QuanticaPrealgebraOpen in the courseIn 8.1 every radicand was a perfect square. Now suppose a tile setter needs a square panel covering exactly \(50\) square inches. The side is \(\sqrt{50}\), an irrational number whose decimal never ends or repeats, so we estimate it instead.
Problem
A square patch covers \(54\) sq m, so the side is \(\sqrt{54}\). Since \(7^2 = 49\) and \(8^2 = 64\), the root lies between \(7\) and \(8\). What is the largest whole number less than \(\sqrt{54}\)?
Show a hint
- Look for the two perfect squares that sandwich \(54\). Which perfect square sits just below \(54\), and which sits just above?
- You found that \(49 < 54 < 64\). The roots keep that same order, so \(7 < \sqrt{54} < 8\). The floor is the smaller of those two whole numbers.
Show the full solution
Since \(7^2 = 49\) and \(8^2 = 64\), we have \(49 < 54 < 64\), so \(7 < \sqrt{54} < 8\). The largest whole number below the side length is \(\boxed{7}\). A bigger number always has a bigger square root, so trapping \(54\) between two perfect squares traps the root between the two whole numbers.
Problem
A square solar farm covers \(6800\) sq m. Given \(82^2 = 6724\) and \(83^2 = 6889\), what is the largest whole number of meters less than \(\sqrt{6800}\)?
Show a hint
- You want two perfect squares with \(6800\) caught between them, and you are already handed the candidates. Since \(82^2 = 6724\) and \(83^2 = 6889\), which two consecutive whole numbers do those squares belong to?
- From \(6724 \le 6800 \le 6889\) you can read off \(82 < \sqrt{6800} < 83\). The largest whole number that stays below \(\sqrt{6800}\) is the smaller of those two bounds.
Show the full solution
From \(82^2 = 6724\) and \(83^2 = 6889\) we get \(6724 < 6800 < 6889\). Taking roots keeps that order, so \(82 < \sqrt{6800} < 83\) and the largest whole number below the side is \(\boxed{82}\). The numbers are bigger but the method is the same, since the two perfect squares around the radicand are all you ever need.
Problem
A skylight has area \(83\) sq in. The edge \(\sqrt{83}\) is between \(9\) and \(10\). The halfway square is \(9.5^2 = 90.25\). To the nearest inch, what is \(\sqrt{83}\)?
Show a hint
- The two candidates are \(9\) and \(10\), and the halfway point between them is \(9.5\). Whichever side of \(9.5\) the root falls on is the side you round toward.
- Square the halfway point instead of the root, since squaring keeps the order for nonnegative numbers. The boundary is \(9.5^2 = 90.25\). Because \(83 < 90.25\), the root is less than \(9.5\), so it rounds down.
Show the full solution
The halfway point between \(9\) and \(10\) is \(9.5\), and \(9.5^2 = 90.25\). Since \(83 < 90.25\), the root is below \(9.5\), so it rounds down. To the nearest inch, \(\sqrt{83} \approx \boxed{9}\). Squaring the midpoint is the whole trick, since it turns a question about an endless decimal into one comparison of plain numbers.
Problem
A skylight has area \(21\) sq in. The edge \(\sqrt{21}\) is between \(4\) and \(5\). The halfway square is \(4.5^2 = 20.25\). To the nearest inch, what is \(\sqrt{21}\)?
Show a hint
- The halfway point between 4 and 5 is 4.5. A root rounds up exactly when it is bigger than 4.5, which happens exactly when the number under the root is bigger than \(4.5^2\).
- Compare \(21\) to \(4.5^2 = 20.25\). Since \(21\) is larger, \(\sqrt{21}\) is larger than \(4.5\), so it has already passed the halfway mark.
Show the full solution
The midpoint between \(4\) and \(5\) is \(4.5\), and \(4.5^2 = 20.25\). Since \(21 > 20.25\), the root is past \(4.5\), so it rounds up. To the nearest inch, \(\sqrt{21} \approx \boxed{5}\). Rounding down to \(4\) is the usual slip here, because \(21\) looks close to \(16\). The midpoint square settles it instead of eyeballing.
Problem
A pendulum is \(\sqrt{23}\) ft long. The root is between \(4.7\) and \(4.8\). Does it pass the halfway square \(4.75^2 = 22.5625\)? To the nearest tenth, what is \(\sqrt{23}\)?
Show a hint
- The halfway mark between \(4.7\) and \(4.8\) is \(4.75\). Whichever side of \(4.75^2\) the number \(23\) falls on tells you which tenth the root is nearer to.
