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Approximate irrational numbers

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5143708
Let \(a=-\sqrt{13}\) and \(b=-3.5\). Which number lies farther to the right on the number line? Justify your answer by comparing squares without using a calculator.

Hints

- Compare the magnitudes by squaring the positive values. - Recall how the order changes when positive numbers are replaced by their opposites. - Compute \(3.5^2\) exactly.

Solution

1. Compare the positive magnitudes. Since \((\sqrt{13})^2=13\) and \(3.5^2=12.25\), \(\sqrt{13}>3.5\). 2. For negative numbers, the number with the smaller magnitude is greater. Therefore, \(-3.5>-\sqrt{13}\), so \(b\) lies farther to the right.

Answer

\(b=-3.5\) lies farther to the right because \(-3.5>-\sqrt{13}\).
5143718
For each number \(150\), \(180\), and \(210\), decide whether its square root lies between \(13\) and \(14\). Justify each decision by calculating \(13^2\) and \(14^2\) and comparing.

Hints

- How does comparing radicands to perfect squares help compare their square roots? - If a square root lies between two numbers, where must its radicand lie relative to their squares? - Calculate the two boundary squares first.

Solution

1. The boundary squares are \(13^2 = 169\) and \(14^2 = 196\). 2. Since \(150 < 169\), \(\sqrt{150} < 13\). 3. Since \(169 < 180 < 196\), \(13 < \sqrt{180} < 14\). 4. Since \(210 > 196\), \(\sqrt{210} > 14\).

Answer

Only \(\sqrt{180}\) lies between \(13\) and \(14\), because \(169 < 180 < 196\).
5245118
Use \(\sqrt{15}\approx3.8729833\ldots\). a) Give lower and upper decimal bounds for \(\sqrt{15}\) at increments of \(0.1\), \(0.01\), and \(0.0001\). b) Which interval is the tightest, and what is its width? c) Round \(\sqrt{15}\) to the nearest thousandth.

Hints

- Truncate the decimal at the requested place for the lower bound. - The upper bound is the next decimal at that increment. - Use the digit immediately after the thousandths place when rounding.

Solution

1. At increments of \(0.1\), \(3.8<\sqrt{15}<3.9\). 2. At increments of \(0.01\), \(3.87<\sqrt{15}<3.88\). 3. At increments of \(0.0001\), \(3.8729<\sqrt{15}<3.8730\). 4. The tightest interval is \((3.8729,3.8730)\), and its width is \(3.8730-3.8729=0.0001\). 5. The thousandths digit is \(2\), and the next digit is \(9\), so \(\sqrt{15}\approx3.873\) to the nearest thousandth.

Answer

a) \(3.8<\sqrt{15}<3.9\); \(3.87<\sqrt{15}<3.88\); \(3.8729<\sqrt{15}<3.8730\) b) The tightest interval has width \(0.0001\). c) \(\sqrt{15}\approx3.873\)
5247138
Use \(\sqrt{13}\approx3.606\) and \(\sqrt{2}\approx1.414\). Estimate \(A=2\sqrt{13}\) and \(B=5\sqrt{2}\) to the nearest tenth. Which value is greater?

Hints

- Use the supplied square-root approximations. - Multiply before rounding the final results. - Round both values to the same place before comparing.

Solution

1. \(A\approx2\cdot3.606=7.212\), so \(A\approx7.2\). 2. \(B\approx5\cdot1.414=7.070\), so \(B\approx7.1\). 3. Since \(7.2>7.1\), \(A>B\).

Answer

\(A \approx 7.2\) and \(B \approx 7.1\). Therefore, \(A > B\).
5248138
Use \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\) to evaluate each expression. Round the result to the nearest hundredth. a) \(1.5\sqrt{2} + 2\) b) \(2\sqrt{3} - \sqrt{2}\) c) \(\frac{\sqrt{3} + \sqrt{2}}{2}\)

Hints

- Substitute the given square-root approximations first. - Follow the order of operations carefully. - Round only after completing the calculation. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. For a), \(1.5 \cdot 1.414 + 2 = 2.121 + 2 = 4.121\), so the result is approximately \(4.12\). 2. For b), \(2 \cdot 1.732 - 1.414 = 3.464 - 1.414 = 2.050\), so the result is approximately \(2.05\). 3. For c), \(\frac{1.732 + 1.414}{2} = \frac{3.146}{2} = 1.573\), so the result is approximately \(1.57\).

Answer

a) \(\approx 4.12\) b) \(\approx 2.05\) c) \(\approx 1.57\)
5248298
Use \(\sqrt{3}\approx1.7321\), \(\sqrt{10}\approx3.1623\), \(\sqrt{8}\approx2.8284\), \(\sqrt{2}\approx1.4142\), and \(\sqrt{5}\approx2.2361\). Estimate each expression and round to the nearest hundredth. 1) \(2.45+\sqrt{3}\) 2) \(7.18-\sqrt{10}\) 3) \(\sqrt{8}+\sqrt{2}\) 4) \(3\sqrt{5}-4\)

Hints

- Substitute the supplied approximation for each square root. - Complete all addition, subtraction, and multiplication before rounding. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. Using \(\sqrt{3} \approx 1.7321\), \(2.45 + \sqrt{3} \approx 4.1821\), which rounds to \(4.18\). 2. Using \(\sqrt{10} \approx 3.1623\), \(7.18 - \sqrt{10} \approx 4.0177\), which rounds to \(4.02\). 3. Using \(\sqrt{8} \approx 2.8284\) and \(\sqrt{2} \approx 1.4142\), the sum is approximately \(4.2426\), which rounds to \(4.24\). 4. Using \(\sqrt{5} \approx 2.2361\), \(3\sqrt{5} - 4 \approx 6.7083 - 4 = 2.7083\), which rounds to \(2.71\).

Answer

1) \(\approx 4.18\) 2) \(\approx 4.02\) 3) \(\approx 4.24\) 4) \(\approx 2.71\)
5280898
Use \(\sqrt{7}\approx2.64575\), \(\sqrt{2}\approx1.41421\), and \(\sqrt{15}\approx3.87298\). Evaluate each expression and round to the nearest hundredth. 1) \(4\sqrt{7}\) 2) \(9\sqrt{2}\) 3) \(2\sqrt{15}\)

Hints

- Substitute the supplied approximation for each square root. - Multiply before rounding. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. \(4\sqrt{7} \approx 4 \cdot 2.64575 = 10.583\), so the result is approximately \(10.58\). 2. \(9\sqrt{2} \approx 9 \cdot 1.41421 = 12.72789\), so the result is approximately \(12.73\). 3. \(2\sqrt{15} \approx 2 \cdot 3.87298 = 7.74596\), so the result is approximately \(7.75\).

Answer

1) \(\approx 10.58\) 2) \(\approx 12.73\) 3) \(\approx 7.75\)
5498578
Two intervals are marked on the number line. Which interval contains \(\sqrt{13}\)? Justify the choice by squaring the endpoints.
Figure for problem 549857

Hints

- Identify the endpoints of each marked interval. - Compare the squares of nearby decimals with the radicand. - The square root lies between endpoints whose squares surround the radicand.

