1.3 Square Roots and Cube Roots Practice

Bronze Level Concepts and Solutions

  • Evaluation of Basic Powers and Exponents:

    • Calculation of 727^2:
    • Expression: 72=7×7=497^2 = 7 \times 7 = 49
    • Calculation of (−5)2(-5)^2:
    • Expression: (−5)2=(−5)×(−5)=25(-5)^2 = (-5) \times (-5) = 25
    • Property: Multiplying two negative factors produces a positive product.
    • Calculation of 232^3:
    • Expression: 23=2×2×2=82^3 = 2 \times 2 \times 2 = 8
    • Calculation of (−4)3(-4)^3:
    • Expression: (−4)3=(−4)×(−4)×(−4)=−64(-4)^3 = (-4) \times (-4) \times (-4) = -64
    • Property: Multiplying an odd number of negative factors produces a negative product.

  • Square Roots and Cube Roots Fundamentals:

    • Two square roots of 8181:
    • Principal (positive) square root: 81=9\sqrt{81} = 9 because 92=819^2 = 81.
    • Negative square root: −81=−9-\sqrt{81} = -9 because (−9)2=81(-9)^2 = 81.
    • Roots: The two square roots of 8181 are 99 and −9-9 (represented as ±9\pm 9).
    • Cube root of 6464:
    • Evaluation: 643=4\sqrt[3]{64} = 4 because 43=4×4×4=644^3 = 4 \times 4 \times 4 = 64.

  • Solving Basic Equations:

    • Solving x2=49x^2 = 49:
    • Taking the square root of both sides gives x=±49x = \pm \sqrt{49}.
    • Solutions: x=7x = 7 or x=−7x = -7.
    • Solving x3=−27x^3 = -27:
    • Taking the cube root of both sides gives x=−273x = \sqrt[3]{-27}.
    • Solution: x=−3x = -3 because (-3)^3 = -27$.\n\n- **Properties of Roots and Integers**:\n - Existence of square roots for 81:\n - Yes, 81hasrealsquareroots,whicharehas real square roots, which are9andand-9,because, because9 \times 9 = 81andand(-9) \times (-9) = 81$.
    • Non-real roots for negative values: In the real number system, negative numbers (such as −81-81) do not have real square roots because squaring any real number results in a non-negative value (x2≥0x^2 \ge 0 for all real xx).
    • Determining integer bases from a given square 144144:
    • Equation: x2=144x^2 = 144 where xx is an integer.
    • Possible integers: x=12x = 12 or x=−12x = -12, since 122=14412^2 = 144 and (-12)^2 = 144$.\n\n# Silver Level Concepts and Solutions\n\n- **Intermediate Powers and Exponents**:\n - Calculation of 12^2:\n - Expression: 12^2 = 12 \times 12 = 144\n - Calculation of (-11)^2:\n - Expression: (-11)^2 = (-11) \times (-11) = 121\n - Calculation of 6^3:\n - Expression: 6^3 = 6 \times 6 \times 6 = 216\n - Calculation of (-7)^3:\n - Expression: (-7)^3 = (-7) \times (-7) \times (-7) = -343\n\n- **Evaluation of Square Roots**:\n - Principal square root of 196:\n - Evaluation: \sqrt{196} = 14becausebecause14^2 = 196.\n - Principal square root of 400:\n - Evaluation: \sqrt{400} = 20becausebecause20^2 = 400.\n\n- **Solving Quadratic and Cubic Equations**:\n - Solving x^2 = 324:\n - Taking square roots: x = \pm \sqrt{324}.\n - Solutions: x = 18ororx = -18.\n - Solving x^2 + 25 = 0:\n - Rearranged equation: x^2 = -25.\n - Real number solution: No real solution exists because x^2 \ge 0forallrealnumbersfor all real numbersx.\n - Complex number solution: x = \pm \sqrt{-25} = \pm 5i,where, wherei = \sqrt{-1}.\n - Solving x^3 = 512:\n - Taking cube root: x = \sqrt[3]{512}.\n - Solution: x = 8becausebecause8^3 = 512$.
    • Solving x3+64=0x^3 + 64 = 0:
    • Rearranged equation: x3=−64x^3 = -64.
    • Taking cube root: x=−643x = \sqrt[3]{-64}.
    • Solution: x=−4x = -4 because (-4)^3 = -64$.\n\n- **Order of Operations with Exponents and Negation**:\n - Comparison of -7^2andand(-7)^2:\n - Calculation of -7^2:Byorderofoperations(BIDMAS/BODMAS),exponentiationprecedesunarynegation.Therefore,: By order of operations (BIDMAS/BODMAS), exponentiation precedes unary negation. Therefore,-7^2 = -(7^2) = -(49) = -49$.
    • Calculation of (−7)2(-7)^2: The parentheses group the negative sign with the base prior to exponentiation. Therefore, (-7)^2 = (-7) \times (-7) = 49$.\n - Key distinction: -7^2 = -49(negative),whereas(negative), whereas(-7)^2 = 49 (positive).\n\n- **Geometric Applications of Squares and Cubes**:\n - Side length of a square with area 1225\,cm^2:\n - Formula: \text{Area} = s^2,where, wheres represents side length.\n - Calculation: s = \sqrt{1225} = 35\,cm$.
    • Physical constraint: Side length must be a positive value, so only the positive root 35 cm35\,cm applies.
    • Edge length of a cube with volume 1000 cm31000\,cm^3:
    • Formula: Volume=e3\text{Volume} = e^3, where ee represents edge length.
    • Calculation: e = \sqrt[3]{1000} = 10\,cm$.\n\n# Gold Level Concepts and Solutions\n\n- **Advanced Geometric Word Problems**:\n - Side length of a square garden with area 2500\,m^2:\n - Formula: \text{Area} = s^2\n - Calculation: s = \sqrt{2500} = 50\,m$.
    • Result: Each side of the garden measures 50\,m$.\n - Edge length of a cube-shaped box with volume 4913\,cm^3:\n - Formula: \text{Volume} = e^3\n - Calculation: e = \sqrt[3]{4913} = 17\,cm$.
    • Result: Each edge of the box measures 17\,cm$.\n\n- **Pythagorean Triples Verification and Generation**:\n - Verification of 8^2 + 15^2 = 17^2:\n - Left-hand side evaluation: 8^2 + 15^2 = 64 + 225 = 289.\n - Right-hand side evaluation: 17^2 = 289.\n - Equivalence: 289 = 289, verifying the relationship.\n - Additional Pythagorean Triples satisfying a^2 + b^2 = c^2:\n - Example 1: (3, 4, 5)sincesince3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
    • Example 2: (5,12,13)(5, 12, 13) since 5^2 + 12^2 = 25 + 144 = 169 = 13^2$.\n - Example 3: (7, 24, 25)sincesince7^2 + 24^2 = 49 + 576 = 625 = 25^2$.

  • Tabular Analysis of Powers, Roots, and Patterns:

  Table of x, x squared, x cubed, and square root of x squared

  • Filled Values for Table Columns xx, x2x^2, x3x^3, and x2\sqrt{x^2}:

    • Row 1 (x=3x = 3):
      • x2=32=9x^2 = 3^2 = 9
      • x3=33=27x^3 = 3^3 = 27
      • x2=9=3\sqrt{x^2} = \sqrt{9} = 3
    • Row 2 (x=5x = 5):
      • x2=52=25x^2 = 5^2 = 25
      • x3=53=125x^3 = 5^3 = 125
      • x2=25=5\sqrt{x^2} = \sqrt{25} = 5
    • Row 3 (x=−4x = -4):
      • x2=(−4)2=16x^2 = (-4)^2 = 16
      • x3=(−4)3=−64x^3 = (-4)^3 = -64
      • x2=16=4\sqrt{x^2} = \sqrt{16} = 4
    • Row 4 (x=7x = 7):
      • x2=72=49x^2 = 7^2 = 49
      • x3=73=343x^3 = 7^3 = 343
      • x2=49=7\sqrt{x^2} = \sqrt{49} = 7

  • Pattern Identifications:

    • x2x^2 is non-negative for all real values of xx.

