Free Printable WorksheetMathAlgebra IIMath CurriculumALG2.149

ALG2.149 Checking Extraneous Solutions Systematically Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Algebra II Math worksheet provides structured practice on ALG2.149 Checking Extraneous Solutions Systematically. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

ALG2.149 Checking Extraneous Solutions Systematically - Practice questions for Checking Extraneous Solutions Systematically.

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Algebra II

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related ALG2.149 Checking Extraneous Solutions Systematically skill assessments:

Problem 1Difficulty: Mastery

Equation: √(x+16)=∛(x+8)+2 Candidates: x=0, x=-7, x=-20. After checking, x=-20 is rejected. Which statement best explains why?

ABecause cube roots are not real for negative inputs.

BBecause x=-20 makes x+8 negative, so ∛(x+8) is not real.

CBecause x=-20 makes the right side equal to 0.

DBecause x=-20 makes x+16 negative, so √(x+16) is not a real number.

Solution & Explanation:Substitute x=-20 into the original equation. A required radical is not real (a radicand becomes negative), so x cannot be a solution. The best explanation is D: Because x=-20 makes x+16 negative, so √(x+16) is not a real number.

Problem 2Difficulty: Mastery

Original equation: ∜(x+81)=∛(x+8)+1 Candidate values: x=0, x=56. Which candidate must be rejected as extraneous? Enter the value.

Solution & Explanation:x=0: LHS=3, RHS=3 → solution. x=56: LHS=3.421, RHS=5 → not a solution. Therefore, the extraneous value is 56.

Problem 3Difficulty: Mastery

Equation: √(x+36)=∛(x+27)+3 Candidates: x=0, x=-27, x=8. After checking, x=8 is rejected. Which statement best explains why?

ABecause x=8 makes x+27 negative.

BAt x=8, LHS=√44 while RHS=∛35+3, so the sides are not equal.

CBecause x=8 is not in the domain of the equation.

DBecause cube roots must be integers to be valid.

Solution & Explanation:Substitute x=8 into the original equation. You get LHS=6.633 and RHS=6.271, which are not equal, so x=8 must be rejected. The best explanation is B: At x=8, LHS=√44 while RHS=∛35+3, so the sides are not equal.

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Algebra II Math students. It aligns with educational standard ALG2.149 (Practice questions for Checking Extraneous Solutions Systematically.).