Solution

1. \(3.6^2=12.96<13\). 2. \(3.7^2=13.69>13\). 3. Therefore, \(3.6<\sqrt{13}<3.7\). 4. Interval B is the interval from \(3.6\) to \(3.7\).

Answer

Interval B
5501588
The number line is marked in tenths. Between which two consecutive tenths does \(\sqrt{29}\) lie? Justify your answer by squaring both endpoints.
Figure for problem 550158

Hints

- First identify the consecutive whole numbers that contain the square root. - Test neighboring tenths by squaring them. - The lower endpoint must have a square below \(29\), and the upper endpoint must have a square above \(29\).

Solution

1. Since \(5^2=25<29<36=6^2\), \(\sqrt{29}\) lies between \(5\) and \(6\). 2. \(5.3^2=28.09<29\). 3. \(5.4^2=29.16>29\). 4. Therefore, \(5.3<\sqrt{29}<5.4\).

Answer

\(5.3<\sqrt{29}<5.4\)
5545738
The number line shows two candidate tenths and their midpoint for \(\sqrt[3]{-20}\). First cube the two candidate tenths to verify that \(\sqrt[3]{-20}\) lies between them. Then cube the displayed midpoint and use that comparison to decide which tenth is closer. Your work must include the midpoint and its cube. Round \(\sqrt[3]{-20}\) to the nearest tenth.
Figure for problem 554573

Hints

- Cube the two candidate tenths shown on the number line and compare those cubes with the radicand. - Remember that the real cube function preserves order, including for negative inputs. - Cube the displayed midpoint and use that comparison as the rounding test rather than comparing distances between the endpoint cubes and the radicand.

Solution

1. Cube the candidate tenths: \((-2.8)^3=-21.952\) and \((-2.7)^3=-19.683\). 2. Since \(-21.952<-20<-19.683\) and cubing is increasing on the real numbers, \(-2.8<\sqrt[3]{-20}<-2.7\). 3. Cube the midpoint: \((-2.75)^3=-20.796875\). 4. Since \(-20>-20.796875\), \(\sqrt[3]{-20}>-2.75\), so it is closer to \(-2.7\). 5. Therefore, \(\sqrt[3]{-20}\approx-2.7\) to the nearest tenth.

Answer

\((-2.8)^3=-21.952<-20<-19.683=(-2.7)^3\), so \(-2.8<\sqrt[3]{-20}<-2.7\). The midpoint satisfies \((-2.75)^3=-20.796875<-20\), so \(\sqrt[3]{-20}>-2.75\) and \(\sqrt[3]{-20}\approx-2.7\) to the nearest tenth.
5126908
A circle has radius \(r\). An inscribed square has area \(2r^2\), and a square circumscribed about the circle has area \(4r^2\). The circle has area \(\pi r^2\). a) Use the three areas to give lower and upper bounds for \(\pi\). b) An inscribed regular dodecagon has area \(3r^2\), and a circumscribed regular dodecagon has area approximately \(3.215r^2\). Give the new bounds for \(\pi\) and find the width of the interval. c) Which pair of polygons gives the tighter bounds? Explain using the interval widths.

Hints

- The area of the circle is between the areas of the inscribed and circumscribed polygons. - Divide each area inequality by the positive quantity \(r^2\). - Compare the widths by subtracting each lower bound from its upper bound.

Solution

1. The circle lies between the two squares, so \(2r^2<\pi r^2<4r^2\). Since \(r^2>0\), divide by \(r^2\) to obtain \(2<\pi<4\). 2. The dodecagon areas give \(3r^2<\pi r^2<3.215r^2\), so \(3<\pi<3.215\). 3. The width of the dodecagon interval is \(3.215-3=0.215\). 4. The square interval has width \(4-2=2\). Because \(0.215<2\), the dodecagons give tighter bounds.

Answer

a) \(2<\pi<4\) b) \(3<\pi<3.215\); the interval width is \(0.215\). c) The dodecagons give tighter bounds because \(0.215<2\).
5142888
Order the numbers from least to greatest without using a calculator: \(\sqrt{18},\ 4.1,\ \sqrt{16},\ 4,\ \sqrt{20}\)

Hints

- Evaluate any square roots of perfect squares first. - To compare a positive decimal with a square root, compare their squares. - Use the order of the radicands to compare positive square roots.

Solution

1. Since \(\sqrt{16}=4\), those two values are equal. 2. Because \(4.1^2=16.81<18\), \(4.1<\sqrt{18}\). 3. Since \(18<20\), \(\sqrt{18}<\sqrt{20}\). 4. Therefore, \(4=\sqrt{16}<4.1<\sqrt{18}<\sqrt{20}\).

Answer

\(4=\sqrt{16}<4.1<\sqrt{18}<\sqrt{20}\)
5142988
Without using a calculator, insert \(<\), \(>\), or \(=\) to make each statement true. Briefly justify each answer. a) \(\sqrt{0.09}\) ___ \(0.09\) b) \(\sqrt{1.21}\) ___ \(1.1\) c) \(\sqrt{20}\) ___ \(4.5\) d) \(\sqrt{\frac{1}{2}}\) ___ \(\frac{1}{2}\)

Hints

- Evaluate square roots of perfect-square decimals when possible. - When both values are nonnegative, compare their squares. - Remember how squaring affects numbers between \(0\) and \(1\).

Solution

1. In a), \(\sqrt{0.09}=0.3\), and \(0.3>0.09\). 2. In b), \(1.1^2=1.21\), so \(\sqrt{1.21}=1.1\). 3. In c), \(4.5^2=20.25>20\), so \(\sqrt{20}<4.5\). 4. In d), both quantities are positive, and \(\left(\frac{1}{2}\right)^2=\frac{1}{4}<\frac{1}{2}\). Therefore, \(\sqrt{\frac{1}{2}}>\frac{1}{2}\).

Answer

a) \(\sqrt{0.09}>0.09\) b) \(\sqrt{1.21}=1.1\) c) \(\sqrt{20}<4.5\) d) \(\sqrt{\frac{1}{2}}>\frac{1}{2}\)
5143068
Estimate each square root without using a calculator. For every part, first locate the root between the two given benchmarks. Then find the midpoint of those benchmarks, square that midpoint, and use the comparison to decide the requested rounding. Your work must include the midpoint and its square. a) Estimate \(\sqrt{500}\) to the nearest whole number. Use \(22^2=484\) and \(23^2=529\). b) Estimate \(\sqrt{0.05}\) to the nearest tenth. Use \(0.2^2=0.04\) and \(0.3^2=0.09\). c) Estimate \(\sqrt{50{,}000}\) to the nearest ten. Use \(220^2=48{,}400\) and \(230^2=52{,}900\).

Hints

- Use the two given square benchmarks to establish an interval for each root. - The rounding boundary is the midpoint of the two candidate rounded values. - Square that midpoint and compare its square with the radicand before choosing the rounded value.

