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

### Question 1(a) - Solving Equations

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

First, we can simplify the equation by eliminating the denominator:
\[
\frac{x^2 - 169}{x^2} = 0 \implies x^2 - 169 = 0
\]
Next, solve for \(x\):
\[
x^2 - 169 = 0 \implies x^2 = 169 \implies x = \pm \sqrt{169} \implies x = \pm 13
\]

**Solution:** \(x = 13\) or \(x = -13\)

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

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \(2x^2 - 7x + 6 = 0\), we have \(a = 2\), \(b = -7\), and \(c = 6\). Plugging these values into the quadratic formula:
\[
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2} = \frac{7 \pm \sqrt{49 - 48}}{4} = \frac{7 \pm 1}{4}
\]
This gives us two solutions:
\[
x = \frac{7 + 1}{4} = 2 \quad \text{and} \quad x = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2}
\]

**Solution:** \(x = 2\) or \(x = \frac{3}{2}\)

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

First, simplify the equation:
\[
x^2 - 12x + 5 = 7 \implies x^2 - 12x + 5 - 7 = 0 \implies x^2 - 12x - 2 = 0
\]
This is a quadratic equation, and it can be solved using the quadratic formula:
\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 + 8}}{2} = \frac{12 \pm \sqrt{152}}{2} = \frac{12 \pm 2\sqrt{38}}{2} = 6 \pm \sqrt{38}
\]

**Solution:** \(x = 6 + \sqrt{38}\) or \(x = 6 - \sqrt{38}\)

### Question 1(b) - Solving Inequalities

#### ii) \(5 \left|\frac{2x+3

Explore step-by-step solutions for solving quadratic equations and inequalities. Learn how to apply the quadratic formula and the square root property to solve complex mathematical problems. Suitable for students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation