Squares, Square Roots and Cube Roots

Squares, Cubes and Roots to Know by Heart

Root questions in SSC are built on a small set of numbers. If the squares up to 30 and the cubes up to 20 are in your head, a question like $\sqrt[3]{0.004913}$ or $\sqrt{218 + \sqrt{37 + \sqrt{144}}}$ takes seconds.

Roots worth remembering.

Two cube roots also help with estimates: $\sqrt[3]{2} \approx 1.26$ and $\sqrt[3]{3} \approx 1.442$.

What the root sign means. $\sqrt{49}$ means the positive root, 7. The equation $x^2 = 49$ has two answers, 7 and −7, but the root sign on its own always gives the positive one. So $\sqrt{(-6)^2} = \sqrt{36} = 6$, not −6. A cube root keeps the sign of the number: $\sqrt[3]{-64} = -4$. The square root of a negative number is not a real number at all.

A check on squares. $n^2$ and $(50 - n)^2$ end in the same two digits, and so do $n^2$ and $(100 - n)^2$. For example, $23^2 = 529$ and $27^2 = 729$ both end in 29, and $18^2 = 324$ and $82^2 = 6724$ both end in 24. It is a quick way to check a square you have just worked out.

Squaring Fast

Numbers ending in 5. Multiply the front part by the next number and write 25 after it: $65^2$ = 4225 (6 × 7 = 42) and $105^2$ = 11025 (10 × 11 = 110). It works with a decimal 5 too: $7.5^2$ = 56.25.

Near 50. Since $(50 + d)^2 = 2500 + 100d + d^2$, write 25 + d as the hundreds and $d^2$ as the last two digits.

• $53^2$: 25 + 3 = 28 and $3^2$ = 09, so 2809.
• $46^2$: 25 − 4 = 21 and $4^2$ = 16, so 2116.
• $62^2$: 25 + 12 = 37 and $12^2$ = 144. Only two digits fit at the end, so the 1 carries: 3700 + 144 = 3844.

Near 100. $(100 + d)^2 = 10000 + 200d + d^2$, so write (the number + d) and then $d^2$ as two digits. Below 100, d is negative, so you subtract.

• $104^2$: 104 + 4 = 108 and $4^2$ = 16, so 10816.
• $97^2$: 97 − 3 = 94 and $3^2$ = 09, so 9409.
• $112^2$: 112 + 12 = 124 and $12^2$ = 144, so 12400 + 144 = 12544.
• $89^2$: 89 − 11 = 78 and $11^2$ = 121, so 7800 + 121 = 7921.

Near 1000. The same idea with three digits at the end: $996^2$ is 992 followed by 016, which is 9,92,016, and $1007^2$ is 1014 followed by 049, which is 10,14,049.

Any other number. Split it into a round number and a small one and use $(a \pm b)^2 = a^2 \pm 2ab + b^2$.

• $67^2 = (70 - 3)^2$ = 4900 − 420 + 9 = 4489.
• $312^2 = (300 + 12)^2$ = 90000 + 7200 + 144 = 97,344.

Decimals. Square the digits without the point, then give the answer twice as many decimal places: $1.3^2$ = 1.69, $0.07^2$ = 0.0049 and $2.05^2$ = 4.2025.

Pairs either side of 50 or 100. These have their own shortcut: $(50 - d)^2 + (50 + d)^2 = 5000 + 2d^2$. So $47^2 + 53^2$ = 5000 + 18 = 5018 without squaring either number. Around 100 it is $20000 + 2d^2$.

Cubes Fast, and Sums of Squares and Cubes

Two-digit cubes. Split the number and use $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

$$23^3 = 8000 + 3 \times 400 \times 3 + 3 \times 20 \times 9 + 27 = 8000 + 3600 + 540 + 27 = 12167$$

For a number just below a round one, use $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$: $29^3$ = 27000 − 2700 + 90 − 1 = 24,389.

The row method. For a two-digit number with digits $a$ and $b$, write $a^3$, $a^2b$, $ab^2$ and $b^3$ in a row (each term is the one before it times $\frac{b}{a}$). Write double the middle two underneath, add the columns, and carry from the right, keeping one digit per column. For $23^3$:

From the right: 27 gives 7, carry 2. Then 54 + 2 = 56 gives 6, carry 5. Then 36 + 5 = 41 gives 1, carry 4. Then 8 + 4 = 12. So $23^3$ = 12167.

Cubes near 100. $(100 + d)^3 = 1000000 + 30000d + 300d^2 + d^3$. Write 1, then $3d$, then $3d^2$, then $d^3$, each as a two-digit group.

