Rational equations, radical branches and systems
| English | Français |
|---|---|
| extraneous solution/ekˈstreɪnɪəs səˈluːʃn/ | extraneous solution |
| feasible region/ˈfiːzɪbl ˈriːdʒn/ | feasible region |
A decision before an answer
- A candidate created by squaring is only a candidate. The original equation can reject it.
- Your goal: Record denominator and radical restrictions before solving.
Read the relationship
- For rational expressions, identify denominator zeros before cancelling factors or multiplying an equation by a denominator. Simplification may remove a visible factor while its original restriction remains. Cancel whole factors, not selected terms in an addition.
- Check transformed-equation candidates in the original.
Which solves √(x+2)=x?
Only nonnegative 2 satisfies the original equation.
Use the defining rule
- Isolate a square root and respect its nonnegative value before squaring. Squaring preserves solutions but can add candidates, so substitute each candidate into the original. The same caution applies to multiplying by an expression that might be zero. A value that satisfies a transformed polynomial but fails the source is extraneous.
- Find and interpret simultaneous solutions or allowed regions.
(x²-1)/(x-1) has the original restriction:
The denominator is zero at 1.
Check the conditions
- A system requires the same ordered pair to satisfy every equation. Substitute a line into a circle or parabola, solve the resulting quadratic and recover the second coordinate. Linear systems may have one intersection, no intersection or coincident lines. Do not confuse two algebraic roots with two separate systems.
- Find and interpret simultaneous solutions or allowed regions.
√(x+6)=x implies x≥0. Squaring gives x²-x-6=0 with candidates 3 and -2; only 3 works. For x²+y²=25 and y=x+1, substitution gives x²+x-12=0, hence (3,4) and (-4,-3). (x²-4)/(x-2) simplifies to x+2 for x≠2, preserving the restriction.
For x+y=9 and x-y=3, x=____.
Adding gives 2x=12; y=3.
Apply the task format
- For systems of inequalities, each condition supplies an allowed region and the solution is their intersection. A point on a strict boundary is excluded even if it satisfies the other inequality. Verify a proposed point by substituting into each original inequality, including its equality sign.
- Find and interpret simultaneous solutions or allowed regions.
Restrictions survive simplification. In a simultaneous system, check both coordinates in every original relation.
Which answer fits this case?
Record denominator and radical restrictions before solving
A point satisfying one inequality must satisfy their whole system.
Every condition must hold simultaneously.
Keep the distinctions
- extraneous solution 增根 — A transformed-equation root that does not satisfy the source equation.
- feasible region 可行域 — The set satisfying all simultaneous inequality conditions.
- Record denominator and radical restrictions before solving.
- Check transformed-equation candidates in the original.
- Find and interpret simultaneous solutions or allowed regions.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.