Tasks 02-04 - Fractional, Radical & Cubic Equations
Section 02: Equations & Problem-Solving Strategies
Instructions
Complete these problems to master solving fractional, radical, and cubic equations. Pay special attention to domain restrictions and checking for extraneous solutions.
Problem 1: Domain Restrictions (x)
For each rational equation, identify the domain restrictions BEFORE solving:
\(\frac{5}{x-3} = 2\)
\(\frac{x}{x+2} + \frac{3}{x-1} = 1\)
\(\frac{2x}{x^2-4} = \frac{1}{x-2}\)
\(\frac{1}{x} + \frac{1}{x^2} = \frac{1}{2}\)
\(\frac{x+1}{x-3} = \frac{x-2}{x+3}\)
- \(\frac{5}{x-3} = 2\)
- Domain restriction: \(x \neq 3\)
- Solve: \(5 = 2(x-3)\)
- \(5 = 2x - 6\)
- \(11 = 2x\)
- \(x = 5.5\) (valid, since \(5.5 \neq 3\))
- \(\frac{x}{x+2} + \frac{3}{x-1} = 1\)
- Domain restrictions: \(x \neq -2, x \neq 1\)
- LCD: \((x+2)(x-1)\)
- \(x(x-1) + 3(x+2) = (x+2)(x-1)\)
- \(x^2 - x + 3x + 6 = x^2 + x - 2\)
- \(2x + 6 = x - 2\)
- \(x = -8\) (valid)
- \(\frac{2x}{x^2-4} = \frac{1}{x-2}\)
- Factor denominator: \(x^2 - 4 = (x-2)(x+2)\)
- Domain restrictions: \(x \neq 2, x \neq -2\)
- Simplify: \(\frac{2x}{(x-2)(x+2)} = \frac{1}{x-2}\)
- Cross multiply: \(2x = x + 2\)
- \(x = 2\) (INVALID - violates domain!)
- No solution
- \(\frac{1}{x} + \frac{1}{x^2} = \frac{1}{2}\)
- Domain restriction: \(x \neq 0\)
- LCD: \(2x^2\)
- \(2x + 2 = x^2\)
- \(x^2 - 2x - 2 = 0\)
- \(x = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}\)
- Both solutions valid (neither equals 0)
- \(\frac{x+1}{x-3} = \frac{x-2}{x+3}\)
- Domain restrictions: \(x \neq 3, x \neq -3\)
- Cross multiply: \((x+1)(x+3) = (x-2)(x-3)\)
- \(x^2 + 4x + 3 = x^2 - 5x + 6\)
- \(9x = 3\)
- \(x = \frac{1}{3}\) (valid)
Problem 2: Solving Rational Equations (x)
Solve each equation, checking domain restrictions:
\(\frac{4}{x} - \frac{3}{2x} = \frac{1}{4}\)
\(\frac{x+2}{x-1} - \frac{x-1}{x+2} = 0\)
\(\frac{2}{x-3} + \frac{x}{x+3} = \frac{12}{x^2-9}\)
\(\frac{1}{x-2} + \frac{2}{x+1} = \frac{3}{x-2}\)
- \(\frac{4}{x} - \frac{3}{2x} = \frac{1}{4}\)
- Domain: \(x \neq 0\)
- LCD: \(4x\)
- \(16 - 6 = x\)
- \(x = 10\) (valid)
- \(\frac{x+2}{x-1} - \frac{x-1}{x+2} = 0\)
- Domain: \(x \neq 1, x \neq -2\)
- \(\frac{x+2}{x-1} = \frac{x-1}{x+2}\)
- Cross multiply: \((x+2)^2 = (x-1)^2\)
- \(x^2 + 4x + 4 = x^2 - 2x + 1\)
- \(6x = -3\)
- \(x = -\frac{1}{2}\) (valid)
- \(\frac{2}{x-3} + \frac{x}{x+3} = \frac{12}{x^2-9}\)
- Note: \(x^2 - 9 = (x-3)(x+3)\)
- Domain: \(x \neq 3, x \neq -3\)
- LCD: \((x-3)(x+3)\)
- \(2(x+3) + x(x-3) = 12\)
- \(2x + 6 + x^2 - 3x = 12\)
- \(x^2 - x - 6 = 0\)
- \((x-3)(x+2) = 0\)
- \(x = 3\) (invalid) or \(x = -2\) (valid)
- Solution: \(x = -2\)
- \(\frac{1}{x-2} + \frac{2}{x+1} = \frac{3}{x-2}\)
- Domain: \(x \neq 2, x \neq -1\)
