Free math worksheets

- Think about which number multiplied by itself gives each radicand. - For decimals, pay attention to the number of decimal places after squaring. - Can you first think about the same digits without the decimal point?

1. Since \(30^2 = 900\), \(\sqrt{900} = 30\). 2. Since \(0.1^2 = 0.01\), \(\sqrt{0.01} = 0.1\). 3. Since \(1.4^2 = 1.96\), \(\sqrt{1.96} = 1.4\). 4. Since \(2.5^2 = 6.25\), \(\sqrt{6.25} = 2.5\).

a) \(30\) b) \(0.1\) c) \(1.4\) d) \(2.5\)

- For each statement, think about which number multiplied by itself gives the radicand. - What happens when a negative number is multiplied by itself? - Recall the difference between rational and irrational numbers. - How does squaring a decimal affect its place value?

1. Statement a) is true because \(12^2 = 12 \cdot 12 = 144\). 2. Statement b) is false because \(\sqrt{-25}\) is not defined in the real numbers. Also, \((-5)^2 = 25\), not \(-25\). 3. Statement c) is true because \(0.2^2 = 0.2 \cdot 0.2 = 0.04\). 4. Statement d) is false. The solutions are \(x = \sqrt{7}\) and \(x = -\sqrt{7}\), and both are irrational.

a) True b) False c) True d) False

- Think about which number multiplied by itself gives each radicand. - Pay close attention to whether a negative sign is inside or outside the radical. - For a decimal radicand, it may help to rewrite the decimal as a fraction. - Use exponent properties when working with powers of \(10\).

1. Since \(1.3^2 = 1.69\), \(\sqrt{1.69} = 1.3\). 2. First, \((-7)^2 = 49\). Therefore, \(\sqrt{(-7)^2} = \sqrt{49} = 7\). 3. The expression \(\sqrt{-0.04}\) is undefined in the real number system because its radicand is negative. 4. \(-\sqrt{\frac{25}{144}} = -\frac{\sqrt{25}}{\sqrt{144}} = -\frac{5}{12}\). 5. Since \(10^4 = 10{,}000\) and \(100^2 = 10{,}000\), \(\sqrt{10^4} = 100\).

a) \(1.3\) b) \(7\) c) Undefined in the real number system because the radicand is negative. d) \(-\frac{5}{12}\) e) \(100\)

- First identify a basic perfect square using the digits without the decimal point or extra zeros. - What happens when a number such as \(10\) or \(0.1\) is squared? - How does taking a square root affect the number of decimal places? - Look for a pattern when the decimal point in the radicand moves two places.

1. Since \(15^2 = 225\), \(\sqrt{225} = 15\). Multiplying a radicand by \(100\) multiplies its square root by \(10\). Therefore, \(\sqrt{2.25} = 1.5\) and \(\sqrt{22{,}500} = 150\). 2. Since \(12^2 = 144\), \(\sqrt{144} = 12\). Dividing a radicand by \(100\) divides its square root by \(10\). Therefore, \(\sqrt{1.44} = 1.2\) and \(\sqrt{0.0144} = 0.12\).

a) \(1.5\); \(15\); \(150\) b) \(0.12\); \(1.2\); \(12\)

- Can you write each decimal as a fraction with a denominator such as \(100\) or \(10{,}000\)? - Use the perfect squares from \(1^2\) through \(20^2\). - Compare the number of zeros in a number and in its square. - Think about how the decimal point moves when a decimal is squared.

1. \(\sqrt{2.25} = \sqrt{\frac{225}{100}} = \frac{15}{10} = 1.5\). 2. \(\sqrt{0.0004} = \sqrt{\frac{4}{10000}} = \frac{2}{100} = 0.02\). 3. \(\sqrt{12{,}100} = \sqrt{121 \cdot 100} = 11 \cdot 10 = 110\). 4. \(\sqrt{0.0169} = \sqrt{\frac{169}{10000}} = \frac{13}{100} = 0.13\). 5. \(\sqrt{90{,}000} = \sqrt{9 \cdot 10{,}000} = 3 \cdot 100 = 300\).

a) \(1.5\) b) \(0.02\) c) \(110\) d) \(0.13\) e) \(300\)

- Relate each decimal to a familiar perfect square. - How can you take the square root of a fraction whose numerator and denominator are perfect squares? - Remember that taking the square root of an area unit gives the corresponding length unit.