- Compare exactly with whole numbers. Multiply both sides by \(10000\), so \(4.75^2\) becomes \(475^2 = 225{,}625\), while \(23\) becomes \(230{,}000\). Since \(230{,}000\) is bigger, \(23\) sits above \(4.75^2\), so the root is on the higher side.
Show the full solution
The midpoint between \(4.7\) and \(4.8\) is \(4.75\), and \(4.75^2 = 22.5625\). Since \(23 > 22.5625\), the root is past the midpoint and rounds up. To the nearest tenth, \(\sqrt{23} = \boxed{4.8}\). If the long decimal feels shaky, scale by \(10{,}000\) and compare \(475^2 = 225{,}625\) with \(230{,}000\), which is exact whole-number work.
Problem
A drone hovers at \(\sqrt{41}\) m. The root is between \(6.4\) and \(6.5\). Does it pass the halfway square \(6.45^2 = 41.6025\)? To the nearest tenth, what is \(\sqrt{41}\)?
Show a hint
- You already know the root is between \(6.4\) and \(6.5\). The only question is which tenth it rounds to, so the tiebreaker is the halfway point \(6.45\).
- Compare \(41\) with \(6.45^2\). To keep it exact, square \(645\) and compare it to \(41 \times 10{,}000\). If \(41\) comes out smaller, the root is below \(6.45\) and rounds down.
Show the full solution
The midpoint between \(6.4\) and \(6.5\) is \(6.45\). Comparing exactly, \(645^2 = 416{,}025\) while \(41 \times 10{,}000 = 410{,}000\), so \(41 < 6.45^2\) and \(\sqrt{41} < 6.45\). The root sits below the midpoint, so it rounds down and \(\sqrt{41} \approx \boxed{6.4}\). Scaling both sides by \(10{,}000\) keeps every comparison in whole numbers, so no rounding error can creep in.
Problem
How many whole numbers \(n\) satisfy \(7 < \sqrt{n} < 11\)? Square the bounds, so \(n\) must satisfy \(49 < n < 121\). Count the whole numbers strictly between \(49\) and \(121\).
Show a hint
- Squaring is fair on both sides because for nonnegative numbers a bigger value always has a bigger square root, so the order never flips. From \(\sqrt{n} > 7\) you get \(n > 49\), and from \(\sqrt{n} < 11\) you get \(n < 121\).
- Now \(n\) is a whole number with \(49 < n < 121\). Both ends are strict, so \(49\) and \(121\) themselves are out, since their roots are exactly \(7\) and \(11\), not between. The list runs from \(50\) up through \(120\). To count a solid run of whole numbers, take last minus first plus one.
Show the full solution
Squaring both bounds turns \(7 < \sqrt{n} < 11\) into \(49 < n < 121\), so \(n\) runs from \(50\) through \(120\). That is \(120 - 50 + 1 = \boxed{71}\) whole numbers. Both ends drop out because \(\sqrt{49} = 7\) and \(\sqrt{121} = 11\) exactly, and the inequalities are strict.
Problem
How many whole numbers \(k\) satisfy \(\sqrt{130} < k < \sqrt{320}\)? The smallest valid \(k\) is \(12\) and the largest is \(17\). Count them.
Show a hint
- A whole number \(k\) fits when its square lands strictly above \(130\) and strictly below \(320\). Find the smallest such \(k\) and the largest such \(k\) by checking squares.
- The smallest fit is \(12\) because \(12^2 = 144 \gt 130\) while \(11^2 = 121\) is too small. The largest fit is \(17\) because \(17^2 = 289 \lt 320\) while \(18^2 = 324\) is too big. Now count from \(12\) through \(17\), and remember to include both ends.
Show the full solution
A whole number \(k\) fits exactly when \(130 < k^2 < 320\). Since \(11^2 = 121\) is too small and \(12^2 = 144\) works, the smallest is \(12\). Since \(17^2 = 289\) works and \(18^2 = 324\) is too big, the largest is \(17\). Counting \(12\) through \(17\) gives \(17 - 12 + 1 = \boxed{6}\). Squaring \(k\) instead of estimating the two roots keeps the whole problem in exact arithmetic.
Problem
Cable A is \(6\sqrt{7}\) m, Cable B is \(4\sqrt{15}\) m. Square each to compare. Option 1 is Cable A, Option 2 is Cable B. Which is longer?