- Simplify: \(\frac{2}{x+1} = \frac{3-1}{x-2} = \frac{2}{x-2}\)
- Therefore: \(x + 1 = x - 2\)
- \(1 = -2\) (impossible)
- No solution
Problem 3: Basic Radical Equations (x)
Solve each radical equation and check for extraneous solutions:
\(\sqrt{x + 5} = 3\)
\(\sqrt{2x - 1} = x - 2\)
\(\sqrt{x^2 + 3} = 2\)
\(x = \sqrt{x + 6}\)
\(\sqrt{4x + 1} = 2x - 1\)
- \(\sqrt{x + 5} = 3\)
- Square: \(x + 5 = 9\)
- \(x = 4\)
- Check: \(\sqrt{4 + 5} = \sqrt{9} = 3\) ✓
- \(\sqrt{2x - 1} = x - 2\)
- Square: \(2x - 1 = x^2 - 4x + 4\)
- \(x^2 - 6x + 5 = 0\)
- \((x - 5)(x - 1) = 0\)
- \(x = 5\) or \(x = 1\)
- Check \(x = 5\): \(\sqrt{9} = 3\) and \(5 - 2 = 3\) ✓
- Check \(x = 1\): \(\sqrt{1} = 1\) but \(1 - 2 = -1\) ✗
- Solution: \(x = 5\)
- \(\sqrt{x^2 + 3} = 2\)
- Square: \(x^2 + 3 = 4\)
- \(x^2 = 1\)
- \(x = \pm 1\)
- Check both: \(\sqrt{1 + 3} = 2\) ✓
- Solutions: \(x = 1\) or \(x = -1\)
- \(x = \sqrt{x + 6}\)
- Square: \(x^2 = x + 6\)
- \(x^2 - x - 6 = 0\)
- \((x - 3)(x + 2) = 0\)
- \(x = 3\) or \(x = -2\)
- Check \(x = 3\): \(3 = \sqrt{9} = 3\) ✓
- Check \(x = -2\): \(-2 \neq \sqrt{4} = 2\) ✗
- Solution: \(x = 3\)
- \(\sqrt{4x + 1} = 2x - 1\)
- Square: \(4x + 1 = 4x^2 - 4x + 1\)
- \(4x^2 - 8x = 0\)
- \(4x(x - 2) = 0\)
- \(x = 0\) or \(x = 2\)
- Check \(x = 0\): \(\sqrt{1} = 1\) but \(-1 \neq 1\) ✗
- Check \(x = 2\): \(\sqrt{9} = 3\) and \(4 - 1 = 3\) ✓
- Solution: \(x = 2\)
Problem 4: Multiple Radicals (xx)
Solve equations with multiple radical terms:
\(\sqrt{x + 3} + \sqrt{x} = 3\)
\(\sqrt{2x + 5} - \sqrt{x + 2} = 1\)
\(\sqrt{x + 8} = 2 + \sqrt{x}\)
- \(\sqrt{x + 3} + \sqrt{x} = 3\)
- Isolate: \(\sqrt{x + 3} = 3 - \sqrt{x}\)
- Square: \(x + 3 = 9 - 6\sqrt{x} + x\)
- \(3 = 9 - 6\sqrt{x}\)
- \(6\sqrt{x} = 6\)
- \(\sqrt{x} = 1\), so \(x = 1\)
- Check: \(\sqrt{4} + \sqrt{1} = 2 + 1 = 3\) ✓
- \(\sqrt{2x + 5} - \sqrt{x + 2} = 1\)
- Isolate: \(\sqrt{2x + 5} = 1 + \sqrt{x + 2}\)
- Square: \(2x + 5 = 1 + 2\sqrt{x + 2} + x + 2\)
- \(x + 2 = 2\sqrt{x + 2}\)
- \(\frac{x + 2}{2} = \sqrt{x + 2}\)
- Square again: \(\frac{(x + 2)^2}{4} = x + 2\)
- \((x + 2)^2 = 4(x + 2)\)
- \((x + 2)(x + 2 - 4) = 0\)
- \((x + 2)(x - 2) = 0\)
- \(x = -2\) (makes \(\sqrt{x + 2} = 0\)) or \(x = 2\)
- Check \(x = 2\): \(\sqrt{9} - \sqrt{4} = 3 - 2 = 1\) ✓
- Check \(x = -2\): \(\sqrt{1} - 0 = 1\) ✓
- Solutions: \(x = -2\) or \(x = 2\)
- \(\sqrt{x + 8} = 2 + \sqrt{x}\)
- Square: \(x + 8 = 4 + 4\sqrt{x} + x\)
- \(4 = 4\sqrt{x}\)
- \(\sqrt{x} = 1\)
- \(x = 1\)
- Check: \(\sqrt{9} = 3\) and \(2 + 1 = 3\) ✓
Problem 5: Cubic Factoring (xx)
Factor and solve each cubic equation:
\(x^3 + x^2 - 6x = 0\)
\(x^3 - 3x^2 - 4x + 12 = 0\)
\(x^3 + 3x^2 - 13x - 15 = 0\)
- \(x^3 + x^2 - 6x = 0\)
- Factor out x: \(x(x^2 + x - 6) = 0\)