1. Since \(2.3^2 = 5.29\), the side length is \(a = 2.3\,\text{cm}\). 2. \(a = \sqrt{\frac{144}{289}\,\text{dm}^2} = \frac{\sqrt{144}}{\sqrt{289}}\,\text{dm} = \frac{12}{17}\,\text{dm}\). 3. Since \(1.1^2 = 1.21\), the side length is \(a = 1.1\,\text{m}\).

a) \(a = 2.3\,\text{cm}\) b) \(a = \frac{12}{17}\,\text{dm}\) c) \(a = 1.1\,\text{m}\)

- Break each expression into simpler parts and evaluate those first. - Recall that \(\sqrt{x^2}\) always gives a nonnegative result. - Pay close attention to sign rules when multiplying.

1. \(\sqrt{12^2} = 12\) and \(\sqrt{(-12)^2} = |-12| = 12\), so the difference is \(12 - 12 = 0\). 2. \((\sqrt{15})^2 = 15\) and \((-\sqrt{15})^2 = 15\), so the sum is \(15 + 15 = 30\). 3. \(\sqrt{(-8)^2} = 8\) and \(-\sqrt{8^2} = -8\), so the product is \(8 \cdot (-8) = -64\).

a) \(0\) b) \(30\) c) \(-64\)

- What is \(\sqrt{144}\)? - How does squaring a decimal affect its number of decimal places? - Besides a positive number, what other number can have the same positive square?

1. Since \(12^2 = 144\), use place value to find the decimal square root. 2. Because \(0.12^2 = 0.0144\), \(\sqrt{0.0144} = 0.12\). 3. The equation \(x^2 = 0.0144\) has the two real solutions \(x = 0.12\) and \(x = -0.12\).

\(\sqrt{0.0144} = 0.12\). The real numbers whose square is \(0.0144\) are \(0.12\) and \(-0.12\).

- Determine which number multiplied by itself gives the value on the right side. - Can the square of a real number be negative? - It may help to rewrite a decimal such as \(1.44\) as a fraction. - An equation of the form \(x^2 = a\) with \(a > 0\) has two real solutions.

1. For a), \(\sqrt{1.44} = 1.2\), so \(x = 1.2\) or \(x = -1.2\). 2. For b), a real number’s square cannot be negative, so there is no real solution. 3. For c), \(z = \pm\sqrt{\frac{49}{100}} = \pm\frac{7}{10}\), so \(z = 0.7\) or \(z = -0.7\). 4. For d), \(w^2 = 2500\), so \(w = 50\) or \(w = -50\).

a) \(x = 1.2\) or \(x = -1.2\) b) No real solution c) \(z = 0.7\) or \(z = -0.7\) d) \(w = 50\) or \(w = -50\)

- Which operation reverses taking a square root? - How can you square a decimal mentally? - How many decimal places does the square of a number with one decimal place usually have?

1. For a), \(14^2 = 196\), so the missing digits are \(9\) and \(6\). 2. For b), \(0.6^2 = 0.36\), so the missing digits are \(3\) and \(6\). 3. For c), \(25^2 = 625\), so the missing digits are \(2\) and \(5\). 4. For d), \(1.1^2 = 1.21\), so the missing digits are \(2\) and \(1\).

a) \(\sqrt{196} = 14\) b) \(\sqrt{0.36} = 0.6\) c) \(\sqrt{625} = 25\) d) \(\sqrt{1.21} = 1.1\)

- What exactly does the symbol \(\sqrt{\phantom{x}}\) represent? - Besides a positive number, what other number can have the same positive square? - Distinguish evaluating an expression from finding every solution to an equation.

1. For \(a \geq 0\), \(\sqrt{a}\) means the unique nonnegative number whose square is \(a\). Thus, \(\sqrt{5}\) is one positive value. 2. The equation \(x^2 = 5\) asks for every real number whose square is \(5\). 3. Both \((\sqrt{5})^2 = 5\) and \((-\sqrt{5})^2 = 5\). 4. Therefore, the equation has two solutions: \(x = \sqrt{5}\) and \(x = -\sqrt{5}\). Jordan correctly describes the principal square root but incorrectly applies that fact to the equation’s full solution set.

Jordan’s claim is incorrect. The expression \(\sqrt{5}\) represents the nonnegative principal square root, but the equation \(x^2 = 5\) has two solutions. Its solution set is \(\{-\sqrt{5}, \sqrt{5}\}\).

- Determine which number multiplied by itself gives each radicand. - Convert a mixed number to an improper fraction before taking its square root. - Pay close attention to decimal place value.

