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.