Show a hint
- Comparing the radicals head on is hard, but squaring clears the roots away. Square each length using \((a\sqrt{b})^2 = a^2 \times b\), which leaves you two whole numbers, then compare those.
- Cable A squared is \(6^2 \times 7 = 36 \times 7 = 252\). Cable B squared is \(4^2 \times 15 = 16 \times 15 = 240\). Since both cables are positive, the one with the larger square is the longer cable.
Show the full solution
Square each length. Cable A gives \((6\sqrt{7})^2 = 6^2 \times 7 = 252\), and Cable B gives \((4\sqrt{15})^2 = 4^2 \times 15 = 240\). Since \(252 > 240\) and both lengths are positive, Cable A is longer, option \(\boxed{1}\). The number out front squares too, which is the step people skip when they try to compare \(6\sqrt{7}\) and \(4\sqrt{15}\) by eye.
Problem
Evaluate \(\left(\sqrt{6}\right)^4\). Group the four factors into two pairs. Each pair \(\sqrt{6} \times \sqrt{6} = 6\). What is the result?
Show a hint
- A square root undoes a square. Multiply two copies of the same root together, \(\sqrt{6} \times \sqrt{6}\), and you are left with just the inside, \(6\). No decimals needed.
- The fourth power is two of those pairs side by side, so \(\left(\sqrt{6}\right)^4 = \left(\sqrt{6}\times\sqrt{6}\right)\times\left(\sqrt{6}\times\sqrt{6}\right) = 6 \times 6\). Now just multiply those two sixes.
Show the full solution
Pair the factors. Each pair gives \(\left(\sqrt{6}\right)^2 = 6\), so \(\left(\sqrt{6}\right)^4 = 6^2 = \boxed{36}\). Any even power of a root collapses this way, since the copies pair off and each pair leaves just the number underneath.
Problem
A hiker covers two trails, \(\sqrt{18}\) miles then \(\sqrt{33}\) miles. What is the largest whole number less than \(\sqrt{18} + \sqrt{33}\)?
Show a hint
- Bracket each root by itself first. Since \(4^2 = 16\) and \(5^2 = 25\), \(\sqrt{18}\) sits between 4 and 5. Since \(5^2 = 25\) and \(6^2 = 36\), \(\sqrt{33}\) sits between 5 and 6. That only tells you the sum is between 9 and 11, which is too wide to name a single floor.
- Push the lower edge up with tenths. Check \(4.2^2 = 17.64\), which is below 18, so \(\sqrt{18} > 4.2\). Check \(5.7^2 = 32.49\), which is below 33, so \(\sqrt{33} > 5.7\). Adding the safe lower edges gives more than \(4.2 + 5.7\). Now find a matching upper edge to trap the sum between two neighbors.
Show the full solution
Sharpen each root to tenths. Since \(4.2^2 = 17.64 < 18\), we know \(\sqrt{18} > 4.2\), and since \(5.7^2 = 32.49 < 33\), we know \(\sqrt{33} > 5.7\), so the sum is more than \(4.2 + 5.7 = 9.9\). Going the other way, \(4.25^2 = 18.0625 > 18\) and \(5.75^2 = 33.0625 > 33\) make the sum less than \(4.25 + 5.75 = 10\). The total is caught in \(9.9 < \sqrt{18} + \sqrt{33} < 10\), so the largest whole number below it is \(\boxed{9}\). Whole-number brackets alone only narrow the sum to between \(9\) and \(11\), which is why the tenths are needed.
Problem
A drone flies \(\sqrt{40}\) mi north then \(\sqrt{75}\) mi east. The total is between \(14\) and \(16\). Does \(\sqrt{40} + \sqrt{75}\) exceed \(15\)? What is the largest whole number below the total?
Show a hint
- The whole question is whether the sum slips below 15 or not. Rewrite the test \(\sqrt{40} + \sqrt{75} < 15\) by moving one root across, so you are asking whether \(\sqrt{40} < 15 - \sqrt{75}\). Both sides are nonnegative, since \(\sqrt{75} < 9 < 15\), so you are allowed to square both sides without flipping the comparison.
- Squaring \(\sqrt{40} < 15 - \sqrt{75}\) gives \(40 < 225 - 30\sqrt{75} + 75\), which tidies up to \(30\sqrt{75} < 260\). There is still one root left, so square one more time to land on plain whole numbers. Compare \(30^2 \cdot 75\) against \(260^2\).