1. The needed square roots are \(\sqrt{0.81} = 0.9\), \(\sqrt{0.01} = 0.1\), \(\sqrt{0.49} = 0.7\), \(\sqrt{1\frac{7}{9}} = \sqrt{\frac{16}{9}} = \frac{4}{3}\), and \(\sqrt{1.44} = 1.2\). 2. For a), \(0.9 - 0.2 = 0.7\). 3. For b), \(0.1 + 0.7 = 0.8\). 4. For c), \(\frac{4}{3} - \frac{1}{3} = 1\). 5. For d), \(0.5 \cdot 1.2 = 0.6\).

a) \(0.7\) b) \(0.8\) c) \(1\) d) \(0.6\)

- Evaluate the expression inside the radical first. - Squaring reverses taking the principal square root. - Trace horizontally from \(y = T\) to the graph.

1. The radicand is \(\frac{1}{4} \cdot 2^4 = \frac{1}{4} \cdot 16 = 4\). Therefore, \(T = \sqrt{4} = 2\). 2. Solve \(\sqrt{x} = 2\). Squaring both sides gives \(x = 4\). 3. On the graph, the horizontal line through \(y = 2\) meets \(y = \sqrt{x}\) at \(x = 4\).

1. \(T = 2\) 2. \(x = 4\)

- Translate the square’s area into an equation involving its side. - Consider every algebraic solution before applying the context. - Use the meaning of a length to eliminate an invalid value.

1. The area equation is \(x^2=324\). 2. The equation has solutions \(x=18\) and \(x=-18\). 3. A physical side length must be positive. 4. Therefore, \(x=18\,\text{cm}\).

\(x=18\,\text{cm}\). The algebraic solution \(x=-18\) is rejected because a side length cannot be negative.

- How do you find the side length of a square when its area is known? - Find the area of the second garden before finding its side length. - If the area is \(100\) times as large, is the side length also \(100\) times as long? - Which operation reverses squaring? - What happens to a square root when its radicand is multiplied by \(100\)?

1. The first side length is \(s_1 = \sqrt{64\,\text{ft}^2} = 8\,\text{ft}\). 2. The second area is \(A_2 = 100 \cdot 64\,\text{ft}^2 = 6400\,\text{ft}^2\), so its side length is \(s_2 = \sqrt{6400\,\text{ft}^2} = 80\,\text{ft}\). 3. The side-length ratio is \(\frac{80}{8} = 10\). 4. When the area of a square is multiplied by a factor \(k\), its side length is multiplied by \(\sqrt{k}\). Here, \(\sqrt{100} = 10\).

a) \(s_1 = 8\,\text{ft}\); \(s_2 = 80\,\text{ft}\) b) The larger side length is \(10\) times the smaller side length. An area scale factor of \(100\) gives a side-length scale factor of \(\sqrt{100} = 10\).

- How can you rewrite a mixed number before taking its square root? - Compare a negative sign outside a radical with a negative sign in the radicand. - Recall the perfect squares through \(20^2\).

1. Since \(144 = 12^2\) and \(289 = 17^2\), \(\sqrt{\frac{144}{289}} = \frac{12}{17}\). 2. Convert the mixed number: \(1\frac{13}{36} = \frac{49}{36}\). Then \(\sqrt{\frac{49}{36}} = \frac{7}{6}\). 3. \(-\sqrt{\frac{1}{10000}} = -\frac{1}{100} = -0.01\). 4. The expression \(\sqrt{-1.44}\) is undefined in the real number system because its radicand is negative. 5. Convert the mixed number: \(3\frac{1}{16} = \frac{49}{16}\). Then \(\sqrt{\frac{49}{16}} = \frac{7}{4}\).

a) \(\frac{12}{17}\) b) \(\frac{7}{6}\), or \(1\frac{1}{6}\) c) \(-0.01\), or \(-\frac{1}{100}\) d) Undefined in the real number system e) \(\frac{7}{4}\), or \(1.75\)

- Evaluate each square root before performing the other operation. - Convert the mixed number in C to an improper fraction. - Compare all four results in the same form.

1. \(A=0.8+0.6=1.4\). 2. \(B=1.5-0.2=1.3\). 3. \(1\frac{11}{25}=\frac{36}{25}\), so \(C=\sqrt{\frac{36}{25}}=\frac{6}{5}=1.2\). 4. \(D=12\cdot0.1=1.2\). 5. Since \(1.4>1.3>1.2\), expression A has the greatest value.

