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 x³ = 125, taking the cube root gives x = ∛125, and since 125 is a perfect cube (5³ = 125), x = 5. This is correct because only one real number, 5, satisfies the equation, unlike square roots which have two solutions. A common error is confusing it with square roots and including ±5. To solve, (1) identify the operation as x³, (2) apply the cube root ∛, (3) check if 125 is a perfect cube (memorize up to 6³ = 216), (4) no ± needed for cube roots, (5) recognize it's rational since it's perfect. Mistakes include adding ± like for square roots, using the wrong root symbol, or misidentifying it as irrational.

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). The true statement is that √2 is irrational, as 2 is not a perfect square. This is correct because √2 cannot be expressed as a fraction and has a non-repeating decimal, unlike √9 = 3 or √4 = 2 which are rational. A common error is misidentifying perfect squares like claiming √3 is rational or √9 is irrational. To solve, (1) identify square roots, (2) apply √, (3) check if the number is a perfect square (memorize up to 12² = 144), (4) include ± only for equations, (5) recognize irrational if non-perfect gives non-repeating decimal. Mistakes include claiming √2 rational, confusing rational and irrational, or forgetting which are perfect squares.

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). Comparing √50 and 7, √50 is greater since 7² = 49 and 50 > 49, so √50 > 7. This is correct because the square root function is increasing, and a larger input gives a larger output. A common error is thinking they are equal since 7² is close to 50, or confusing with cubes. To solve, (1) identify square roots, (2) apply √, (3) check squares (7² = 49 < 50, memorize up to 12² = 144), (4) no ± for comparison, (5) recognize √50 irrational and greater than 7. Mistakes include claiming equality, forgetting √50 > √49 = 7, or misidentifying rationality.

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 x² = 49, taking the square root gives x = ±√49, and since 49 is a perfect square (7² = 49), x = ±7. This is correct because both 7² and (-7)² equal 49, so both solutions apply. A common error is forgetting the negative and choosing only x = 7, or miscalculating √49 as 6. To solve, (1) identify x², (2) apply √, (3) check perfect square (7² = 49, memorize up to 12² = 144), (4) include ±, (5) recognize rational. Mistakes include omitting negative, using wrong root, or claiming 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 $√p$ 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 $x^2 = p$ (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 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). √50 lies between 7 and 8, since 7² = 49 and 8² = 64, and 50 is between 49 and 64. This is correct because √50 is slightly more than 7 but less than 8. A common error is miscalculating the squares, like thinking it's between 6 and 7 since 6² = 36 and 7² = 49, but 50 > 49. To solve, (1) identify it's a square root, (2) apply √, (3) find perfect squares around 50 (7² = 49, 8² = 64, memorize up to 12² = 144), (4) no ± for evaluation, (5) recognize it's irrational since 50 is non-perfect. Mistakes include wrong interval due to calculation error, confusing with cube roots, or claiming it's 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 $\sqrt[3]{216}$ 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.

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 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. Among the statements, √4 is rational because √4 = 2, which is an integer. This is correct because 4 is a perfect square, yielding a rational root, while the others are false: √2 is irrational despite approximations, √9 = 3 is rational, and √16 = 4 is rational. A common error is misidentifying irrational roots, like claiming √9 is irrational or √2 is rational due to its decimal approximation. The strategy is to (1) identify if the number under the root is a perfect square or cube, (2) apply the root, (3) check perfectness (memorize up to 12² = 144 and 6³ = 216), (4) note ± only for solving square equations, and (5) recognize irrationality for non-perfect squares. Common mistakes include claiming √2 rational, confusing √ and ∛, or forgetting negative solutions in equations.