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Let's go through each problem one by one:

### Problem 3:
You are given a right triangle \( \triangle ABC \) with:
- \( AB = 2mn \),
- \( BC = m^2 - n^2 \),
- \( AC = m^2 + n^2 \).

This setup uses a parameterization common in generating Pythagorean triples, where:
\[
AB^2 + BC^2 = AC^2
\]
You can verify if this holds by showing that \( (2mn)^2 + (m^2 - n^2)^2 = (m^2 + n^2)^2 \).

Let's check it:
1. \( AB^2 = (2mn)^2 = 4m^2n^2 \).
2. \( BC^2 = (m^2 - n^2)^2 = m^4 - 2m^2n^2 + n^4 \).
3. \( AC^2 = (m^2 + n^2)^2 = m^4 + 2m^2n^2 + n^4 \).

Adding \( AB^2 \) and \( BC^2 \):
\[
AB^2 + BC^2 = 4m^2n^2 + m^4 - 2m^2n^2 + n^4 = m^4 + 2m^2n^2 + n^4 = AC^2
\]
Thus, the triangle is indeed a right triangle as expected.

### Problem 4:
Solve the inequality:
\[
\frac{3x - 7}{x - 4} < 2
\]
for \( x \in \mathbb{R} \) and \( x \neq 4 \).

1. Rewrite the inequality:
   \[
   \frac{3x - 7}{x - 4} - 2 < 0
   \]
2. Combine terms to get a single fraction:
   \[
   \frac{3x - 7 - 2(x - 4)}{x - 4} < 0
   \]
3. Simplify the numerator:
   \[
   = \frac{3x - 7 - 2x + 8}{x - 4} = \frac{x + 1}{x - 4} < 0
   \]
4. Now, find intervals where \( \frac{x + 1}{x - 4} < 0 \):
   - The critical points are \( x = -1 \) and \( x = 4 \).
   - Test intervals around these points to find where the expression is negative.

The solution is:
\[
-1 < x < 4
\]

### Problem 5:
Verify that the quadratic equation \( 2x^2 - x + 3 = 0 \) has no real roots.

1. Use the discriminant, \( \Delta = b^2 - 4ac \):
   \[
   \Delta = (-1)^2 - 4 \cdot 2 \cdot 3 = 1 - 24 = -23
   \]
2. Since the discriminant is negative (\( \Delta < 0 \)), the quadratic equation has no real roots.

Would you like more details on any of these steps, or do you have specific questions?

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### Related Questions:
1. How can you derive the formulas for Pythagorean triples?
2. What method could you use to check inequalities with more complex terms?
3. How does the discriminant help determine the nature of roots in quadratic equations?
4. What are the possible solutions if an inequality involves an absolute value?
5. Can Pythagorean triples be generalized for non-integer solutions?

### Tip:
For inequalities with fractions, consider setting up intervals based on the points where the denominator or numerator becomes zero.

Solve the problems involving right triangles, inequalities, and quadratic equations.

Explore solutions to problems involving right triangles with Pythagorean triples, solving inequalities, and verifying the absence of real roots for a quadratic equation. Includes step-by-step methods using algebraic techniques and the discriminant for quadratic equations. Ideal for high school students in Grades 9-12.

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