A: \(1.4\) B: \(1.3\) C: \(1.2\) D: \(1.2\) Expression A has the greatest value.

- How are the area and side length of a square related? - Which operation reverses squaring? - Recall how to convert meters to centimeters.

1. Find the side length in meters: \(s = \sqrt{0.49\,\text{m}^2} = 0.7\,\text{m}\). 2. Convert to centimeters: \(0.7\,\text{m} = 70\,\text{cm}\).

The side length of the glass panel is \(70\,\text{cm}\).

- First find the side length of the smaller square lot. - Find the area of the larger square lot. - Then find the new side length and compare the two lengths.

1. The smaller lot has side length \(s_1 = \sqrt{3600\,\text{ft}^2} = 60\,\text{ft}\). 2. The larger lot has area \(A_2 = 4 \cdot 3600\,\text{ft}^2 = 14{,}400\,\text{ft}^2\). 3. Its side length is \(s_2 = \sqrt{14{,}400\,\text{ft}^2} = 120\,\text{ft}\). 4. The difference is \(120\,\text{ft} - 60\,\text{ft} = 60\,\text{ft}\).

The larger lot’s side length is \(60\,\text{ft}\) greater than the smaller lot’s side length.

- What stays the same when the metal is reshaped? - How can you find the side length of a square from its area? - Use the perimeter formulas for a rectangle and a square.

1. The rectangle’s area is \(A = 2\,\text{cm} \cdot 18\,\text{cm} = 36\,\text{cm}^2\). 2. The side length of the equal-area square is \(s = \sqrt{36\,\text{cm}^2} = 6\,\text{cm}\). 3. The rectangle’s perimeter is \(P_{\text{rectangle}} = 2(2\,\text{cm} + 18\,\text{cm}) = 40\,\text{cm}\). 4. The square’s perimeter is \(P_{\text{square}} = 4 \cdot 6\,\text{cm} = 24\,\text{cm}\). 5. The difference is \(40\,\text{cm} - 24\,\text{cm} = 16\,\text{cm}\).

The square’s perimeter is \(24\,\text{cm}\), which is \(16\,\text{cm}\) shorter than the rectangle’s perimeter of \(40\,\text{cm}\).

- How can you find a square’s side length from its area? - If a square’s perimeter doubles, what happens to its side length? - How does doubling the side length affect the area?

1. The first sign’s side length is \(a_1 = \sqrt{7.84\,\text{ft}^2} = 2.8\,\text{ft}\). 2. Because \(P = 4a\), doubling a square’s perimeter doubles its side length. 3. The second sign’s side length is \(a_2 = 2 \cdot 2.8\,\text{ft} = 5.6\,\text{ft}\). 4. Its area is \(A_2 = (5.6\,\text{ft})^2 = 31.36\,\text{ft}^2\).

The second sign has side length \(5.6\,\text{ft}\) and area \(31.36\,\text{ft}^2\).

- First convert the mixed number to an improper fraction. - What happens to a square root when the radicand is divided by \(4\)? - Express the side length as a fraction before converting it to a decimal.

1. Convert the mixed number: \(6\frac{19}{25} = \frac{6 \cdot 25 + 19}{25} = \frac{169}{25}\). 2. The side length is \(a = \sqrt{\frac{169}{25}}\,\text{cm} = \frac{13}{5}\,\text{cm} = 2.6\,\text{cm}\). 3. If the area is multiplied by \(\frac{1}{4}\), the side length is multiplied by \(\sqrt{\frac{1}{4}} = \frac{1}{2}\). Therefore, the side length is halved. 4. The new side length is \(a_{\text{new}} = \frac{1}{2} \cdot 2.6\,\text{cm} = 1.3\,\text{cm}\).

a) \(a = 2.6\,\text{cm}\) b) The side length is multiplied by \(\frac{1}{2}\), so it is halved. The new side length is \(1.3\,\text{cm}\).

- Evaluate each expression one step at a time. - Consider when squaring and taking a square root reverse each other. - Pay attention to the location of each negative sign and whether it is included in a square.

