Find All Values of x Satisfying the Given Conditions
When you’re handed a problem that asks you to “find all values of x” that satisfy certain conditions, you’re essentially looking for the set of numbers that make every part of the statement true at the same time. Whether the conditions involve algebraic equations, inequalities, or a mix of both, the process is systematic: isolate x, simplify, check for extraneous solutions, and then express the result in the most useful form—intervals, set notation, or a list of numbers. Below is a complete walkthrough that walks through the entire workflow, complete with examples, common pitfalls, and tips for keeping your solutions clean and error‑free Not complicated — just consistent. Which is the point..
1. Understand the Problem
Before you even touch a pencil, read the statement carefully. Identify:
- The type of conditions – Are they equations, inequalities, or both?
- The domain – Is x restricted to integers, real numbers, or a specific interval?
- Any hidden constraints – To give you an idea, a square root demands a non‑negative radicand, a denominator cannot be zero, or a logarithm requires a positive argument.
Tip: Rewrite the conditions in your own words. “Find all x such that x² − 4 ≤ 0” becomes “All real x whose square minus four is not greater than zero.”
2. Isolate x Wherever Possible
2.1. Linear Equations
For a simple linear equation like (3x + 5 = 20), move constants to the other side and divide:
[ 3x = 15 \quad\Rightarrow\quad x = 5 ]
2.2. Quadratic Equations
When you have (ax^2 + bx + c = 0), use factoring, completing the square, or the quadratic formula:
[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} ]
Example: (x^2 - 5x + 6 = 0) factors to ((x-2)(x-3)=0), giving (x=2) or (x=3).
2.3. Rational Expressions
If a fraction equals something, cross‑multiply after ensuring the denominator isn’t zero. Take this case: (\frac{2x}{x-1} = 4):
[ 2x = 4(x-1) ;\Rightarrow; 2x = 4x-4 ;\Rightarrow; -2x = -4 ;\Rightarrow; x = 2 ]
Remember to check (x \neq 1) (the original denominator).
2.4. Radical and Logarithmic Equations
- Radicals: Square both sides to eliminate the square root, then simplify. Always check for extraneous solutions introduced by squaring.
- Logarithms: Use logarithm properties to combine terms. If (\log_a(x) = \log_a(y)), then (x = y) provided (x, y > 0).
3. Solve Inequalities
Inequalities require a bit more care because you’re dealing with ranges of numbers.
3.1. Linear Inequalities
Solve like an equation, but keep the inequality sign:
[ 5x - 3 < 12 ;\Rightarrow; 5x < 15 ;\Rightarrow; x < 3 ]
If you multiply or divide by a negative number, flip the sign: (-2x \ge 6 \Rightarrow x \le -3) Worth keeping that in mind..
3.2. Quadratic Inequalities
Factor the quadratic and use a sign chart or test points:
[ x^2 - 4x - 5 \le 0 ;\Rightarrow; (x-5)(x+1) \le 0 ]
The critical points are (x = -1) and (x = 5). Test intervals:
- (x < -1): product positive → not allowed
- (-1 \le x \le 5): product non‑positive → allowed
- (x > 5): product positive → not allowed
Result: (-1 \le x \le 5).
3.3. Absolute Value Inequalities
Rewrite (|f(x)| < a) as (-a < f(x) < a), then solve the two resulting inequalities.
Example: (|2x-3| \le 5 \Rightarrow -5 \le 2x-3 \le 5)
Add 3: (-2 \le 2x \le 8)
Divide by 2: (-1 \le x \le 4).
4. Combine Multiple Conditions
When a problem presents several equations or inequalities, each must be satisfied simultaneously. Use intersection of solution sets.
Example
Find all real x such that: [ \begin{cases} x^2 - 4 \le 0 \ x - 1 > 0 \end{cases} ]
- First condition: (-2 \le x \le 2).
- Second condition: (x > 1).
The intersection is (1 < x \le 2).