Show the full solution
The sum already beats \(14\), since \(\sqrt{40} > 6\) and \(\sqrt{75} > 8\). So the real test is \(15\). Ask whether \(\sqrt{40} < 15 - \sqrt{75}\), which is safe to square because \(\sqrt{75} < 9\) keeps both sides positive. Squaring gives \(40 < 300 - 30\sqrt{75}\), which rearranges to \(30\sqrt{75} < 260\). Square again and compare \(30^2 \cdot 75 = 67{,}500\) with \(260^2 = 67{,}600\). Since \(67{,}500 < 67{,}600\), the sum really is under \(15\), so it sits between \(14\) and \(15\) and the answer is \(\boxed{14}\). Every squaring step reverses cleanly here because both sides stay nonnegative throughout.
Practice these ideas
Practice
A square sticker covers \(34\) sq cm. Since \(5^2 = 25\) and \(6^2 = 36\), the side \(\sqrt{34}\) is between \(5\) and \(6\). What is the largest whole number less than \(\sqrt{34}\)?
Show the solution
Since \(5^2 = 25\) and \(6^2 = 36\), we have \(25 < 34 < 36\), so \(5 < \sqrt{34} < 6\). The largest whole number less than \(\sqrt{34}\) is \(\boxed{5}\). Taking roots keeps the order, so bracketing \(34\) between perfect squares brackets the root between whole numbers.
Practice
Rope 1 is \(\sqrt{63}\) m and Rope 2 is \(\sqrt{57}\) m. Both lie between \(7\) and \(8\). Use the order rule to decide. Option 1 is \(\sqrt{63}\), Option 2 is \(\sqrt{57}\). Which is longer?
Show the solution
Squaring each length gives \(63\) and \(57\). Since \(63 > 57\), we get \(\sqrt{63} > \sqrt{57}\), so Rope 1 is longer, option \(\boxed{1}\). Both roots have floor \(7\), so the floors tie and only the numbers under the radicals tell them apart.
Practice
A square window has area \(75\) sq in. The edge \(\sqrt{75}\) is between \(8\) and \(9\). The halfway square is \(8.5^2 = 72.25\). To the nearest inch, what is \(\sqrt{75}\)?
Show the solution
The midpoint between \(8\) and \(9\) is \(8.5\), and \(8.5^2 = 72.25\). Since \(75 > 72.25\), the root is past the midpoint and rounds up, so \(\sqrt{75} \approx \boxed{9}\) inches. To skip the decimal entirely, double both sides and compare \(4 \times 75 = 300\) with \(17^2 = 289\), which gives the same verdict.
Practice
A ramp rises \(\sqrt{44}\) feet. Since \(6^2 = 36\) and \(7^2 = 49\), the height is between \(6\) and \(7\). The halfway square is \(6.5^2 = 42.25\). To the nearest whole foot, what is \(\sqrt{44}\)?
Show the solution
The midpoint between \(6\) and \(7\) is \(6.5\), and \(6.5^2 = 42.25\). Since \(44 > 42.25\), the root is past the midpoint, so \(\sqrt{44}\) rounds up and the height is \(\boxed{7}\) feet. Doubling first keeps it all in whole numbers, comparing \(4 \times 44 = 176\) with \(13^2 = 169\).
Practice
A wire is \(\sqrt{31}\) inches long. The root is between \(5.5\) and \(5.6\). The halfway square is \(5.55^2 = 30.8025\). To the nearest tenth, what is \(\sqrt{31}\)?
Show the solution
The midpoint between \(5.5\) and \(5.6\) is \(5.55\), and \(5.55^2 = 30.8025\). Since \(31 > 30.8025\), the root is above \(5.55\) and rounds up, so \(\sqrt{31} = \boxed{5.6}\) to the nearest tenth. Rounding to tenths works exactly like rounding to whole numbers, only the midpoint you square is a halfway tenth.
Practice
A flagpole is \(\sqrt{88}\) units tall. The root is between \(9.3\) and \(9.4\). The halfway square is \(9.35^2 = 87.4225\). To the nearest tenth, what is \(\sqrt{88}\)?
Show the solution
The midpoint between \(9.3\) and \(9.4\) is \(9.35\), and \(9.35^2 = 87.4225\). Since \(88 > 87.4225\), the root is past the midpoint and rounds up, so \(\sqrt{88} \approx \boxed{9.4}\). Comparing \(935^2 = 874{,}225\) with \(880{,}000\) reaches the same answer using only whole numbers.