1. \(A = \sqrt{\left(\frac{4}{9}\right)^2} = \frac{4}{9}\). 2. \(B = \left(\sqrt{\frac{4}{9}}\right)^2 = \frac{4}{9}\). 3. Since \(\left(-\frac{4}{9}\right)^2 = \left(\frac{4}{9}\right)^2\), \(C = \sqrt{\left(-\frac{4}{9}\right)^2} = \frac{4}{9}\). 4. Squaring removes the negative sign, so \(D = \left(-\sqrt{\frac{4}{9}}\right)^2 = \frac{4}{9}\). 5. In \(E\), the negative sign is outside the radical, so \(E = -\sqrt{\left(\frac{4}{9}\right)^2} = -\frac{4}{9}\). 6. Therefore, the common value is \(\frac{4}{9}\), and expression \(E\) does not match.

The common value is \(\frac{4}{9}\). Expression \(E\) does not match the others.

- How are a square’s area and side length related? - How many equal sides does a square have? - Pay attention to decimal place value when taking each square root.

1. For a square, \(a = \sqrt{A}\) and \(P = 4a\). 2. For (1), \(a = \sqrt{1.44\,\text{m}^2} = 1.2\,\text{m}\), so \(P = 4 \cdot 1.2\,\text{m} = 4.8\,\text{m}\). 3. For (2), \(a = \sqrt{0.0025\,\text{km}^2} = 0.05\,\text{km}\), so \(P = 4 \cdot 0.05\,\text{km} = 0.2\,\text{km}\). 4. For (3), \(a = \sqrt{169\,\text{cm}^2} = 13\,\text{cm}\), so \(P = 4 \cdot 13\,\text{cm} = 52\,\text{cm}\).

(1) \(a = 1.2\,\text{m}\); \(P = 4.8\,\text{m}\) (2) \(a = 0.05\,\text{km}\); \(P = 0.2\,\text{km}\) (3) \(a = 13\,\text{cm}\); \(P = 52\,\text{cm}\)

- How many square faces make up a cube’s surface? - Once you know the area of one face, how can you find the edge length? - Which formula gives the volume of a cube?

1. A cube has \(6\) congruent square faces, so one face has area \(486\,\text{cm}^2 \div 6 = 81\,\text{cm}^2\). 2. The edge length is \(a = \sqrt{81\,\text{cm}^2} = 9\,\text{cm}\). 3. The volume is \(V = a^3 = (9\,\text{cm})^3 = 729\,\text{cm}^3\).

The cube’s volume is \(V = 729\,\text{cm}^3\).

- Use \(a = \sqrt{\frac{S}{6}}\) to find both edge lengths. - Compare the two edge lengths by forming their ratio. - For the general case, compare \(a^2=\frac{S}{6}\) with the new squared edge length after \(S\) is replaced by \(4S\).

1. The edge lengths are \(a_1 = \sqrt{\frac{150}{6}}\,\text{cm} = \sqrt{25}\,\text{cm} = 5\,\text{cm}\) and \(a_2 = \sqrt{\frac{600}{6}}\,\text{cm} = \sqrt{100}\,\text{cm} = 10\,\text{cm}\). 2. The ratio is \(\frac{a_2}{a_1} = \frac{10}{5} = 2\). 3. For any cube, \(a^2=\frac{S}{6}\). If the surface area becomes \(4S\), then the new edge length satisfies \(a_{\text{new}}^2=\frac{4S}{6}=4a^2=(2a)^2\). 4. Edge lengths are positive, so \(a_{\text{new}}=2a\). Therefore, multiplying the surface area by \(4\) doubles the edge length.

a) \(a_1 = 5\,\text{cm}\); \(a_2 = 10\,\text{cm}\) b) \(\frac{a_2}{a_1} = 2\) c) The edge length doubles because \(\sqrt{4} = 2\).

- Which number multiplied by itself gives \(1.69\)? - What happens to the sign when a negative number is squared? - Can the square of a real number ever be negative?

1. When \(a = 1.69\), the equation is \(x^2 = 1.69\). Since \(1.3^2 = 1.69\), the solutions are \(x = 1.3\) and \(x = -1.3\). 2. The square of every real number is nonnegative, so \(x^2 \geq 0\) for every real \(x\). 3. Because \(-1.69\) is negative, no real number can have that square. The fact that \((-1.3)^2 = 1.69\) confirms that squaring a negative number gives a positive result.

a) \(x = 1.3\) and \(x = -1.3\) b) A real number’s square cannot be negative, so \(x^2 = -1.69\) has no real solution.

- In part a), follow the order of operations: which operation comes first? - What condition must a radicand meet for its square root to be real? - Which numbers have a square of \(121\)?

