This question tests solving x² = p and x³ = p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x² = p (for example, if x² = 49, then x = √49 = 7 or x = -√49 = -7, both since (±7)² = 49), while the cube root ∛p solves x³ = p (for example, if x³ = 64, then x = ∛64 = 4 since 4³ = 64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as √2 ≈ 1.414... with non-repeating decimals. For x² = 1/16, taking square roots gives x = ±√(1/16), and since √(1/16) = 1/4 (because (1/4)² = 1/16), x = ±1/4. This is correct because 1/16 is a perfect square (fractionally), and both positive and negative satisfy the equation. A common error is choosing ±1/16, confusing the root with squaring again, or forgetting the negative solution. The strategy is to (1) identify x² operation, (2) apply √ to both sides, (3) check if 1/16 is a perfect square (1/4² = 1/16), (4) include ± for solutions, and (5) recognize rational since perfect. Common mistakes include forgetting negative for x² = p, using ∛ wrongly, or claiming √2 rational.

This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (for example, if $x^2 = 49$, then $x = \sqrt{49} = 7$ or $x = -\sqrt{49} = -7$, both since $(\pm 7)^2 = 49$), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (for example, if $x^3 = 64$, then $x = \sqrt[3]{64} = 4$ since $4^3 = 64$, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as $\sqrt{2} \approx 1.414\ldots$ with non-repeating decimals. For a square with area 50 cm$^2$, the side is $\sqrt{50}$ cm, and since $\sqrt{49} = 7$ and $50 > 49$, $\sqrt{50} > 7$. This is correct because the square root function is increasing, so a larger input gives a larger output, and 50 is not a perfect square, making $\sqrt{50}$ irrational. A common error is thinking $\sqrt{50} < 7$ because of miscomparing 50 and 49, or approximating incorrectly like saying it's greater than 8. The strategy is to (1) identify the square root context, (2) apply $\sqrt{}$ to 50, (3) compare to nearby perfect squares ($7^2 = 49$, $8^2 = 64$), (4) note no $\pm$ for length, and (5) recognize irrational for non-perfect. Common mistakes include forgetting irrationality, using wrong comparisons, or claiming $\sqrt{2}$ rational.

This question tests solving x² = p and x³ = p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x² = p (if x² = 49, then x = √49 = 7 or x = -√49 = -7, both since (±7)² = 49), while the cube root ∛p solves x³ = p (if x³ = 64, then x = ∛64 = 4 since 4³ = 64, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots (√2 ≈ 1.414... non-repeating). For a square with area 20 cm², the side length is √20 cm, the positive square root. This is correct because side length is a physical measurement that must be positive, and area = side², so side = √area. A common error is including the negative ±√20 or confusing with √10. To solve, (1) identify it's x² = 20 for side x, (2) apply √, (3) check 20 not perfect (between 4²=16 and 5²=25), (4) use positive for length, (5) recognize irrational since non-perfect. Mistakes include adding negative for length, using wrong value like 20, or claiming rational.

This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (for example, if $x^2 = 49$, then $x = \sqrt{49} = 7$ or $x = -\sqrt{49} = -7$, both since $(\pm 7)^2 = 49$), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (for example, if $x^3 = 64$, then $x = \sqrt[3]{64} = 4$ since $4^3 = 64$, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as $\sqrt{2} \approx 1.414\ldots$ with non-repeating decimals. For a cube with volume 216 cm³, the side length is $\sqrt[3]{216}$, and since $6^3 = 216$, it is 6 cm. This is correct because volume of a cube is side³, so cube root gives the side, and 216 is a perfect cube. A common error is confusing with square root and choosing 36 or miscalculating $\sqrt[3]{216}$ as 8 (since $8^3 = 512$), or adding ± for length. The strategy is to (1) identify $x^3$ for volume, (2) apply $\sqrt[3]{}$ to 216, (3) check if perfect cube (memorize up to $6^3 = 216$), (4) note no ± for lengths, and (5) recognize rational for perfect cubes. Common mistakes include using $\sqrt{}$ instead of $\sqrt[3]{}$, forgetting negatives in equations, or claiming $\sqrt{2}$ rational.

This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $√p$ solves $x^2 = p$ (if $x^2 = 49$, then $x = √49 = 7$ or $x = -√49 = -7$, both since $(±7)^2 = 49$), while the cube root $∛p$ solves $x^3 = p$ (if $x^3 = 64$, then $x = ∛64 = 4$ since $4^3 = 64$, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots ($√2 ≈ 1.414...$ non-repeating). The irrational expression is $√12$, since 12 is not a perfect square. This is correct because $√12$ simplifies to $2√3$, which is irrational, unlike $√16 = 4$, $√144 = 12$, or $∛64 = 4$ which are rational integers. A common error is misidentifying perfect squares, like thinking $√12$ is rational or choosing $∛64$ as irrational. To solve, (1) identify roots, (2) apply $√$ or $∛$, (3) check perfect (12 not square, but 16=$4^2$, 144=$12^2$, 64=$4^3$; memorize up to 12$^2$=144, 6$^3$=216), (4) no ± for evaluation, (5) recognize irrational if non-perfect square. Mistakes include claiming $√2$ rational equivalent for others, confusing square and cube roots, or forgetting non-perfect means irrational.

