Math Problem Statement

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

Solution

Let's go through each problem one by one:

Problem 3:

You are given a right triangle ABC\triangle ABC with:

  • AB=2mnAB = 2mn,
  • BC=m2n2BC = m^2 - n^2,
  • AC=m2+n2AC = m^2 + n^2.

This setup uses a parameterization common in generating Pythagorean triples, where: AB2+BC2=AC2AB^2 + BC^2 = AC^2 You can verify if this holds by showing that (2mn)2+(m2n2)2=(m2+n2)2(2mn)^2 + (m^2 - n^2)^2 = (m^2 + n^2)^2.

Let's check it:

  1. AB2=(2mn)2=4m2n2AB^2 = (2mn)^2 = 4m^2n^2.
  2. BC2=(m2n2)2=m42m2n2+n4BC^2 = (m^2 - n^2)^2 = m^4 - 2m^2n^2 + n^4.
  3. AC2=(m2+n2)2=m4+2m2n2+n4AC^2 = (m^2 + n^2)^2 = m^4 + 2m^2n^2 + n^4.

Adding AB2AB^2 and BC2BC^2: AB2+BC2=4m2n2+m42m2n2+n4=m4+2m2n2+n4=AC2AB^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: 3x7x4<2\frac{3x - 7}{x - 4} < 2 for xRx \in \mathbb{R} and x4x \neq 4.

  1. Rewrite the inequality: 3x7x42<0\frac{3x - 7}{x - 4} - 2 < 0
  2. Combine terms to get a single fraction: 3x72(x4)x4<0\frac{3x - 7 - 2(x - 4)}{x - 4} < 0
  3. Simplify the numerator: =3x72x+8x4=x+1x4<0= \frac{3x - 7 - 2x + 8}{x - 4} = \frac{x + 1}{x - 4} < 0
  4. Now, find intervals where x+1x4<0\frac{x + 1}{x - 4} < 0:
    • The critical points are x=1x = -1 and x=4x = 4.
    • Test intervals around these points to find where the expression is negative.

The solution is: 1<x<4-1 < x < 4

Problem 5:

Verify that the quadratic equation 2x2x+3=02x^2 - x + 3 = 0 has no real roots.

  1. Use the discriminant, Δ=b24ac\Delta = b^2 - 4ac: Δ=(1)2423=124=23\Delta = (-1)^2 - 4 \cdot 2 \cdot 3 = 1 - 24 = -23
  2. Since the discriminant is negative (Δ<0\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?


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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Algebra
Inequalities
Quadratic Equations

Formulas

Pythagorean theorem: a^2 + b^2 = c^2
Inequality solving techniques
Discriminant formula for quadratic equations: Δ = b^2 - 4ac

Theorems

Pythagorean Theorem
Quadratic Formula

Suitable Grade Level

Grades 9-12