Solution

1. Since \(484<500<529\), \(22<\sqrt{500}<23\). The midpoint is \(22.5\), and \(22.5^2=506.25>500\), so \(\sqrt{500}<22.5\). Therefore, \(\sqrt{500}\approx22\) to the nearest whole number. 2. Since \(0.04<0.05<0.09\), \(0.2<\sqrt{0.05}<0.3\). The midpoint is \(0.25\), and \(0.25^2=0.0625>0.05\), so \(\sqrt{0.05}<0.25\). Therefore, \(\sqrt{0.05}\approx0.2\) to the nearest tenth. 3. Since \(48{,}400<50{,}000<52{,}900\), \(220<\sqrt{50{,}000}<230\). The midpoint is \(225\), and \(225^2=50{,}625>50{,}000\), so \(\sqrt{50{,}000}<225\). Therefore, \(\sqrt{50{,}000}\approx220\) to the nearest ten.

Answer

a) Midpoint \(22.5\); \(22.5^2=506.25>500\), so \(\sqrt{500}\approx22\). b) Midpoint \(0.25\); \(0.25^2=0.0625>0.05\), so \(\sqrt{0.05}\approx0.2\). c) Midpoint \(225\); \(225^2=50{,}625>50{,}000\), so \(\sqrt{50{,}000}\approx220\).
5143548
A square storage room has a floor area of \(32\,\text{m}^2\). a) Find the side length \(s\). Give the exact value as a square root and an approximation rounded to the nearest hundredth. b) Between which two consecutive whole numbers of meters does the side length lie? Justify your answer by comparing perfect squares. c) If the side length were doubled, what would the new floor area be?
Figure for problem 514354

Hints

- Is \(32\) a perfect square? - Which familiar perfect squares are closest to \(32\)? - How does a square’s area change when its side length is multiplied by a factor?

Solution

1. The side length is \(s = \sqrt{32}\,\text{m} \approx 5.66\,\text{m}\). 2. Since \(5^2 = 25\) and \(6^2 = 36\), and \(25 < 32 < 36\), \(\sqrt{32}\) lies between \(5\) and \(6\). 3. Doubling the side length multiplies the area by \(2^2 = 4\). The new area is \(4 \cdot 32\,\text{m}^2 = 128\,\text{m}^2\).

Answer

a) \(s = \sqrt{32}\,\text{m} \approx 5.66\,\text{m}\) b) The side length lies between \(5\,\text{m}\) and \(6\,\text{m}\). c) The new floor area is \(128\,\text{m}^2\).
5144038
Consider \(\sqrt{7}\). a) Find two rational numbers with exactly one decimal place between which \(\sqrt{7}\) lies. b) A student claims, “If I keep narrowing rational intervals, eventually I will find two rational endpoints so close together that \(\sqrt{7}\) is exactly one of them.” Explain why this claim contradicts the definition of an irrational number.

Hints

- Square decimal values to compare them with \(7\). - What property separates rational and irrational numbers? - Think about the decimal expansion of an irrational number.

Solution

1. Since \(2.6^2 = 6.76\) and \(2.7^2 = 7.29\), \(2.6 < \sqrt{7} < 2.7\). 2. Every terminating decimal is rational because it can be written as a fraction of integers. 3. If the interval process ended with \(\sqrt{7}\) equal to a rational endpoint, then \(\sqrt{7}\) would be rational. 4. But \(\sqrt{7}\) is irrational. Its decimal expansion is nonterminating and nonrepeating. Rational intervals can approximate it as closely as desired, but no terminating rational endpoint equals it exactly.

Answer

a) \(2.6 < \sqrt{7} < 2.7\) b) The claim is false. Every terminating decimal endpoint is rational, but \(\sqrt{7}\) is irrational, so no such endpoint can equal \(\sqrt{7}\) exactly.
5144058
Let \(x = \sqrt{45}\) and \(y = 6.71\). a) Find an interval of width \(0.01\) that contains \(x\). b) Compare \(x\) and \(y\). Which number is greater? Justify your answer using the interval from part a).

Hints

- Square nearby hundredths to determine whether they are below or above \(45\). - What does the upper endpoint of the interval tell you when comparing \(x\) with \(6.71\)?

Solution

1. Since \(6.70^2 = 44.89\) and \(6.71^2 = 45.0241\), \(44.89 < 45 < 45.0241\). 2. Therefore, \(6.70 < \sqrt{45} < 6.71\), so an interval of width \(0.01\) containing \(x\) is \([6.70, 6.71]\). 3. Because \(x < 6.71 = y\), \(y\) is greater than \(x\).

Answer

a) \(x \in [6.70, 6.71]\) b) \(y > x\), because \(6.71^2 = 45.0241 > 45\), so \(\sqrt{45} < 6.71\).
5144068
A real number \(z\) lies in the interval \(I_3 = [5.38, 5.39]\). Determine whether each number could be the value of \(z\). Justify each answer with a calculation. a) \(\sqrt{29}\) b) \(\frac{43}{8}\) c) \(5.382\)

Hints

- A number is in \([a, b]\) when it is at least \(a\) and at most \(b\). - Compare a square root with positive bounds by squaring the bounds. - Convert the fraction to a decimal for a direct comparison.

Solution

1. For a), square the positive interval endpoints: \(5.38^2 = 28.9444\) and \(5.39^2 = 29.0521\). Since \(28.9444 < 29 < 29.0521\), it follows that \(5.38 < \sqrt{29} < 5.39\). Therefore, \(\sqrt{29}\) is in the interval. 2. For b), \(\frac{43}{8} = 5.375\). Since \(5.375 < 5.38\), the number is not in the interval. 3. For c), \(5.38 < 5.382 < 5.39\), so \(5.382\) is in the interval.

Answer

a) Yes, because \(5.38^2 < 29 < 5.39^2\). b) No, because \(\frac{43}{8} = 5.375 < 5.38\). c) Yes, because \(5.38 < 5.382 < 5.39\).
5144078
The Babylonian method for approximating square roots can be viewed geometrically as repeatedly changing a rectangle into a more square-like rectangle with the same area. A rectangle has area \(18\,\text{cm}^2\) and one side length \(a_0=6\,\text{cm}\). a) Find the other side length \(b_0\). b) The next approximation \(a_1\) is the mean of \(a_0\) and \(b_0\). Find \(a_1\). c) Find the new side length \(b_1\) that keeps the area equal to \(18\,\text{cm}^2\). d) Calculate \(b_1^2\) and \(a_1^2\). Use those comparisons with \(18\,\text{cm}^2\) to place the square's side length \(\sqrt{18}\,\text{cm}\) between \(b_1\) and \(a_1\). Then find the absolute square error \(\left|a_1^2-18\,\text{cm}^2\right|\).

Hints

- Relate the side lengths of a rectangle to its area. - The next Babylonian approximation uses the mean of the two current side lengths. - After finding \(a_1\) and \(b_1\), compare their squares with the target area. - Use the order of the positive squares to infer the order of the corresponding positive side lengths and \(\sqrt{18}\).

Solution

1. Since \(A=a_0b_0\), \(b_0=18\div6=3\,\text{cm}\). 2. The mean is \(a_1=\frac{6+3}{2}=4.5\,\text{cm}\). 3. To preserve the area, \(b_1=18\div4.5=4\,\text{cm}\). 4. \(b_1^2=(4\,\text{cm})^2=16\,\text{cm}^2\), while \(a_1^2=(4.5\,\text{cm})^2=20.25\,\text{cm}^2\). 5. Since \(16\,\text{cm}^2<18\,\text{cm}^2<20.25\,\text{cm}^2\) and \(b_1\) and \(a_1\) are positive, \(4\,\text{cm}<\sqrt{18}\,\text{cm}<4.5\,\text{cm}\). 6. The absolute square error of \(a_1\) is \(\left|20.25-18\right|\,\text{cm}^2=2.25\,\text{cm}^2\).