1. First square \(-8\): \((-8)^2 = 64\). Then \(A = \sqrt{64} = 8\). 2. A real square root is defined only for a nonnegative radicand. Since \(-8 < 0\), \(\sqrt{-8}\) is not a real number, so \(B\) is undefined in the real number system. 3. Since \(\sqrt{x^2} = |x|\), the equation becomes \(|x| = 11\). Therefore, \(x = 11\) or \(x = -11\).

a) \(A = 8\) b) \(B\) is undefined in the real number system because its radicand is negative. c) \(x = 11\) and \(x = -11\)

- Break each large number into smaller prime factors systematically. - What happens to exponents when you take a square root? - Use divisibility rules to test small prime factors such as \(2\), \(3\), and \(5\).

1. \(1764 = 2^2 \cdot 3^2 \cdot 7^2\). Pairing equal prime factors gives \(\sqrt{1764} = 2 \cdot 3 \cdot 7 = 42\). 2. \(3136 = 2^6 \cdot 7^2\). Halving the even exponents gives \(\sqrt{3136} = 2^3 \cdot 7 = 8 \cdot 7 = 56\). 3. A positive integer is a perfect square exactly when every exponent in its prime factorization is even.

a) \(1764 = 2^2 \cdot 3^2 \cdot 7^2\); \(\sqrt{1764} = 42\) b) \(3136 = 2^6 \cdot 7^2\); \(\sqrt{3136} = 56\) c) A positive integer is a perfect square exactly when all exponents in its prime factorization are even.

- Examine the exponents in each prime factorization. - What must be true of every exponent for the square root to be a positive integer? - What does the prime factorization of a prime number look like?

1. A positive integer is a perfect square when every exponent in its prime factorization is even. The exponents in \(a\) are \(4\), \(2\), and \(2\), so \(a\) is a perfect square. The exponent \(3\) in \(b\) is odd, so \(b\) is not a perfect square. 2. \(\sqrt{a} = \sqrt{2^4 \cdot 3^2 \cdot 5^2} = 2^2 \cdot 3 \cdot 5 = 60\). 3. A prime number \(p\) has prime factorization \(p^1\). Since the exponent \(1\) is odd, \(p\) is not the square of a positive integer, so \(\sqrt{p}\) is not a positive integer.

a) \(a\) is a perfect square because all its exponents are even. \(b\) is not because the exponent of \(2\) is odd. b) \(\sqrt{a} = 60\) c) A prime number has prime factorization \(p^1\), whose exponent is odd, so its square root cannot be a positive integer.

- How are the area and side length of a square related? - Use divisibility rules to break \(4356\) into prime factors. - Nearby perfect squares can help you check the size of your answer.

1. The prime factorization is \(4356 = 2^2 \cdot 3^2 \cdot 11^2\). 2. Therefore, \(s = \sqrt{4356}\,\text{ft} = 2 \cdot 3 \cdot 11\,\text{ft} = 66\,\text{ft}\). 3. Since \(60^2 = 3600\) and \(70^2 = 4900\), \(3600 < 4356 < 4900\). Thus, \(\sqrt{4356}\) must be between \(60\) and \(70\), so \(66\) is reasonable.

a) \(s = 66\,\text{ft}\), because \(4356 = 2^2 \cdot 3^2 \cdot 11^2\) b) Since \(60^2 = 3600\) and \(70^2 = 4900\), the value \(66\) is in the expected range.

- Isolate the squared term before taking a square root. - Track the signs carefully as you undo addition or subtraction. - Can a real number have a negative square?

1. For a), divide by \(4\): \(x^2 = 16\). Therefore, \(x = 4\) or \(x = -4\). 2. For b), subtract \(0.7\): \(x^2 = 0.09\). Therefore, \(x = 0.3\) or \(x = -0.3\). 3. For c), add \(5\): \(y^2 = 16\). Therefore, \(y = 4\) or \(y = -4\). 4. For d), subtract \(10\): \(z^2 = -6\). A real number’s square cannot be negative, so there is no real solution.

a) \(x = 4\) or \(x = -4\) b) \(x = 0.3\) or \(x = -0.3\) c) \(y = 4\) or \(y = -4\) d) No real solution

- Compare \((-a)^2\) with \(-a^2\). Is the negative sign included in the square? - Simplify each equation before taking a square root. - For fractions, rewrite terms with a common denominator.