- \(x(x + 3)(x - 2) = 0\)
- Solutions: \(x = 0, -3, 2\)
- \(x^3 - 3x^2 - 4x + 12 = 0\)
- Group: \(x^2(x - 3) - 4(x - 3) = 0\)
- \((x - 3)(x^2 - 4) = 0\)
- \((x - 3)(x - 2)(x + 2) = 0\)
- Solutions: \(x = -2, 2, 3\)
- \(x^3 + 3x^2 - 13x - 15 = 0\)
- Try: Test \(x = 3\)
- \(27 + 27 - 39 - 15 = 0\) ✓
- Factor: \((x - 3)(x^2 + 6x + 5) = 0\)
- \((x - 3)(x + 5)(x + 1) = 0\)
- Solutions: \(x = -5, -1, 3\)
Problem 6: Work Rate Problem (xx)
Two pipes can fill a tank together in 4 hours. The larger pipe alone can fill it 3 hours faster than the smaller pipe alone.
- Set up the equation with appropriate variables
- Identify domain restrictions
- Solve for the individual filling times
- Verify your answer makes practical sense
- Setup:
- Let \(x\) = hours for smaller pipe alone
- Then \(x - 3\) = hours for larger pipe alone
- Smaller pipe rate: \(\frac{1}{x}\) tanks/hour
- Larger pipe rate: \(\frac{1}{x-3}\) tanks/hour
- Combined rate: \(\frac{1}{4}\) tanks/hour
- Equation: \(\frac{1}{x} + \frac{1}{x-3} = \frac{1}{4}\)
- Domain restrictions:
- \(x \neq 0\) and \(x \neq 3\)
- Also need \(x > 3\) for practical sense (larger pipe is faster)
- Solve:
- LCD: \(4x(x-3)\)
- \(4(x-3) + 4x = x(x-3)\)
- \(4x - 12 + 4x = x^2 - 3x\)
- \(x^2 - 11x + 12 = 0\)
- Using quadratic formula: \(x = \frac{11 \pm \sqrt{121 - 48}}{2} = \frac{11 \pm \sqrt{73}}{2}\)
- \(x \approx 9.77\) or \(x \approx 1.23\)
- Since we need \(x > 3\): \(x \approx 9.77\) hours
- Verify:
- Smaller pipe: 9.77 hours alone
- Larger pipe: 6.77 hours alone
- Combined rate: \(\frac{1}{9.77} + \frac{1}{6.77} \approx 0.102 + 0.148 = 0.25 = \frac{1}{4}\) ✓
Problem 7: Investment Returns (xxx)
An investor divides €10,000 between two funds. Fund A returns interest at rate \(r\)%, while Fund B returns at rate \((r+2)\)%. After one year, the total interest earned is €650, if the amount in Fund B is €2,000 more than in Fund A:
- Set up equations for the investment amounts and returns
- Find the interest rates for both funds
- Calculate the exact amounts invested in each fund
- What would the total return be if all money was in Fund B?
- Setup:
- Let \(x\) = amount in Fund A
- Then \(x + 2000\) = amount in Fund B
- Total: \(x + (x + 2000) = 10000\), so \(x = 4000\)
- Fund A has €4,000, Fund B has €6,000
- Interest equation: \(\frac{r}{100}(4000) + \frac{r+2}{100}(6000) = 650\)
- Find rates:
- \(40r + 60(r + 2) = 650\)
- \(40r + 60r + 120 = 650\)
- \(100r = 530\)
- \(r = 5.3\%\)
- Fund A rate: 5.3%
- Fund B rate: 7.3%
- Verify amounts:
- Fund A: €4,000 at 5.3% = €212
- Fund B: €6,000 at 7.3% = €438
- Total interest: €212 + €438 = €650 ✓
- All in Fund B:
- €10,000 at 7.3% = €730
- Additional return: €730 - €650 = €80 more