Answer

a) \(b_0=3\,\text{cm}\) b) \(a_1=4.5\,\text{cm}\) c) \(b_1=4\,\text{cm}\) d) \(b_1^2=16\,\text{cm}^2<18\,\text{cm}^2<20.25\,\text{cm}^2=a_1^2\), so \(4\,\text{cm}<\sqrt{18}\,\text{cm}<4.5\,\text{cm}\). The absolute square error of \(a_1\) is \(2.25\,\text{cm}^2\).
5144098
Two students estimate \(\sqrt{2}\). Student A uses \(1.41\), and Student B uses \(1.42\). a) Use \(1.41^2=1.9881\) and \(1.42^2=2.0164\) to show that \(\sqrt{2}\) lies between the two estimates. b) The midpoint of the estimates is \(1.415\), and \(1.415^2=2.002225\). Which student has the closer estimate? Explain.

Hints

- Compare the given squares with \(2\) to locate the square root. - The midpoint separates values closer to \(1.41\) from values closer to \(1.42\). - Square the midpoint to decide on which side of it \(\sqrt{2}\) lies.

Solution

1. Since \(1.9881<2<2.0164\), it follows that \(1.41<\sqrt{2}<1.42\). 2. Because \(1.415^2=2.002225>2\), \(\sqrt{2}<1.415\). The exact value lies below the midpoint between \(1.41\) and \(1.42\), so it is closer to \(1.41\). Student A has the closer estimate.

Answer

a) \(1.41<\sqrt{2}<1.42\) b) Student A; \(1.41\) is closer to \(\sqrt{2}\).
5144178
The number line shows an interval \(I\). a) Show that \(\sqrt{10}\) lies in \(I\) by squaring the endpoints. b) Compare \(\sqrt{10}\) with \(3.165\). c) Round \(\sqrt{10}\) to the nearest hundredth.
Figure for problem 514417

Hints

- Square the two endpoints of the displayed interval. - Use \(3.165\) as the midpoint between the hundredths. - The side of the midpoint determines the nearest hundredth.

Solution

1. The interval is \(I=[3.16,3.17]\). Since \(3.16^2=9.9856<10\) and \(3.17^2=10.0489>10\), \(3.16<\sqrt{10}<3.17\). 2. Since \(3.165^2=10.017225>10\), \(\sqrt{10}<3.165\). 3. The value lies below the midpoint between \(3.16\) and \(3.17\), so \(\sqrt{10}\approx3.16\) to the nearest hundredth.

Answer

a) \(3.16<\sqrt{10}<3.17\) b) \(\sqrt{10}<3.165\) c) \(\sqrt{10}\approx3.16\)
5144528
Find three nested decimal intervals that contain \(\sqrt{11}\): first an interval with consecutive whole-number endpoints, then one with consecutive tenth endpoints, and then one with consecutive hundredth endpoints.

Hints

- Which perfect squares are closest to \(11\)? - Square consecutive tenths to narrow the interval. - Continue by testing consecutive hundredths.

Solution

1. Since \(3^2 = 9\) and \(4^2 = 16\), \(3 < \sqrt{11} < 4\). The first interval is \([3, 4]\). 2. Since \(3.3^2 = 10.89\) and \(3.4^2 = 11.56\), \(3.3 < \sqrt{11} < 3.4\). The second interval is \([3.3, 3.4]\). 3. Since \(3.31^2 = 10.9561\) and \(3.32^2 = 11.0224\), \(3.31 < \sqrt{11} < 3.32\). The third interval is \([3.31, 3.32]\).

Answer

1. \([3, 4]\) 2. \([3.3, 3.4]\) 3. \([3.31, 3.32]\)
5144538
A nested-interval approximation for \(\sqrt{2}\) begins with \(I_1 = [1, 2]\). Each new interval is formed by dividing the previous interval into \(10\) equal parts and selecting the part that contains \(\sqrt{2}\). a) What is the width of the fourth interval \(I_4\)? b) Use squaring to determine \(I_4\), and decide whether \(1.414\) belongs to it.

Hints

- What happens to an interval’s width when it is divided into \(10\) equal parts? - Square the proposed endpoints to check whether \(\sqrt{2}\) lies between them. - A closed interval includes both endpoints.

Solution

1. The width of \(I_1\) is \(1\). Each subdivision divides the width by \(10\), so the widths of \(I_2\), \(I_3\), and \(I_4\) are \(0.1\), \(0.01\), and \(0.001\), respectively. 2. Since \(1.4^2 = 1.96\) and \(1.5^2 = 2.25\), \(I_2 = [1.4, 1.5]\). 3. Since \(1.41^2 = 1.9881\) and \(1.42^2 = 2.0164\), \(I_3 = [1.41, 1.42]\). 4. Since \(1.414^2 = 1.999396\) and \(1.415^2 = 2.002225\), \(I_4 = [1.414, 1.415]\). The number \(1.414\) is the lower endpoint, so it belongs to this closed interval.

Answer

a) The width of \(I_4\) is \(0.001\). b) \(I_4 = [1.414, 1.415]\), and \(1.414\) belongs to the interval.
5144548
Consider \(\sqrt{18}\). a) Find an interval of the form \([n, n+1]\), where \(n\) is a natural number, that contains \(\sqrt{18}\). b) Without directly evaluating the square root on a calculator, decide whether \(\sqrt{18}\) is closer to \(4.2\) or \(4.3\). Justify your answer by comparing squares.

Hints

- Begin with neighboring perfect squares. - The midpoint determines which of two approximations is closer. - Find the midpoint of \(4.2\) and \(4.3\), then compare its square with \(18\).

Solution

1. Since \(4^2 = 16 < 18 < 25 = 5^2\), \(4 < \sqrt{18} < 5\). Thus, \(\sqrt{18}\) lies in \([4, 5]\). 2. The midpoint of \(4.2\) and \(4.3\) is \(4.25\). Since \(4.2^2 = 17.64 < 18\) and \(4.25^2 = 18.0625 > 18\), \(4.2 < \sqrt{18} < 4.25\). 3. Because \(\sqrt{18}\) is below the midpoint, it is closer to \(4.2\) than to \(4.3\).

Answer

a) \(\sqrt{18} \in [4, 5]\) b) \(\sqrt{18}\) is closer to \(4.2\), because \(4.2 < \sqrt{18} < 4.25\), and \(4.25\) is the midpoint of \(4.2\) and \(4.3\).
5144788
Consider \(\sqrt{13}\). a) Find the two consecutive natural numbers between which \(\sqrt{13}\) lies. b) Use squaring to determine whether \(\sqrt{13}\) is closer to \(3.6\) or \(3.61\). c) Without a calculator, explain why \(\frac{11}{3}\) must be greater than \(\sqrt{13}\).