Express it as:
- Interval notation: ((1, 2])
- Set notation: ({x \in \mathbb{R}\mid 1 < x \le 2})
5. Check for Extraneous Solutions
Operations like squaring, taking square roots, or cross‑multiplying can introduce numbers that satisfy the transformed equation but not the original one That's the whole idea..
Common Scenarios
| Operation | Potential Issue | Checklist |
|---|---|---|
| Squaring both sides | Introduces negative roots | Verify each candidate in the original equation |
| Cross‑multiplying | Denominator zero | Ensure no candidate makes any denominator zero |
| Taking logs | Logarithm domain | Candidate must keep argument positive |
Example: Solve (\sqrt{x+3} = x-1).
- Square: (x+3 = (x-1)^2 = x^2 - 2x + 1).
- Rearrange: (x^2 - 3x - 2 = 0).
- Factor: ((x-2)(x+1) = 0) → (x = 2) or (x = -1).
- Check:
- (x = 2): (\sqrt{5} ≈ 2.236 \neq 1) → extraneous.
- (x = -1): (\sqrt{2} ≈ 1.414) vs (-2) → extraneous.
No real solutions exist Practical, not theoretical..
6. Express the Final Answer Clearly
Depending on the context, choose the most useful format:
- Interval notation for continuous ranges.
- Set notation for precise definitions.
- List of numbers when the solution set is finite.
- Graphical depiction (optional) to illustrate the solution on the number line.
Always include any domain restrictions you identified earlier That alone is useful..
7. Worked Example: A Multi‑Step Problem
Problem: Find all real x that satisfy: [ \begin{cases} 2x^2 - 8x + 6 \ge 0 \ \sqrt{x-1} \le 3 \end{cases} ]
Step 1: Solve the quadratic inequality
Factor: [ 2x^2 - 8x + 6 = 2(x^2 - 4x + 3) = 2(x-1)(x-3) ]
Critical points: (x = 1) and (x = 3).
Test intervals:
- (x < 1): choose (x=0): (2(0-1)(0-3) = 6 > 0) → satisfies.
- (1 < x < 3): choose (x=2): (2(2-1)(2-3) = -2 < 0) → does not satisfy.
- (x > 3): choose (x=4): (2(4-1)(4-3) = 6 > 0) → satisfies.
Thus first condition → (x \in (-\infty, 1] \cup [3, \infty)).
Step 2: Solve the radical inequality
[ \sqrt{x-1} \le 3 \quad\Rightarrow\quad x-1 \le 9 \quad\Rightarrow\quad x \le 10 ]
Also, domain of the square root: (x-1 \ge 0 \Rightarrow x \ge 1) And that's really what it comes down to. But it adds up..
So second condition → (x \in [1, 10]).
Step 3: Intersect the two sets
[ (-\infty, 1] \cup [3, \infty) ;\cap; [1, 10] = [1, 1] \cup [3, 10] = {1} \cup [3, 10] ]
Step 4: Final answer
[ x = 1 \quad\text{or}\quad 3 \le x \le 10 ]
In interval notation: ({1} \cup [3,10]).
8. FAQ
| Question | Answer |
|---|---|
| How do I solve a system of inequalities? | Treat each inequality separately, find its solution set, then intersect all sets. Because of that, |
| **What if the solution set is empty? In practice, ** | State “No real solutions” or “∅”. |
| Can I use a graphing calculator? | Yes, but always verify algebraically; a graph can mislead near points of discontinuity. |
| What if the problem asks for integer solutions only? | After finding the continuous solution set, list the integers that fall within it. In practice, |
| **How to handle absolute value inequalities? ** | Split into two inequalities, solve each, then intersect. |
9. Conclusion
Finding all values of x that satisfy given conditions is a blend of algebraic manipulation, careful domain checking, and logical intersection of solution sets. By following a structured approach—understanding the problem, isolating x, solving each condition, checking for extraneous solutions, and combining results—you can tackle even the most complex systems with confidence. Remember to always double‑check your work; the most elegant algebraic path can still lead to a subtle mistake if a domain restriction slips through. Armed with these strategies, you’ll be ready to solve any “find all x” problem that comes your way.