This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (for example, if $x^2 = 49$, then $x = \sqrt{49} = 7$ or $x = -\sqrt{49} = -7$, both since $(±7)^2 = 49$), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (for example, if $x^3 = 64$, then $x = \sqrt[3]{64} = 4$ since $4^3 = 64$, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as $\sqrt{2} \approx 1.414\ldots$ with non-repeating decimals. Evaluating $\sqrt[3]{125}$, since $5^3 = 125$, $\sqrt[3]{125} = 5$. This is correct because 125 is a perfect cube, and cube roots have a single real value. A common error is confusing with square roots and choosing $\pm 5$, or miscalculating as 25 since $5^2 = 25$. The strategy is to (1) identify the operation as cube root, (2) apply $\sqrt[3]{}$ to 125, (3) check if 125 is a perfect cube (memorize up to $12^2 = 144$ and $6^3 = 216$), (4) note no $\pm$ for cube roots, and (5) recognize rational for perfect cubes. Common mistakes include adding $\pm$ like for $x^2 = p$, using $\sqrt{}$ instead of $\sqrt[3]{}$, or claiming $\sqrt{2}$ is rational.

This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (for example, if $x^2 = 49$, then x = $\sqrt{49}$ = 7 or x = -$\sqrt{49}$ = -7, both since $(\pm 7)^2 = 49$), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (for example, if $x^3 = 64$, then x = $\sqrt[3]{64}$ = 4 since $4^3 = 64$, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as $\sqrt{2} \approx 1.414\ldots$ with non-repeating decimals. Evaluating $\sqrt{81}$, since $9^2 = 81$ and the principal square root is the positive value, $\sqrt{81} = 9$. This is correct because 81 is a perfect square, and the square root symbol denotes the non-negative root. A common error is including the negative root like -9 or $\pm 9$, forgetting that $\sqrt{}$ refers to the principal (positive) root. The strategy is to (1) identify the operation as square root, (2) apply $\sqrt{}$ to 81, (3) check if 81 is a perfect square (memorize up to $12^2 = 144$ and $6^3 = 216$), (4) use only the positive for principal square root, and (5) recognize rational for perfect squares. Common mistakes include adding a negative like for solving $x^2 = p$, confusing with $\sqrt[3]{}$, or claiming $\sqrt{2}$ is rational.

This question tests solving x² = p and x³ = p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x² = p (for example, if x² = 49, then x = √49 = 7 or x = -√49 = -7, both since (±7)² = 49), while the cube root ∛p solves x³ = p (for example, if x³ = 64, then x = ∛64 = 4 since 4³ = 64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as √2 ≈ 1.414... with non-repeating decimals. For x² = 49, taking the square root of both sides gives x = ±√49, and since 49 is 7², √49 = 7, so x = ±7. This is correct because both 7² = 49 and (-7)² = 49 satisfy the equation, providing all real solutions. A common error is forgetting the negative solution and choosing only x = 7, or confusing it with cube roots which have only one real root. The strategy is to (1) identify the operation as x², (2) apply the square root √ to both sides, (3) check if 49 is a perfect square (memorize up to 12² = 144 and 6³ = 216), (4) include ± for square roots, and (5) recognize that perfect squares yield rational roots. Common mistakes include forgetting the negative solution for x² = p, using the wrong root symbol like ∛ instead of √, or claiming √2 is rational.

This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (if $x^2 = 49$, then x = $\sqrt{49}$ = 7 or x = -$\sqrt{49}$ = -7, both since ($\pm 7$)$^2$ = 49), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (if $x^3 = 64$, then x = $\sqrt[3]{64}$ = 4 since $4^3$ = 64, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots ($\sqrt{2} \approx 1.414\ldots$ non-repeating). Evaluating $\sqrt{121}$ gives 11, since $11^2$ = 121 and the principal square root is the positive value. This is correct because the square root symbol denotes the non-negative root, not the negative or both. A common error is including the negative or both, like $\pm 11$, confusing evaluation with solving equations. To solve, (1) identify it's a square root, (2) apply $\sqrt{}$, (3) check if 121 is a perfect square ($11^2$ = 121, memorize up to $12^2$ = 144), (4) no $\pm$ for evaluation of $\sqrt{}$, (5) recognize it's rational since it's perfect. Mistakes include adding negative solutions when just evaluating $\sqrt{p}$, confusing with cube roots, or claiming it's irrational.

This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (if $x^2 = 49$, then $x = \sqrt{49} = 7$ or $x = -\sqrt{49} = -7$, both since $(\pm 7)^2 = 49$), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (if $x^3 = 64$, then $x = \sqrt[3]{64} = 4$ since $4^3 = 64$, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots ($\sqrt{2} \approx 1.414\ldots$ non-repeating). Evaluating $\sqrt[3]{216}$ gives 6, since $6^3 = 216$. This is correct because cube roots have a single real value, and 216 is a perfect cube. A common error is confusing with square roots and adding $\pm 6$ or choosing a square like 36. To solve, (1) identify it's a cube root, (2) apply $\sqrt[3]{}$, (3) check if 216 is a perfect cube ($6^3 = 216$, memorize up to $6^3 = 216$), (4) no $\pm$ for cube roots, (5) recognize it's rational since it's perfect. Mistakes include adding $\pm$ like for square roots, using the wrong root symbol, or miscalculating as 8.