Hints

- Use neighboring perfect squares. - To compare \(3.6\) and \(3.61\), square their midpoint. - A positive number is greater than \(\sqrt{13}\) when its square is greater than \(13\).

Solution

1. Since \(3^2 = 9 < 13 < 16 = 4^2\), \(3 < \sqrt{13} < 4\). 2. The midpoint of \(3.6\) and \(3.61\) is \(3.605\). Since \(3.605^2 = 12.996025 < 13\), \(\sqrt{13}\) lies above the midpoint. Therefore, it is closer to \(3.61\). 3. \(\left(\frac{11}{3}\right)^2 = \frac{121}{9} > 13\). Since \(\frac{11}{3}\) is positive, \(\frac{11}{3} > \sqrt{13}\).

Answer

a) \(3 < \sqrt{13} < 4\) b) \(\sqrt{13}\) is closer to \(3.61\). c) Since \(\left(\frac{11}{3}\right)^2 > 13\), \(\frac{11}{3} > \sqrt{13}\).
5144818
A square lot has an area of \(18\,\text{m}^2\), so its side length is \(s = \sqrt{18}\,\text{m}\). a) Between which two consecutive whole numbers does \(s\) lie? b) Determine the tenths interval containing \(s\) by comparing the squares of \(4.1\), \(4.2\), and \(4.3\). c) Briefly explain why this process never produces a terminating decimal equal to \(\sqrt{18}\).

Hints

- Which familiar perfect squares are nearest to \(18\)? - Compare how the square changes as the decimal input increases by tenths. - Recall the difference between rational and irrational numbers.

Solution

1. Since \(4^2 = 16\) and \(5^2 = 25\), \(4 < \sqrt{18} < 5\). 2. \(4.1^2 = 16.81\), \(4.2^2 = 17.64\), and \(4.3^2 = 18.49\). Therefore, \(4.2 < \sqrt{18} < 4.3\), so the tenths digit is \(2\). 3. The integer \(18\) is not a perfect square, so \(\sqrt{18}\) is irrational. Its decimal expansion is nonterminating and nonrepeating, so increasingly narrow decimal intervals never end at an exact terminating decimal.

Answer

a) \(4\,\text{m} < s < 5\,\text{m}\) b) \(4.2 < \sqrt{18} < 4.3\), so the tenths digit is \(2\). c) \(\sqrt{18}\) is irrational, so its decimal expansion is nonterminating and nonrepeating.
5155588
Let \(A=\sqrt{0.36}+\sqrt{0.25}\) and \(B=\sqrt{0.36+0.25}\). Which expression has the greater value? Justify your answer without approximating the square root in \(B\).

Hints

- Find the exact value of \(A\). - Combine the numbers inside \(B\). - Compare two positive values by comparing their squares.

Solution

1. Evaluate \(A\): \(A=0.6+0.5=1.1\). 2. Combine the numbers under the radical in \(B\): \(B=\sqrt{0.61}\). 3. Both values are positive. Since \(1.1^2=1.21>0.61\), \(1.1>\sqrt{0.61}\). Therefore, \(A>B\).

Answer

\(A>B\)
5245148
Order the real numbers from least to greatest without using a calculator: \(\sqrt{5},\ 2.23,\ 2.\overline{2},\ \frac{9}{4},\ 2.24\)

Hints

- Convert the repeating decimal and fraction to comparable decimal forms. - Compare a positive decimal with \(\sqrt{5}\) by squaring the decimal. - Place the two decimal bounds on opposite sides of \(\sqrt{5}\).

Solution

1. Write the rational values in comparable form: \(2.\overline{2}=2.2222\ldots\) and \(\frac{9}{4}=2.25\). 2. Since \(2.23^2=4.9729<5\), \(2.23<\sqrt{5}\). 3. Since \(2.24^2=5.0176>5\), \(\sqrt{5}<2.24\). 4. Therefore, \(2.\overline{2}<2.23<\sqrt{5}<2.24<\frac{9}{4}\).

Answer

\(2.\overline{2}<2.23<\sqrt{5}<2.24<\frac{9}{4}\)
5245258
The area \(A\) of a circle is given. Use a calculator to find the radius \(r\) and round to the nearest hundredth. a) \(A = 10\,\text{cm}^2\) b) \(A = 35.5\,\text{m}^2\) c) \(A = 0.8\,\text{dm}^2\)

Hints

- Which formula relates a circle’s area and radius? - Rearrange the formula so the squared radius is isolated before taking a square root. - Use the third decimal place to round to the nearest hundredth. - Include the correct length unit in each answer.

Solution

1. From \(A = \pi r^2\), solve for the radius: \(r = \sqrt{\frac{A}{\pi}}\). 2. For a), \(r = \sqrt{\frac{10}{\pi}}\,\text{cm} \approx 1.78\,\text{cm}\). 3. For b), \(r = \sqrt{\frac{35.5}{\pi}}\,\text{m} \approx 3.36\,\text{m}\). 4. For c), \(r = \sqrt{\frac{0.8}{\pi}}\,\text{dm} \approx 0.50\,\text{dm}\).

Answer

a) \(r \approx 1.78\,\text{cm}\) b) \(r \approx 3.36\,\text{m}\) c) \(r \approx 0.50\,\text{dm}\)
5245268
The area of a circle is \(A = \pi r^2\). a) Use a calculator to find the radius of a circle with area \(50\,\text{cm}^2\). Round to the nearest hundredth. b) Find the new area \(A_{\text{new}}\) needed to make the radius exactly twice the original circle’s radius. c) In general, how does a circle’s radius change when its area is multiplied by \(4\)? Justify your answer using the formula.

Hints

- Solve the circle-area formula for the positive radius in part a). - Substitute \(2r\) for the radius to see how the area changes. - Use the same relationship in reverse when the area is multiplied by \(4\).

Solution

1. For part a), \(r=\sqrt{\frac{50}{\pi}}\,\text{cm}\approx3.99\,\text{cm}\). 2. If the radius doubles, \(A_{\text{new}}=\pi(2r)^2=4\pi r^2=4A\). Therefore, \(A_{\text{new}}=4\cdot50\,\text{cm}^2=200\,\text{cm}^2\). 3. Conversely, a circle with radius \(2r\) has area \(4A\). Because a radius is positive, multiplying a circle's area by \(4\) doubles its radius.

Answer

a) \(r \approx 3.99\,\text{cm}\) b) \(A_{\text{new}} = 200\,\text{cm}^2\) c) The radius doubles when the area is multiplied by \(4\).
5247058
Estimate \(15\sqrt{2}\) to the nearest tenth using two methods. a) First round \(\sqrt{2}\) to the nearest tenth, and then multiply by \(15\). b) Instead, use \(\sqrt{2}\approx1.414\), multiply first, and round only the final product to the nearest tenth. c) The correctly rounded value is \(21.2\). Which method is more accurate, and why?

Hints

- Follow the stated order of operations in each method. - Keep all three given decimal places in method b). - Consider what happens to an early approximation error when it is multiplied by \(15\).

