Irrational Equations
Key Definitions
- Irrational Equation: An equation where the unknown variable appears under a radical sign (√, ∛, etc.) or is raised to a fractional power.
- Domain of Definition: The set of real values for which expressions under even radicals are non-negative. For √f(x), we must have f(x) ≥ 0.
- Extraneous Roots: Roots found during algebraic steps (like squaring) that do not satisfy the original equation. Note: a² = b² does not always mean a = b.
- Principal Root: In real numbers, the symbol √f(x) always denotes the non-negative square root.
Theory and Concepts
1. Method of Squaring (Radical Removal)
To solve √f(x) = g(x):
- Constraint Check: Ensure f(x) ≥ 0 and g(x) ≥ 0.
- Squaring: Solve f(x) = [g(x)]².
- Verification: You must check solutions in the original equation to discard extraneous roots.
2. Use of Substitution
If radicals repeat, let t = √f(x) (where t ≥ 0) to transform the irrational equation into a standard quadratic or polynomial.
3. Important Inequalities
For √f(x) + √g(x) = h(x):
- The domain is the intersection of f(x) ≥ 0 and g(x) ≥ 0.
- If h(x) < 0, there are no real solutions.
Figure 14.1: Solving √x = x – 2 graphically. Only one intersection exists.
Solved Examples
Solved Example 1.1
Problem: Solve the equation √(x + 5) = x – 1.
Solution
Square both sides: x + 5 = (x – 1)² ⇒ x + 5 = x² – 2x + 1.
x² – 3x – 4 = 0 ⇒ (x – 4)(x + 1) = 0 ⇒ x = 4, -1.
Check x = 4: √9 = 3 (Valid).
Check x = -1: √4 ≠ -2 (Invalid).
Answer: x = 4.
Solved Example 1.2
Problem: Solve x² – 5x + 2√(x² – 5x + 3) = 9.
Solution
Equation: (y² – 3) + 2y = 9 ⇒ y² + 2y – 12 = 0.
y = -1 ± √13. Since y ≥ 0, y = √13 – 1.
Substitute back: x² – 5x + 3 = (√13 – 1)² = 14 – 2√13.
x² – 5x + (2√13 – 11) = 0.
Solved Example 1.3
Problem: Solve √(3x² – 7x – 30) + √(2x² – 7x – 5) = x – 5.
Solution
Let A = √(3x² – 7x – 30) and B = √(2x² – 7x – 5).
Note: A² – B² = x² – 25 = (x – 5)(x + 5).
Given A + B = x – 5, then A – B = (A² – B²) / (A + B) = x + 5.
Adding: 2A = 2x ⇒ A = x. Squaring: x² = 3x² – 7x – 30 ⇒ 2x² – 7x – 30 = 0.
x = 6 or -2.5. Verification shows neither satisfies the original equation.
Answer: No Solution.
Solved Example 1.4
Problem: Solve ∛(x+1) + ∛(x-1) = ∛(5x).
Solution
2x + 3∛(x²-1)∛(5x) = 5x ⇒ 3∛(5x³-5x) = 3x.
5x³ – 5x = x³ ⇒ 4x³ – 5x = 0.
x = 0, ±√5/2.
Solved Example 1.5
Problem: Find the number of real roots of √(x+3-4√(x-1)) + √(x+8-6√(x-1)) = 1.
Solution
|a-2| + |a-3| = 1. This is an identity for 2 ≤ a ≤ 3.
4 ≤ x-1 ≤ 9 ⇒ 5 ≤ x ≤ 10.
Answer: Infinite roots.
Solved Example 1.6
Problem: Solve √(2x+1) + √(x-3) = 4.
Solution
Squaring: (3x-2) + 2√(2x²-5x-3) = 16 ⇒ 2√(2x²-5x-3) = 18 – 3x.
Squaring again and solving leads to x = 4 and x = 84.
Check x = 4: √9 + √1 = 3 + 1 = 4 (Correct).
Check x = 84: √169 + √81 = 13 + 9 ≠ 4 (Extraneous).
Answer: x = 4.
Solved Example 1.7
Problem: Solve √x + √(x-√(1-x)) = 1.
Solution
√(x-√(1-x)) = 1 – √x. Square: x – √(1-x) = 1 + x – 2√x.
2√x – 1 = √(1-x). Square: 4x + 1 – 4√x = 1 – x.
5x = 4√x ⇒ x(25x – 16) = 0.
Answer: x = 16/25 (x=0 is invalid).
Solved Example 1.8
Problem: Solve √(x-2) + √(4-x) = x² – 6x + 11.
Solution
Domain: x ∈ [2, 4]. Maximum of LHS is 2 (at x=3).
RHS is (x-3)² + 2. Minimum of RHS is 2 (at x=3).
Equality holds only when both sides are 2. Answer: x = 3.
Solved Example 1.9
Problem: Solve √(x²+1) – √(x²-1) = 1.
Solution
2x² – 1 = 2√(x⁴-1) ⇒ 4x⁴ – 4x² + 1 = 4x⁴ – 4.
4x² = 5 ⇒ x = ±√5/2.
Solved Example 1.10
Problem: Solve √(x+1) – √(x-1) = √(4x-1).
Solution
For x ≥ 1, (1-2x) is always negative, while 2√(x²-1) is non-negative.
Answer: No real solution.
Concept Application Exercise 1.1
Solve for x: √(2x+8) + √(x+5) = 7.
Find all real roots of x² + √(x² + 20) = 22.
Solve √(x-1) = x-3.
Solve √(5x² – 6x + 8) – √(5x² – 6x – 7) = 1.
Solve ∛x + ∛(2x-3) = ∛(3x-3).
Find the number of real solutions of √(x²-4x+3) + √(x²-9x+14) = 0.
Solve the equation √(x²-x+1) + 1/√(x²-x+1) = 2-x².
Solve √(2x²+3x+5) + √(2x²+3x+20) = 15.
Solve √(x²+ax+b) = x+c. Find the condition for a unique solution.
Solve the system: √(x+y) = 3 and √(x-y) = 1.