Practice
A square field covers \(250\) sq yd. Since \(15^2 = 225\) and \(16^2 = 256\), the side \(\sqrt{250}\) is between \(15\) and \(16\). What is the largest whole number less than \(\sqrt{250}\)?
Show the solution
Since \(15^2 = 225\) and \(16^2 = 256\), we have \(225 < 250 < 256\), so \(15 < \sqrt{250} < 16\). The largest whole number below \(\sqrt{250}\) is \(\boxed{15}\).
Practice
A square reservoir has footprint \(9000\) sq m. Find the largest whole number less than \(\sqrt{9000}\).
Show the solution
Testing squares in the nineties, \(94^2 = 8836\) and \(95^2 = 9025\). Since \(8836 < 9000 < 9025\), the root sits between \(94\) and \(95\), so the floor is \(\boxed{94}\). When no perfect square is handed to you, size it up first. Here \(90^2 = 8100\) and \(100^2 = 10{,}000\) show the answer is in the nineties.
Practice
Find the largest whole number less than \(\sqrt{3{,}000{,}000}\). Since the root is in the seventeen-hundreds, test \(1{,}732^2\) and \(1{,}733^2\).
Show the solution
Testing \(1{,}732^2 = 2{,}999{,}824\), which is under \(3{,}000{,}000\), and \(1{,}733^2 = 3{,}003{,}289\), which is over, traps the root between the two. So the largest whole number below \(\sqrt{3{,}000{,}000}\) is \(\boxed{1732}\). Only the two squares straddling the target matter, no matter how large the numbers get.
Practice
How many whole numbers \(n\) satisfy \(\sqrt{50} < \sqrt{n} < \sqrt{130}\)? Square the bounds and count whole numbers strictly between \(50\) and \(130\).
Show the solution
Squaring both bounds turns the question into \(50 < n < 130\), so \(n\) runs from \(51\) through \(129\). That is \(129 - 51 + 1 = \boxed{79}\) whole numbers. Both ends are strict, so \(50\) and \(130\) themselves are left out of the count.
Practice
How many whole numbers \(k\) satisfy \(\sqrt{40} < k < \sqrt{210}\)? Find the smallest whole number above \(\sqrt{40}\) and the largest below \(\sqrt{210}\), then count.
Show the solution
Since \(6^2 = 36 < 40 < 49 = 7^2\), the smallest whole number above \(\sqrt{40}\) is \(7\). Since \(14^2 = 196 < 210 < 225 = 15^2\), the largest below \(\sqrt{210}\) is \(14\). Counting \(7\) through \(14\) gives \(14 - 7 + 1 = \boxed{8}\). Both \(7\) and \(14\) do count here, which is why the run is \(8\) long and not \(7\).
Practice
Beam 1 is \(7\sqrt{5}\) m and Beam 2 is \(4\sqrt{15}\) m. Square each to compare. Option 1 is Beam 1, Option 2 is Beam 2. Which is longer?
Show the solution
Square each. Beam 1 gives \(\left(7\sqrt{5}\right)^2 = 49 \times 5 = 245\), and Beam 2 gives \(\left(4\sqrt{15}\right)^2 = 16 \times 15 = 240\). Since \(245 > 240\) and both lengths are positive, Beam 1 is longer, option \(\boxed{1}\).
Practice
Evaluate \(\sqrt{3}^6\). Group the six factors into three pairs. Each pair \(\sqrt{3} \cdot \sqrt{3} = 3\). What whole number is the result?
Show the solution
Pair the six factors. Each \(\sqrt{3}^2\) is \(3\), so \(\sqrt{3}^6 = \left(\sqrt{3}^2\right)^3 = 3^3 = \boxed{27}\). Any even power of a root comes out a whole number this way, since the copies pair off and each pair leaves the number underneath.
Practice
Two trails are \(\sqrt{28}\) miles and \(\sqrt{52}\) miles long. What is the largest whole number less than \(\sqrt{28} + \sqrt{52}\)?
Show the solution
Bracket each root with tenths. Since \(5.2^2 = 27.04 < 28 < 28.09 = 5.3^2\), we get \(5.2 < \sqrt{28} < 5.3\), and since \(7.2^2 = 51.84 < 52 < 53.29 = 7.3^2\), we get \(7.2 < \sqrt{52} < 7.3\). Adding the low ends and the high ends separately, the sum lies between \(5.2 + 7.2 = 12.4\) and \(5.3 + 7.3 = 12.6\). The largest whole number below the total is \(\boxed{12}\). Whole-number brackets would only give \(12\) to \(14\), so the tenths are what pin the floor down.