Question 1:( 3 marks each)     

	Solve the equations:  

	(x^2- 169)/x^2 =0

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


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

	Solve the following inequalities

     ii)    5 |(2x+3)/3|  ≤ 10 

iii)    |2x+1|≥ 13


### Question 1

#### Solve the equations:

1. \(\frac{x^2 - 169}{x^2} = 0\)

For the equation \(\frac{x^2 - 169}{x^2} = 0\), we need to set the numerator equal to zero since a fraction is zero when its numerator is zero:

\[ x^2 - 169 = 0 \]

This is a difference of squares, which can be factored:

\[ (x - 13)(x + 13) = 0 \]

Setting each factor to zero gives:

\[ x - 13 = 0 \quad \text{or} \quad x + 13 = 0 \]
\[ x = 13 \quad \text{or} \quad x = -13 \]

So, the solutions are \( x = 13 \) and \( x = -13 \).

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

The quadratic formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the equation \(2x^2 - 7x + 6 = 0\), the coefficients are \( a = 2 \), \( b = -7 \), and \( c = 6 \). Calculate the discriminant:

\[ b^2 - 4ac = (-7)^2 - 4(2)(6) = 49 - 48 = 1 \]

Applying the quadratic formula:

\[ x = \frac{-(-7) \pm \sqrt{1}}{2(2)} \]
\[ x = \frac{7 \pm 1}{4} \]

This gives two solutions:

\[ x = \frac{7 + 1}{4} = \frac{8}{4} = 2 \]
\[ x = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2} \]

So, the solutions are \( x = 2 \) and \( x = \frac{3}{2} \).

3. \(x^2 - 12x + 5 = 7\) (apply the square root property)

First, simplify the equation:

\[ x^2 - 12x + 5 = 7 \]
\[ x^2 - 12x + 5 - 7 = 0 \]
\[ x^2 - 12x - 2 = 0 \]

Complete the square:

\[ x^2 - 12x = 2 \]
\[ x^2 - 12x + 36 = 2 + 36 \]
\[ (x - 6)^2 = 38 \]

Take the square root of both sides:

\[ x - 6 = \pm \sqrt{38} \]

Thus,

\[ x = 6 \pm \sqrt{38} \]

So, the solutions are \( x = 6 + \sqrt{38} \) and \( x = 6 - \sqrt{38} \).

#### Solve the following inequalities:

1. \(5 \left| \frac{2x + 3}{3} \right| \leq 10\)

Isolate the absolute value:

\[ \left| \frac{2x + 3}{3} \right| \leq \frac{10}{5} \]
\[ \left| \frac{2x + 3}{3} \right| \leq 2 \]

This gives two inequalities to solve:

\[ -2 \leq \frac{2x + 3}{3} \leq 2 \]

Multiply through by 3:

\[ -6 \leq 2x + 3 \leq 6 \]

Subtract 3 from all parts:

\[ -9 \leq 2x \leq 3 \]

Divide by 2:

\[ -\frac{9}{2} \leq x \leq \frac{3}{2} \]

So, the solution is:

\[ -4.5 \leq x \leq 1.5 \]

2. \(|2x + 1| \geq 13\)

This gives two inequalities to solve:

\[ 2x + 1 \geq 13 \quad \text{or} \quad 2x + 1 \leq -13 \]

Solve each one:

\[ 2x + 1 \geq 13 \]
\[ 2x \geq 12 \]
\[ x \geq 6 \]

And,

\[ 2x + 1 \leq -13 \]
\[ 2x \leq -14 \]
\[ x \leq -7 \]

So, the solution is:

\[ x \leq -7 \quad \text{or} \quad x \geq 6 \]

Do you have any questions or need further details on any part?

Here are some related questions to practice:

1. Solve \( \frac{x^2 - 49}{x^2} = 0 \).
2. Solve \(3x^2 - 5x + 2 = 0\) using the quadratic formula.
3. Solve \(x^2 - 10x + 4 = 5\) using the square root property.
4. Solve \(4 \left| \frac{x - 1}{2} \right| \leq 8\).
5. Solve \(|3x - 4| \geq 11\).
6. Determine the intervals where \(2x^2 - 3x + 1 > 0\).
7. Solve the inequality \(x^2 - 4x + 3 \leq 0\).
8. Prove that the equation \(x^2 - 6x + 8 = 0\) has two real solutions.

**Tip:** When solving absolute value inequalities, always split the inequality into two separate cases, one for the positive and one for the negative scenario.

Learn how to solve quadratic equations and inequalities step-by-step. Includes solutions for equations like (x^2-169)/x^2=0, 2x^2-7x+6=0, and x^2-12x+5=7 using the quadratic formula and square root property. Also covers solving inequalities such as 5 |(2x+3)/3| ≤ 10 and |2x+1|≥ 13. Suitable for Grades 9-12.

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