Let's solve the given inequations step by step using both algebraic and graphical methods.
### i) Solve \( x^2 - 8x + 15 > 0 \)
**Step 1: Factor the quadratic expression.**
We can factor \( x^2 - 8x + 15 \) as:
\[
(x - 5)(x - 3) > 0
\]
**Step 2: Find the critical points.**
The critical points are \( x = 3 \) and \( x = 5 \).
**Step 3: Determine the intervals.**
The intervals to test are:
- \( (-\infty, 3) \)
- \( (3, 5) \)
- \( (5, \infty) \)
**Step 4: Test each interval.**
- For \( x < 3 \) (e.g., \( x = 0 \)):
\((0 - 5)(0 - 3) = 15 > 0\) (True)
- For \( 3 < x < 5 \) (e.g., \( x = 4 \)):
\((4 - 5)(4 - 3) = -1 < 0\) (False)
- For \( x > 5 \) (e.g., \( x = 6 \)):
\((6 - 5)(6 - 3) = 3 > 0\) (True)
**Step 5: Write the solution.**
The solution is:
\[
x \in (-\infty, 3) \cup (5, \infty)
\]
**Graphical Method:**
- The parabola opens upwards (since the coefficient of \( x^2 \) is positive).
- The roots are at \( x = 3 \) and \( x = 5 \).
- The regions where the parabola is above the x-axis are \( (-\infty, 3) \) and \( (5, \infty) \).
### ii) Solve \( 2x^2 + 3x - 2 < 0 \)
**Step 1: Factor the quadratic expression.**
We can factor \( 2x^2 + 3x - 2 \) as:
\[
(2x - 1)(x + 2) < 0
\]
**Step 2: Find the critical points.**
The critical points are \( x = -2 \) and \( x = \frac{1}{2} \).
**Step 3: Determine the intervals.**
The intervals to test are:
- \( (-\infty, -2) \)
- \( (-2, \frac{1}{2}) \)
- \( (\frac{1}{2}, \infty) \)
**Step 4: Test each interval.**
- For \( x < -2 \) (e.g., \( x = -3 \)):
\((2(-3) - 1)(-3 + 2) = (-7)(-1) = 7 > 0\) (False)
- For \( -2 < x < \frac{1}{2} \) (e.g., \( x = 0 \)):
\((2(0) - 1)(0 + 2) = (-1)(2) = -2 < 0\) (True)
- For \( x > \frac{1}{2} \) (e.g., \( x = 1 \)):
\((2(1) - 1)(1 + 2) = (1)(3) = 3 > 0\) (False)
**Step 5: Write the solution.**
The solution is:
\[
x \in (-2, \frac{1}{2})
\]
**Graphical Method:**
- The parabola opens upwards.
- The roots are at \( x = -2 \) and \( x = \frac{1}{2} \).
- The region where the parabola is below the x-axis is \( (-2, \frac{1}{2}) \).
### iii) Solve \( x^2 - 4x + 5 > 0 \)
**Step 1: Determine the discriminant.**
The discriminant \( D = b^2 - 4ac = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0 \).
This means there are no real roots.
**Step 2: Analyze the expression.**
Since the parabola opens upwards and has no real roots, it is always positive.
**Step 3: Write the solution.**
The solution is:
\[
x \in \mathbb{R} \quad \text{(all real numbers)}
\]
**Graphical Method:**
- The parabola opens upwards and does not intersect the x-axis, remaining above it for all \( x \).
### iv) Solve \( 15x^2 + 4x - 5 \leq 0 \)
**Step 1: Find the roots using the quadratic formula.**
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 300}}{30} = \frac{-4 \pm \sqrt{316}}{30}
\]
This simplifies to:
\[
x = \frac{-4 \pm 2\sqrt{79}}{30} = \frac{-2 \pm \sqrt{79}}{15}
\]
**Step 2: Determine the intervals.**
The critical points are \( x_1 = \frac{-2 - \sqrt{79}}{15} \) and \( x_2 = \frac{-2 + \sqrt{79}}{15} \).
**Step 3: Test the intervals.**
- For \( x < x_1 \) (e.g., \( x = -2 \)):
Positive value (True)
- For \( x_1 < x < x_2 \):
Negative value (True)
- For \( x > x_2 \):
Positive value (False)
**Step 4: Write the solution.**
The solution is:
\[
x \in \left[ \frac{-2 - \sqrt{79}}{15}, \frac{-2 + \sqrt{79}}{15} \right]
\]
**Graphical Method:**
- The parabola opens upwards.
- The roots are at \( x_1 \) and \( x_2 \).
- The region where the parabola is below or on the x-axis is between the roots.