    • x3x^3 retains the original algebraic sign of xx (positive when x>0x > 0, negative when x<0x < 0).

    • x2=∣x∣\sqrt{x^2} = |x| (the absolute value of xx), ensuring that principal square roots always yield non-negative results regardless of whether xx is positive or negative.

    • Classification of Numbers in Number Systems:

  • Definitions of Number Sets:

    • Natural Numbers (N\mathbb{N}): Positive integers used for counting, i.e., {1,2,3,4,… }\{1, 2, 3, 4, \dots\}.
    • Integers (Z\mathbb{Z}): Whole numbers including negative integers, zero, and positive integers, i.e., {…,−3,−2,−1,0,1,2,3,… }\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}.
    • Rational Numbers (Q\mathbb{Q}): Numbers expressible as a ratio ab\frac{a}{b} of two integers where b≠0b \neq 0.

  • Number Classifications:

    • a) −11-11: Integer (Z\mathbb{Z}) and Rational Number (Q\mathbb{Q}) because −11=−111-11 = \frac{-11}{1}. It is not a Natural Number.

    • b) 00: Integer (Z\mathbb{Z}) and Rational Number (Q\mathbb{Q}) because 0=010 = \frac{0}{1}. It is not a Natural Number.

    • c) 94\frac{9}{4}: Rational Number (Q\mathbb{Q}) because it is a fraction of two integers (2.252.25). It is neither an Integer nor a Natural Number.

    • d) −186-\frac{18}{6}: Simplifies to −3-3. Therefore, it is an Integer (Z\mathbb{Z}) and a Rational Number (Q\mathbb{Q}). It is not a Natural Number.

    • e) 7.757.75: Expressible as 314\frac{31}{4}. Therefore, it is a Rational Number (Q\mathbb{Q}). It is neither an Integer nor a Natural Number.

    • Constructing Equations Involving Radical Expressions with Two Solutions:

  • Formulation of Equation:

    • Example Equation: x2=6\sqrt{x^2} = 6

  • Algebraic Solution:

    • Utilizing the identity x2=∣x∣\sqrt{x^2} = |x| gives the equation |x| = 6$.\n - Solutions: x = 6ororx = -6$.

  • Verification of Both Solutions:

    • Substitute x=6x = 6: 62=36=6\sqrt{6^2} = \sqrt{36} = 6 (Valid).

    • Substitute x=−6x = -6: (−6)2=36=6\sqrt{(-6)^2} = \sqrt{36} = 6 (Valid).

    • Challenge Problem: Integer Difference of Squares:

  • Problem Formulation: Find integers aa and bb satisfying a^2 - b^2 = 39$.\n - Algebraic Factorization:\n - Difference of squares identity: a^2 - b^2 = (a - b)(a + b) = 39$.

  • Factor Pairs Analysis of 3939:

    • Factor Pair 1 (1×39=391 \times 39 = 39):
      • System of equations:
      • a+b=39a + b = 39
      • a−b=1a - b = 1
      • Summing equations: 2a = 40 \implies a = 20$.\n - Subtracting equations: 2b = 38 \implies b = 19$.
      • Verification: 20^2 - 19^2 = 400 - 361 = 39$.\n - Integer coordinate pairs (a, b)::(20, 19),,(-20, 19),,(20, -19),,(-20, -19).\n - Factor Pair 2 (3 \times 13 = 39):\n - System of equations:\n - a + b = 13\n - a - b = 3\n - Summing equations: 2a = 16 \implies a = 8$.
      • Subtracting equations: 2b = 10 \implies b = 5$.\n - Verification: 8^2 - 5^2 = 64 - 25 = 39$.
      • Integer coordinate pairs (a,b)(a, b): (8,5)(8, 5), (−8,5)(-8, 5), (8,−5)(8, -5), (−8,−5)(-8, -5).

  • Complete Solution Set:

    • Positive integer pairs for (a,b)(a, b) are (20,19)(20, 19) and (8,5)(8, 5).
    • All integer pairs for (a,b)(a, b) are {(±20,±19),(±8,±5)}\{(\pm 20, \pm 19), (\pm 8, \pm 5)\}.