1. For a), \((-12)^2 = 144\), so \(x = 12\) or \(x = -12\). 2. For b), the exponent is evaluated before the negative sign: \(-8^2 = -(8^2) = -64\). Thus, \(x^2 = -64\) has no real solution. 3. For c), \(3x^2 = 27\), so \(x^2 = 9\). Therefore, \(x = 3\) or \(x = -3\). 4. For d), \(x^2 = 1 - \frac{7}{16} = \frac{9}{16}\). Therefore, \(x = \frac{3}{4}\) or \(x = -\frac{3}{4}\).

a) \(x = 12\) or \(x = -12\) b) No real solution c) \(x = 3\) or \(x = -3\) d) \(x = \frac{3}{4}\) or \(x = -\frac{3}{4}\)

- Relate the square-root inequality to an inequality involving squares. - When squaring a number, focus on its ones digit. - Which digits do not appear in your list from part b)?

1. The inequality \(4 < \sqrt{n} < 5\) is equivalent to \(4^2 < n < 5^2\), so \(16 < n < 25\). Thus, \(n\) can be \(17, 18, 19, 20, 21, 22, 23\), or \(24\). 2. The digit squares are \(0, 1, 4, 9, 16, 25, 36, 49, 64\), and \(81\). Their ones digits are \(0, 1, 4, 5, 6\), and \(9\). 3. Every positive-integer square ends in \(0, 1, 4, 5, 6\), or \(9\). Since \(157\) ends in \(7\), it cannot be a perfect square.

a) \(n \in \{17, 18, 19, 20, 21, 22, 23, 24\}\) b) The possible ones digits are \(0, 1, 4, 5, 6\), and \(9\). c) The number \(157\) ends in \(7\), but no positive-integer square ends in \(7\). Therefore, \(157\) is not a perfect square.

- List the two-digit perfect squares and test them systematically. - In part b), call \(\sqrt{x}\) a new nonnegative number and compare the square of twice that number with \(4x\). - How many decimal places result when a number with two decimal places is squared?

1. The two-digit perfect squares are \(16, 25, 36, 49, 64\), and \(81\). Only \(81\) works because \(\sqrt{81} = 9\) and \(8 + 1 = 9\). 2. Let \(r=\sqrt{x}\), so \(x=r^2\) and \(r\geq0\). Then \(4x=4r^2=(2r)^2\). Therefore, \(\sqrt{4x}=2r=2\sqrt{x}\), so the square root doubles. 3. Since \(0.09^2 = 0.0081\), the missing digits are \(8\) and \(1\).

a) The number is \(81\), because \(\sqrt{81} = 8 + 1 = 9\). b) The square root doubles: \(\sqrt{4x}=2\sqrt{x}\) for \(x\geq0\). c) The missing digits are \(8\) and \(1\), so \(\sqrt{0.0081} = 0.09\).

- Convert a mixed number to an improper fraction before taking its square root. - Fully simplify the right side before deciding how many solutions the equation has. - Align decimal place values carefully when subtracting. - Evaluate any square root on the right side before solving the resulting equation.

1. For a), \(0.01 \cdot 0.25 = 0.0025\). Since this is positive, there are two solutions: \(x = 0.05\) and \(x = -0.05\). 2. For b), \(3\frac{1}{16} = \frac{49}{16}\). Therefore, \(y = \pm\sqrt{\frac{49}{16}} = \pm\frac{7}{4}\). 3. For c), \(z^2 = 0.01 - 0.1 = -0.09\). Since a real square cannot be negative, there is no real solution. 4. For d), \(\sqrt{625} = 25\), so \(w^2 = 25\). Therefore, \(w = 5\) or \(w = -5\).

a) Two solutions: \(x = 0.05\) and \(x = -0.05\) b) Two solutions: \(y = \frac{7}{4}\) and \(y = -\frac{7}{4}\) c) No real solution d) Two solutions: \(w = 5\) and \(w = -5\)

- Pay close attention to the location of each negative sign and the parentheses. - Which is evaluated first: an exponent or a leading negative sign? - Determine whether each radicand is positive, zero, or negative.

1. \(\sqrt{(-6)^2} = \sqrt{36} = 6\), so expression (1) is defined. 2. \(-\sqrt{6^2} = -\sqrt{36} = -6\), so expression (2) is defined. 3. Exponents are evaluated before the leading negative sign, so \(\sqrt{-6^2} = \sqrt{-(6^2)} = \sqrt{-36}\). This expression is undefined in the real number system because its radicand is negative.

(1) \(\sqrt{(-6)^2} = 6\) (2) \(-\sqrt{6^2} = -6\) (3) \(\sqrt{-6^2}\) is undefined in the real number system because the radicand is \(-36\).