Solution

1. Method a): \(\sqrt{2}\approx1.4\), so \(15\cdot1.4=21.0\). 2. Method b): \(15\cdot1.414=21.21\), which rounds to \(21.2\). 3. Method b) is more accurate. Method a) rounds before multiplying, so the initial rounding error is also multiplied by \(15\). Method b) keeps more information and rounds only once, at the end.

Answer

a) \(21.0\) b) \(21.2\) c) Method b) is more accurate because it delays rounding until the final result.
5247148
Consider \(4\sqrt{5}\). a) Compute \((4\sqrt{5})^2\). Use the result to find two consecutive whole numbers between which \(4\sqrt{5}\) lies. b) Use \(\sqrt{5}\approx2.236\) to estimate \(4\sqrt{5}\) to the nearest tenth.

Hints

- Square the entire positive expression rather than rewriting the radical. - Compare the resulting value with neighboring perfect squares. - Multiply the supplied decimal approximation before rounding.

Solution

1. \((4\sqrt{5})^2=16\cdot5=80\). 2. Since \(8^2=64<80<81=9^2\) and \(4\sqrt{5}\) is positive, \(8<4\sqrt{5}<9\). 3. Using \(\sqrt{5}\approx2.236\), \(4\sqrt{5}\approx4\cdot2.236=8.944\), which rounds to \(8.9\).

Answer

a) \(8<4\sqrt{5}<9\) b) \(4\sqrt{5}\approx8.9\)
5248308
A square has area \(A=7\,\text{cm}^2\). a) Find the side length \(s\) and perimeter \(P\). Round both values to the nearest hundredth. b) Use \(d^2=s^2+s^2\) to find the diagonal length \(d\). Round to the nearest hundredth.

Hints

- Use the area to determine the positive side length. - A square has four equal sides. - For the diagonal, substitute \(s^2=7\) into the given relationship and then take the positive square root.

Solution

1. The side length is \(s=\sqrt{7}\,\text{cm}\approx2.65\,\text{cm}\). 2. The perimeter is \(P=4s=4\sqrt{7}\,\text{cm}\approx10.58\,\text{cm}\). 3. For the diagonal, \(d^2=s^2+s^2=7+7=14\). Since a length is positive, \(d=\sqrt{14}\,\text{cm}\approx3.74\,\text{cm}\).

Answer

a) \(s \approx 2.65\,\text{cm}\); \(P \approx 10.58\,\text{cm}\) b) \(d \approx 3.74\,\text{cm}\)
5281018
Find all approximate solutions to each equation. Round to the nearest hundredth. 1) \(x^2 - 13 = 0\) 2) \(0.5x^2 = 11\) 3) \(4x^2 - 30 = 50\)

Hints

- Isolate \(x^2\) before taking a square root. - Include both the positive and negative solutions. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. \(x^2 = 13\), so \(x = \pm\sqrt{13} \approx \pm 3.61\). 2. Multiply by \(2\): \(x^2 = 22\). Therefore, \(x = \pm\sqrt{22} \approx \pm 4.69\). 3. Add \(30\) and divide by \(4\): \(x^2 = 20\). Therefore, \(x = \pm\sqrt{20} \approx \pm 4.47\).

Answer

1) \(x \approx 3.61\) or \(x \approx -3.61\) 2) \(x \approx 4.69\) or \(x \approx -4.69\) 3) \(x \approx 4.47\) or \(x \approx -4.47\)
5281028
Consider the equation \(x^2=40\). a) Without using a calculator’s square-root key, find two consecutive whole numbers between which the positive solution lies. Explain. b) Use a calculator to find both solutions to the nearest hundredth.

Hints

- Which familiar perfect squares are nearest to \(40\)? - Compare \(40\) with those squares. - If a positive number has square \(40\), what can you say about its opposite?

Solution

1. Since \(6^2 = 36\) and \(7^2 = 49\), and \(36 < 40 < 49\), the positive solution \(\sqrt{40}\) lies between \(6\) and \(7\). 2. \(\sqrt{40} \approx 6.32455\ldots\), so the positive solution is approximately \(6.32\). 3. The negative solution is its opposite, approximately \(-6.32\).

Answer

a) The positive solution lies between \(6\) and \(7\). b) \(x \approx 6.32\) or \(x \approx -6.32\)
5498058
Every number in the list \(1.4,\ 1.41,\ 1.414,\ 1.4142,\ldots\) is a terminating decimal approximation to \(\sqrt{2}\). A student argues that \(\sqrt{2}\) must therefore be rational because all the approximations are rational. Explain why the conclusion does not follow.

Hints

- Separate “close to” from “equal to.” - Ask whether any finite truncation contains the complete decimal expansion. - Classification applies to the exact number, not to its approximations.

Solution

1. Each listed approximation is rational because it terminates. 2. None of the listed decimals equals \(\sqrt{2}\) exactly. 3. A sequence of rational approximations can approach an irrational number arbitrarily closely. 4. The exact decimal expansion of \(\sqrt{2}\) is nonterminating and nonrepeating, so \(\sqrt{2}\) is irrational.

Answer

The argument is invalid. Rational approximations can approach an irrational number without ever equaling it; \(\sqrt{2}\) remains irrational.
5498568
The number line shows three labeled points. Which labeled point is closest to \(\sqrt{5}\)? Justify your choice without using a calculator’s square-root key.
Figure for problem 549856

Hints

- Square the two neighboring decimal values. - Decide which interval contains the irrational number. - Compare with the midpoint of that interval.

Solution

1. \(2.2^2=4.84\) and \(2.3^2=5.29\), so \(2.2<\sqrt{5}<2.3\). 2. The midpoint is \(2.25\), and \(2.25^2=5.0625>5\). 3. Therefore, \(\sqrt{5}<2.25\), so it is closer to \(2.2\) than to \(2.3\). 4. Point A marks \(2.2\), so A is closest.

Answer

Point A
5498588
The number line shows points P and Q. Which point gives the better nearest-tenth approximation to \(\sqrt{70}\)? Use the midpoint to justify your decision.
Figure for problem 549858

Hints

- Find the value halfway between the two candidate tenths. - Compare the square of that midpoint with the radicand. - Use which side of the midpoint contains the root.

Solution

1. The midpoint between \(8.3\) and \(8.4\) is \(8.35\). 2. \(8.35^2=69.7225<70\), so \(\sqrt{70}>8.35\). 3. Also, \(8.4^2=70.56>70\), so \(\sqrt{70}<8.4\). 4. Therefore, \(\sqrt{70}\) is closer to \(8.4\), which is point Q.

Answer

Point Q; \(\sqrt{70}\approx8.4\) to the nearest tenth.
5498598
Use the bound \(3.14<\pi<3.15\) to estimate \(\pi^2\) to the nearest tenth. Show how squaring the bounds supports the rounded value.

Hints

- Apply the same increasing operation to both positive bounds. - Calculate the two endpoint values accurately. - Check whether the entire resulting interval rounds to one tenth value.

Solution

1. Squaring the positive bounds gives \(3.14^2<\pi^2<3.15^2\). 2. \(3.14^2=9.8596\) and \(3.15^2=9.9225\). 3. Every value in this interval rounds to \(9.9\) to the nearest tenth. 4. Therefore, \(\pi^2\approx9.9\).

