x^2 - 12x = 2
   \]
   \[
   x^2 - 12x + \left(\frac{12}{2}\right)^2 = 2 + \left(\frac{12}{2}\right)^2
   \]
   \[
   x^2 - 12x + 36 = 38
   \]
   \[
   (x - 6)^2 = 38
   \]
3. Apply the square root property:
   \[
   x - 6 = \pm \sqrt{38}
   \]
   \[
   x = 6 \pm \sqrt{38}
   \]
So, the solutions are \(x = 6 + \sqrt{38}\) and \(x = 6 - \sqrt{38}\).

### Part b) Solve the inequalities:

#### ii) \(5 \left|\frac{2x+3}{3}\right| \leq 10\)

1. Divide both sides by 5:
   \[
   \left|\frac{2x+3}{3}\right| \leq 2
   \]
2. Set up the two cases for the absolute value:
   - Case 1: \(\frac{2x+3}{3} \leq 2\)
   - Case 2: \(\frac{2x+3}{3} \geq -2\)

3. Solve each case:
   - Case 1:
     \[
     2x + 3 \leq 6
     \]
     \[
     2x \leq 3 \quad \Rightarrow \quad x \leq \frac{3}{2}
     \]
   - Case 2:
     \[
     2x + 3 \geq -6
     \]
     \[
     2x \geq -9 \quad \Rightarrow \quad x \geq -\frac{9}{2}
     \]

4. Combine the results:
   \[
   -\frac{9}{2} \leq x \leq \frac{3}{2}
   \]
   So, the solution is \(x \in \left[-\frac{9}{2}, \frac{3}{2}\right]\).

#### iii) \(\left|2x + 1\right| \geq 13\)

1. Set up the two cases for the absolute value:
   - Case 1: \(2x + 1 \geq 13\)
   - Case 2: \(2x + 1 \leq -13\)

2. Solve each case:
   - Case 1:
     \[
     2x \geq 12 \quad \Rightarrow \quad x \geq 6
     \]
   - Case 2:
     \[
     2x \leq -14 \quad \Rightarrow \quad x \leq -7
     \]

3. Combine the results:
   \[
   x \geq 6 \quad \text{or} \quad x \leq -7
   \]
   So, the solution is \(x \in (-\infty, -7] \cup [6, \infty)\).

---

Do you want any more details or have any questions? Here are some related questions:

1. How do you complete the square in quadratic equations?
2. What is the discriminant, and how does it affect the roots of a quadratic equation?
3. How do you solve quadratic equations by factoring?
4. What is the square root property and how is it used?
5. How do you solve absolute value inequalities?
6. What are the properties of absolute value functions?
7. How do you graph quadratic equations and inequalities?
8. Can all quadratic equations be solved using the quadratic formula?

**Tip:** When solving absolute value inequalities, always consider both cases (positive and negative) to get the complete solution.

Learn how to solve quadratic equations and inequalities step-by-step with detailed explanations. This problem covers quadratic equations using the quadratic formula and solving absolute value inequalities. Suitable for students in Grades 9-12.

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