Answer :
### Part (a): Solve the Rational Inequality [tex]\(\frac{2x}{x+1} \leq \frac{1/2}\)[/tex]
1. Start by bringing all terms to one side to form a single inequality:
[tex]\[ \frac{2x}{x+1} - \frac{1}{2} \leq 0 \][/tex]
2. Find a common denominator to combine the fractions:
[tex]\[ \frac{4x}{2(x+1)} - \frac{x+1}{2(x+1)} \leq 0 \][/tex]
3. Combine the fractions:
[tex]\[ \frac{4x - (x + 1)}{2(x+1)} \leq 0 \][/tex]
[tex]\[ \frac{4x - x - 1}{2(x+1)} \leq 0 \][/tex]
[tex]\[ \frac{3x - 1}{2(x+1)} \leq 0 \][/tex]
4. Determine where the expression [tex]\(\frac{3x - 1}{2(x+1)}\)[/tex] equals zero or is undefined:
- Numerator [tex]\(3x - 1 = 0 \implies x = \frac{1}{3}\)[/tex]
- Denominator [tex]\(2(x + 1) = 0 \implies x = -1\)[/tex]
5. Identify the intervals for testing:
- Interval 1: [tex]\(x < -1\)[/tex]
- Interval 2: [tex]\(-1 < x < \frac{1}{3}\)[/tex]
- Interval 3: [tex]\(x > \frac{1}{3}\)[/tex]
6. Test these intervals in the inequality:
- For [tex]\(x < -1\)[/tex]:
[tex]\[ \text{Pick } x = -2 \implies \frac{3(-2) - 1}{2(-2 + 1)} = \frac{-6 - 1}{-2} = \frac{-7}{-2} > 0 \quad \text{(True for \(\leq 0\))} \][/tex]
- For [tex]\(-1 < x < \frac{1}{3}\)[/tex]:
[tex]\[ \text{Pick } x = 0 \implies \frac{3(0) - 1}{2(0 + 1)} = \frac{-1}{2} \leq 0 \quad \text{(True for \(\leq 0\))} \][/tex]
- For [tex]\(x > \frac{1}{3}\)[/tex]:
[tex]\[ \text{Pick } x = 1 \implies \frac{3(1) - 1}{2(1+1)} = \frac{2}{2} = 1 \quad \text{(False for \(\leq 0\))} \][/tex]
7. Combine the intervals where the inequality holds:
[tex]\[ x \in (-\infty, -1) \cup \left(-1, \frac{1}{3}\right] \][/tex]
So, the solution to the inequality is:
[tex]\[ x \in (-\infty, -1) \cup \left[-1, \frac{1}{3}\right] \][/tex]
### Part (b): Equation with Roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]
1. The roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] of the quadratic equation:
[tex]\[ -2x^2 + 2x - 7 = 0 \][/tex]
2. Solve for [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ x = \frac{1 \pm \sqrt{15} i}{2} \][/tex]
[tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are:
[tex]\[ \alpha = \frac{1}{2} - \frac{\sqrt{13} i}{2}, \quad \beta = \frac{1}{2} + \frac{\sqrt{13} i}{2} \][/tex]
3. Calculate [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
[tex]\[ \alpha^2 = \left(\frac{1}{2} - \frac{\sqrt{13} i}{2}\right)^2 \][/tex]
[tex]\[ \beta^2 = \left(\frac{1}{2} + \frac{\sqrt{13} i}{2}\right)^2 \][/tex]
4. Form the new quadratic equation with the roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
[tex]\[ (x - \alpha^2)(x - \beta^2) \][/tex]
5. Simplify the polynomial:
[tex]\[ x^2 + (6)x + \frac{49}{4} \][/tex]
So, the equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex] is:
[tex]\[ x^2 + 6x + \frac{49}{4} = 0 \][/tex]
1. Start by bringing all terms to one side to form a single inequality:
[tex]\[ \frac{2x}{x+1} - \frac{1}{2} \leq 0 \][/tex]
2. Find a common denominator to combine the fractions:
[tex]\[ \frac{4x}{2(x+1)} - \frac{x+1}{2(x+1)} \leq 0 \][/tex]
3. Combine the fractions:
[tex]\[ \frac{4x - (x + 1)}{2(x+1)} \leq 0 \][/tex]
[tex]\[ \frac{4x - x - 1}{2(x+1)} \leq 0 \][/tex]
[tex]\[ \frac{3x - 1}{2(x+1)} \leq 0 \][/tex]
4. Determine where the expression [tex]\(\frac{3x - 1}{2(x+1)}\)[/tex] equals zero or is undefined:
- Numerator [tex]\(3x - 1 = 0 \implies x = \frac{1}{3}\)[/tex]
- Denominator [tex]\(2(x + 1) = 0 \implies x = -1\)[/tex]
5. Identify the intervals for testing:
- Interval 1: [tex]\(x < -1\)[/tex]
- Interval 2: [tex]\(-1 < x < \frac{1}{3}\)[/tex]
- Interval 3: [tex]\(x > \frac{1}{3}\)[/tex]
6. Test these intervals in the inequality:
- For [tex]\(x < -1\)[/tex]:
[tex]\[ \text{Pick } x = -2 \implies \frac{3(-2) - 1}{2(-2 + 1)} = \frac{-6 - 1}{-2} = \frac{-7}{-2} > 0 \quad \text{(True for \(\leq 0\))} \][/tex]
- For [tex]\(-1 < x < \frac{1}{3}\)[/tex]:
[tex]\[ \text{Pick } x = 0 \implies \frac{3(0) - 1}{2(0 + 1)} = \frac{-1}{2} \leq 0 \quad \text{(True for \(\leq 0\))} \][/tex]
- For [tex]\(x > \frac{1}{3}\)[/tex]:
[tex]\[ \text{Pick } x = 1 \implies \frac{3(1) - 1}{2(1+1)} = \frac{2}{2} = 1 \quad \text{(False for \(\leq 0\))} \][/tex]
7. Combine the intervals where the inequality holds:
[tex]\[ x \in (-\infty, -1) \cup \left(-1, \frac{1}{3}\right] \][/tex]
So, the solution to the inequality is:
[tex]\[ x \in (-\infty, -1) \cup \left[-1, \frac{1}{3}\right] \][/tex]
### Part (b): Equation with Roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]
1. The roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] of the quadratic equation:
[tex]\[ -2x^2 + 2x - 7 = 0 \][/tex]
2. Solve for [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ x = \frac{1 \pm \sqrt{15} i}{2} \][/tex]
[tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are:
[tex]\[ \alpha = \frac{1}{2} - \frac{\sqrt{13} i}{2}, \quad \beta = \frac{1}{2} + \frac{\sqrt{13} i}{2} \][/tex]
3. Calculate [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
[tex]\[ \alpha^2 = \left(\frac{1}{2} - \frac{\sqrt{13} i}{2}\right)^2 \][/tex]
[tex]\[ \beta^2 = \left(\frac{1}{2} + \frac{\sqrt{13} i}{2}\right)^2 \][/tex]
4. Form the new quadratic equation with the roots [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex]:
[tex]\[ (x - \alpha^2)(x - \beta^2) \][/tex]
5. Simplify the polynomial:
[tex]\[ x^2 + (6)x + \frac{49}{4} \][/tex]
So, the equation whose roots are [tex]\(\alpha^2\)[/tex] and [tex]\(\beta^2\)[/tex] is:
[tex]\[ x^2 + 6x + \frac{49}{4} = 0 \][/tex]