- Write all three values as decimals. - Convert the percent by dividing by \(100\). - Evaluate the power and square root before comparing.

1. \(A=0.3^2=0.09\). 2. \(B=\sqrt{0.04}=0.2\). 3. \(C=12.5\%=0.125\). 4. Since \(0.09<0.125<0.2\), the order is \(A<C<B\).

\(A<C<B\), or \(0.09<0.125<0.2\)

- Evaluate every square root that does not contain the unknown first. - Isolate \(x\) or the expression containing \(x\) as you would in other equations. - When \(\sqrt{x}\) remains, which operation reverses the square root?

1. For a), \(0.5 + x = 0.9\), so \(x = 0.4\). 2. For b), \(\frac{1}{2}x = \frac{3}{4}\), so \(x = \frac{3}{2} = 1.5\). 3. For c), \(1.1 - \sqrt{x} = 0.6\), so \(\sqrt{x} = 0.5\). Squaring gives \(x = 0.25\). 4. For d), \(\frac{0.2}{\sqrt{x}} = 0.1\). Thus, \(\sqrt{x} = 2\), so \(x = 4\).

a) \(x = 0.4\) b) \(x = 1.5\) c) \(x = 0.25\) d) \(x = 4\)

- Compare the number of solutions to \(x^2 = a\) with the single value of \(\sqrt{a}\). - Consider both possible signs for numbers whose squares are positive. - What does the definition of the principal square root say about its sign? - List all combinations of the possible values of \(x\) and \(y\).

1. The solutions of \(x^2 = 36\) are \(x = 6\) and \(x = -6\). 2. The solutions of \(y^2 = 4\) are \(y = 2\) and \(y = -2\). 3. The possible sums are \(6 + 2 = 8\), \(6 + (-2) = 4\), \(-6 + 2 = -4\), and \(-6 + (-2) = -8\). 4. The principal square root is the unique nonnegative number whose square is the radicand. Therefore, \(\sqrt{36} + \sqrt{4} = 6 + 2 = 8\), with no sign choices.

a) The possible values are \(-8, -4, 4\), and \(8\). b) The value is \(8\). Each radical denotes its unique nonnegative principal square root.

- Subtract the lower prism's volume from the total. - Divide the upper prism's volume by its height to find the square base area. - Use a square root to find the side length.

1. The lower prism has volume \(8\cdot8\cdot3=192\,\text{cm}^3\). 2. The upper prism has volume \(256-192=64\,\text{cm}^3\). 3. Its square base has area \(64\div4=16\,\text{cm}^2\). 4. Since \(x^2=16\) and \(x\) is a length, \(x=\sqrt{16}=4\,\text{cm}\).

\(x=4\,\text{cm}\)

- Locate \(1500\) between two consecutive perfect squares. - The least addition reaches the next square above it. - Subtract the original number from that square.

1. \(38^2=1444\) and \(39^2=1521\). 2. The first perfect square greater than \(1500\) is \(1521\). 3. The required addition is \(1521-1500=21\).

Add \(21\). The resulting perfect square is \(1521=39^2\).

- Find the whole-number square roots near the area interval. - Apply the multiple condition to the possible side lengths. - Square the remaining side length.

1. The whole-number square roots of areas between \(500\) and \(700\) are \(23,24,25,26\). 2. The only one that is a multiple of \(4\) is \(24\). 3. The area is \(24^2=576\,\text{ft}^2\).

Side length: \(24\,\text{ft}\) Area: \(576\,\text{ft}^2\)

- Test one value between \(0\) and \(1\), the value \(1\), and one value greater than \(1\). - Consider how squaring changes a positive number in each interval. - Include the domain restriction for a real square root.

1. In a), \(\sqrt{0.09}=0.3\), so \(0.09<0.3\). Also, \(\sqrt{1}=1\), and \(\sqrt{1.44}=1.2<1.44\). 2. For b), the square root is defined only when \(x\ge0\). At \(x=0\) and \(x=1\), \(\sqrt{x}=x\). 3. For \(0<x<1\), squaring makes a positive number smaller, so \(x^2<x\), which means \(x<\sqrt{x}\). 4. For \(x>1\), squaring makes the number larger, so \(x^2>x\), which means \(x>\sqrt{x}\). Therefore, \(\sqrt{x}<x\) exactly when \(x>1\).

a) (1) \(<\) (2) \(=\) (3) \(>\) b) \(x>1\)