Answer

\(\pi^2\approx9.9\)
5498608
The number line shows a marked interval and its midpoint for \(\sqrt[3]{30}\). First cube the two endpoints to verify that \(\sqrt[3]{30}\) lies in the marked interval. Then cube the midpoint shown on the number line and use that comparison to round \(\sqrt[3]{30}\) to the nearest hundredth.
Figure for problem 549860

Hints

- Use cubes, not squares, to verify the endpoints of the marked interval. - The midpoint shown on the number line is the rounding boundary between the two hundredths. - Cube that midpoint and compare its cube with \(30\).

Solution

1. \(3.10^3=29.791<30\). 2. \(3.11^3=30.080231>30\). 3. Therefore, \(3.10<\sqrt[3]{30}<3.11\). 4. The midpoint is \(3.105\), and \(3.105^3=29.935382625<30\). 5. Since cubing is increasing on the real numbers, \(\sqrt[3]{30}>3.105\), so it rounds to \(3.11\) to the nearest hundredth.

Answer

\(3.10^3=29.791<30<30.080231=3.11^3\), so \(3.10<\sqrt[3]{30}<3.11\). The midpoint satisfies \(3.105^3=29.935382625<30\), so \(\sqrt[3]{30}>3.105\) and \(\sqrt[3]{30}\approx3.11\) to the nearest hundredth.
5498618
The number line shows a highlighted interval. Use squaring to show that \(\sqrt{27}\) lies in this interval. Then decide whether \(\sqrt{27}\) or \(5.20\) is greater, and state how far apart they are at most.
Figure for problem 549861

Hints

- Square both endpoints of the highlighted interval. - Compare the results with the radicand. - Use the interval width to bound the difference.

Solution

1. \(5.19^2=26.9361<27\). 2. \(5.20^2=27.04>27\). 3. Therefore, \(5.19<\sqrt{27}<5.20\). 4. Thus, \(5.20\) is greater than \(\sqrt{27}\), and their difference is less than \(0.01\).

Answer

\(5.19<\sqrt{27}<5.20\). The number \(5.20\) is greater, and the difference is less than \(0.01\).
5545748
Use the bounds \(3.141<\pi<3.142\) and \(1.414<\sqrt{2}<1.415\) to bound \(2\pi-\sqrt{2}\). Then round \(2\pi-\sqrt{2}\) to the nearest hundredth. Do not use a calculator decimal for \(\pi\) or \(\sqrt{2}\).

Hints

- For the smallest possible difference, pair the smaller possible value of the first term with the larger possible value of the quantity being subtracted. - Reverse those choices to obtain the largest possible difference. - Check whether the entire resulting interval rounds to the same hundredth.

Solution

1. For a lower bound, use the lower bound for \(2\pi\) and the upper bound for \(\sqrt{2}\): \(2\cdot3.141-1.415=4.867\). 2. For an upper bound, use the upper bound for \(2\pi\) and the lower bound for \(\sqrt{2}\): \(2\cdot3.142-1.414=4.870\). 3. Therefore, \(4.867<2\pi-\sqrt{2}<4.870\). 4. Every number in this interval rounds to \(4.87\) to the nearest hundredth.

Answer

\(4.867<2\pi-\sqrt{2}<4.870\), so \(2\pi-\sqrt{2}\approx4.87\) to the nearest hundredth.
5545758
Riley says \(\sqrt[3]{70}\approx8.4\) because \(8.4^2\) is close to \(70\). Explain why Riley's method is invalid. Then show that \(\sqrt[3]{70}\) lies between \(4.1\) and \(4.2\). Finally, cube the midpoint \(4.15\) and use that result to decide which tenth is closer.

Hints

- Match the inverse operation to the root index before checking any decimal estimate. - Test the two proposed tenths using the appropriate power. - Use their midpoint to decide the nearest tenth without needing the exact cube root.

Solution

1. Riley used squares, but a cube root must be checked with cubes. 2. \(4.1^3=68.921\) and \(4.2^3=74.088\), so \(68.921<70<74.088\). Therefore, \(4.1<\sqrt[3]{70}<4.2\). 3. The midpoint is \(4.15\), and \(4.15^3=71.473375\). 4. Since \(70<71.473375\), \(\sqrt[3]{70}<4.15\), so it is closer to \(4.1\). 5. Thus, \(\sqrt[3]{70}\approx4.1\) to the nearest tenth.

Answer

Riley used squares instead of cubes. \(4.1^3=68.921<70<74.088=4.2^3\), so \(4.1<\sqrt[3]{70}<4.2\). Since \(4.15^3=71.473375>70\), \(\sqrt[3]{70}<4.15\) and \(\sqrt[3]{70}\approx4.1\) to the nearest tenth.
5143738
Without a calculator, find two consecutive natural numbers \(n\) and \(n+1\) such that \(T = \sqrt{40 + \sqrt{125}}\) lies between them. Show your reasoning.

Hints

- Begin with the inner square root and bound it between nearby whole numbers. - Use those bounds to estimate the entire radicand of the outer square root. - Which perfect squares surround that resulting interval?

Solution

1. Since \(11^2 = 121\) and \(12^2 = 144\), \(11 < \sqrt{125} < 12\). 2. Add \(40\) to each part: \(51 < 40 + \sqrt{125} < 52\). 3. Because \(7^2 = 49\) and \(8^2 = 64\), \(49 < 40 + \sqrt{125} < 64\). 4. Taking square roots gives \(7 < \sqrt{40 + \sqrt{125}} < 8\). Therefore, \(T\) lies between \(7\) and \(8\).

Answer

The value of \(T\) lies between \(7\) and \(8\).
5144168
Use the Babylonian method to approximate \(\sqrt{2}\), starting with \(x_1=1\). Calculate the next two approximations, \(x_2\) and \(x_3\), as reduced fractions. Then find the difference \(x_3^2-2\).

Hints

- Substitute the current approximation into the iteration formula. - Use common denominators when adding fractions. - Square the third approximation before subtracting \(2\).

Solution

1. The second approximation is \(x_2=\frac{1}{2}\left(x_1+\frac{2}{x_1}\right)=\frac{1}{2}(1+2)=\frac{3}{2}\). 2. The third approximation is \(x_3=\frac{1}{2}\left(\frac{3}{2}+\frac{2}{3/2}\right)=\frac{1}{2}\left(\frac{3}{2}+\frac{4}{3}\right)=\frac{17}{12}\). 3. Squaring gives \(x_3^2=\left(\frac{17}{12}\right)^2=\frac{289}{144}\). 4. Therefore, \(x_3^2-2=\frac{289}{144}-\frac{288}{144}=\frac{1}{144}\).

Answer

\(x_2=\frac{3}{2}\), \(x_3=\frac{17}{12}\), and \(x_3^2-2=\frac{1}{144}\).
5144188
Use one Babylonian step to improve the estimate of \(\sqrt{13}\), starting with \(x_1=3.5\). Use \(x_2=\frac{1}{2}\left(x_1+\frac{13}{x_1}\right)\). a) Calculate \(x_2\) as a fraction and round it to four decimal places. b) Compare \(x_2^2\) with \(13\). Is \(x_2\) above or below \(\sqrt{13}\)? c) Use \(3.60^2\) and \(3.61^2\) to show that \(x_2\) is within \(0.01\) of \(\sqrt{13}\).

Hints

- Substitute \(x_1=3.5\) into the given iteration formula and keep an exact fraction. - Compare the square of the approximation with the radicand. - Locate both values inside the same interval of width \(0.01\).

Solution

1. \(x_2=\frac{1}{2}\left(3.5+\frac{13}{3.5}\right)=\frac{1}{2}\left(\frac{7}{2}+\frac{26}{7}\right)=\frac{101}{28}\approx3.6071\). 2. \(x_2^2=\frac{10201}{784}\), while \(13=\frac{10192}{784}\). Therefore, \(x_2^2>13\), so \(x_2>\sqrt{13}\). 3. Since \(3.60^2=12.96<13\) and \(3.61^2=13.0321>13\), \(3.60<\sqrt{13}<3.61\). Also, \(3.60<x_2<3.61\). Two numbers in this interval differ by less than \(0.01\), so \(x_2\) is within \(0.01\) of \(\sqrt{13}\).

Answer

a) \(x_2=\frac{101}{28}\approx3.6071\) b) \(x_2>\sqrt{13}\) c) \(x_2\) is within \(0.01\) of \(\sqrt{13}\).
5144768
A student proposes \(\frac{161}{72}\) as a rational approximation to \(\sqrt{5}\). a) Compare \(\left(\frac{161}{72}\right)^2\) with \(5\). Is the approximation above or below \(\sqrt{5}\)? b) Use \(2.236^2=4.999696\) and \(2.237^2=5.004169\) to locate \(\sqrt{5}\) in an interval of width \(0.001\). c) Show that \(\frac{161}{72}\) is in the same interval, and give an upper bound for its absolute error.

Hints

- Compare positive values by comparing their squares. - Use the given endpoint squares to locate \(\sqrt{5}\). - If two numbers lie in one interval, their distance is less than the interval width.

Solution

1. \(\left(\frac{161}{72}\right)^2=\frac{25921}{5184}\), while \(5=\frac{25920}{5184}\). The squared approximation is greater than \(5\), so \(\frac{161}{72}>\sqrt{5}\). 2. Since \(4.999696<5<5.004169\), \(2.236<\sqrt{5}<2.237\). 3. \(\frac{161}{72}=2.236111\ldots\), so it also lies between \(2.236\) and \(2.237\). Two numbers in the same interval of width \(0.001\) differ by less than \(0.001\). Therefore, \(\left|\frac{161}{72}-\sqrt{5}\right|<0.001\).

Answer

a) \(\frac{161}{72}>\sqrt{5}\) b) \(2.236<\sqrt{5}<2.237\) c) \(\left|\frac{161}{72}-\sqrt{5}\right|<0.001\)
5144778
Two students propose rational approximations for \(\sqrt{10}\). Lucas proposes \(\frac{19}{6}\), and Sofia proposes \(\frac{22}{7}\). a) Without a calculator, use squaring to decide which approximation is closer to \(\sqrt{10}\). b) Use a calculator to find the absolute error of each approximation. Who is correct?

Hints

- First square each fraction to determine which side of \(\sqrt{10}\) it lies on. - For two values on opposite sides, compare \(\sqrt{10}\) with their midpoint. - Square the midpoint rather than directly evaluating \(\sqrt{10}\).

Solution

1. The approximations lie on opposite sides of \(\sqrt{10}\): \(\left(\frac{22}{7}\right)^2 = \frac{484}{49} < 10\), while \(\left(\frac{19}{6}\right)^2 = \frac{361}{36} > 10\). 2. Their midpoint is \(m = \frac{1}{2}\left(\frac{22}{7} + \frac{19}{6}\right) = \frac{265}{84}\). Since \(m^2 = \frac{70225}{7056} < 10\), the midpoint is below \(\sqrt{10}\). Therefore, \(\sqrt{10}\) is closer to the larger approximation, \(\frac{19}{6}\). 3. Using \(\sqrt{10} \approx 3.162278\), the errors are \(\left|\frac{19}{6} - \sqrt{10}\right| \approx 0.004389\) and \(\left|\frac{22}{7} - \sqrt{10}\right| \approx 0.019421\). Lucas is correct.

Answer

a) \(\frac{19}{6}\) is closer to \(\sqrt{10}\). b) Lucas’s absolute error is approximately \(0.004389\), and Sofia’s is approximately \(0.019421\). Lucas is correct.
5245188
Two irrational numbers are \(a = \sqrt{2}\) and \(b = 2-\sqrt{2}\). a) Determine whether the sum \(s = a+b\) is rational or irrational. b) Find the first three digits after the decimal point of \(d = a-b\) by bounding \(\sqrt{2}\) to four decimal places.

Hints

- Simplify the sum before classifying it. - Simplify the difference before substituting bounds. - Track how multiplying and subtracting change both interval endpoints. - Which decimal digits are fixed throughout the final interval?

Solution

1. The sum is \(s = \sqrt{2} + (2-\sqrt{2}) = 2\), which is rational. 2. The difference is \(d = \sqrt{2}-(2-\sqrt{2}) = 2\sqrt{2}-2\). 3. Since \(1.4142 < \sqrt{2} < 1.4143\), multiplying by \(2\) gives \(2.8284 < 2\sqrt{2} < 2.8286\). 4. Subtracting \(2\) gives \(0.8284 < d < 0.8286\). Throughout this interval, the first three digits after the decimal point are \(828\).

Answer

a) \(s = 2\), so the sum is rational. b) The first three digits after the decimal point are \(828\).
5144088
Use the Babylonian method to approximate \(\sqrt{10}\), starting with \(x_0=3\). a) Calculate \(x_1\) and \(x_2\) as fractions. b) Write \(x_2\) as a decimal rounded to six digits after the decimal point. c) Find the absolute difference between \(x_2^2\) and \(10\).

Hints

- Use \(x_{n+1}=\frac{1}{2}\left(x_n+\frac{A}{x_n}\right)\) with \(A=10\). - Keep the first two iterations as fractions to avoid rounding error. - Dividing by a fraction means multiplying by its reciprocal. - Compare the square of the second approximation with \(10\).

Solution

1. Using \(x_{n+1}=\frac{1}{2}\left(x_n+\frac{10}{x_n}\right)\), the first approximation is \(x_1=\frac{1}{2}\left(3+\frac{10}{3}\right)=\frac{19}{6}\). 2. The second approximation is \(x_2=\frac{1}{2}\left(\frac{19}{6}+\frac{10}{19/6}\right)=\frac{1}{2}\left(\frac{19}{6}+\frac{60}{19}\right)=\frac{721}{228}\). 3. As a decimal, \(x_2=\frac{721}{228}\approx3.162281\). 4. The squared value is \(x_2^2=\frac{519841}{51984}\). Therefore, the absolute difference is \(\left|x_2^2-10\right|=\frac{1}{51984}\approx0.00001924\).

Answer

a) \(x_1=\frac{19}{6}\) and \(x_2=\frac{721}{228}\) b) \(x_2\approx3.162281\) c) \(\left|x_2^2-10\right|=\frac{1}{51984}\approx0.00001924\)

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