• $102^3$: 1 | 06 | 12 | 08, which is 10,61,208.
• $103^3$: 1 | 09 | 27 | 27, which is 10,92,727.
• Below 100 the signs alternate: $98^3$ = 1000000 − 60000 + 1200 − 8 = 9,41,192.

Decimals. Cube the digits, then give the answer three times as many decimal places: $0.3^3$ = 0.027, $1.2^3$ = 1.728 and $0.05^3$ = 0.000125.

Sums of squares and cubes. The formulas $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ and $1^3 + 2^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$ give the full sums. In simplification papers SSC usually gives one sum and asks for a related one.

• Every term doubled. $2^2 + 4^2 + \cdots + 20^2 = 2^2\left(1^2 + 2^2 + \cdots + 10^2\right)$ = 4 × 385 = 1540. For cubes the multiplier is $2^3 = 8$, so $2^3 + 4^3 + \cdots + 20^3$ = 8 × 3025 = 24,200.
• A middle stretch. $11^2 + 12^2 + \cdots + 20^2$ = 2870 − 385 = 2485, where 2870 is the sum of the squares up to $20^2$.
• Odd squares only. $1^2 + 3^2 + \cdots + 19^2$ = (all squares up to $20^2$) − (the even ones) = 2870 − 1540 = 1330.

In general, if the terms go up in steps of $k$ ($k$, $2k$, $3k$, …), take $k^2$ out of a sum of squares and $k^3$ out of a sum of cubes. Steps of 3 mean multiplying by 9 for squares and by 27 for cubes.

Square Roots by Prime Factors and by Long Division

Prime factors. Factorise, pair up equal primes, and take one prime from each pair.

$$7056 = 2^4 \times 3^2 \times 7^2 \quad\Rightarrow\quad \sqrt{7056} = 2^2 \times 3 \times 7 = 84$$

You don't need to multiply a product out first. For $\sqrt{18 \times 32 \times 49}$: 18 = $2 \times 3^2$, 32 = $2^5$ and 49 = $7^2$, so the product is $2^6 \times 3^2 \times 7^2$, and the root is $2^3 \times 3 \times 7$ = 168.

If a prime is left without a partner, the number is not a perfect square and the root stays as a surd: $\sqrt{72} = \sqrt{2^3 \times 3^2} = 6\sqrt{2}$. (Working with surds like this is in the Surds and Rationalisation section.)

Long division. This is the method for big numbers with awkward factors, and for numbers that are not perfect squares.

1. Mark off pairs of digits starting from the decimal point: leftwards for the whole-number part, rightwards for the decimal part. 119716 becomes 11 | 97 | 16.
2. Find the largest digit whose square fits into the first group. $3^2$ = 9 fits into 11, so the root starts with 3. Subtract to get 2.
3. Bring down the next pair to get 297. Double the root so far (3 × 2 = 6) and find the digit $x$ that makes "6$x$" × $x$ as large as possible without going over 297. 64 × 4 = 256 works, so the next digit is 4. Subtract to get 41.
4. Bring down 16 to get 4116. Double the root so far (34 × 2 = 68) and find $x$ again: 686 × 6 = 4116 exactly. The next digit is 6 and nothing is left.

So $\sqrt{119716}$ = 346. If the last remainder is not zero, it is exactly the least number to subtract to get a perfect square.

A number that is not a perfect square. Add pairs of zeros after the decimal point and keep going. For $\sqrt{5}$: $2^2$ = 4 leaves 1. Bringing down 00 gives 100, and 42 × 2 = 84 leaves 16. Bringing down 00 gives 1600, and 443 × 3 = 1329 leaves 271. So $\sqrt{5}$ = 2.23…, and one more step gives 2.236.

How many digits does the root have? One for each pair marked off in step 1. A perfect square with $n$ digits has a root with $\frac{n}{2}$ digits when $n$ is even and $\frac{n+1}{2}$ digits when $n$ is odd. So a seven-digit square has a four-digit root.

[SSC Tip] In an MCQ you rarely need the whole division. Bracket the root first ($300^2$ = 90,000 and $350^2$ = 1,22,500, so $\sqrt{119716}$ is between 300 and 350), then use the last digit (a square ending in 6 has a root ending in 4 or 6) to pick from the options.

Roots of Decimals and Fractions

Decimals. A decimal can be an exact square only if it has an even number of decimal places, and its root has half as many. For an exact cube the number of places must be a multiple of 3, and the root has a third as many.

• $\sqrt{0.0081}$ = 0.09 (four places, so two in the root)
• $\sqrt{0.000169}$ = 0.013 (six places, so three)
• $\sqrt[3]{0.000027}$ = 0.03 (six places, so two)

The classic trap is $\sqrt{0.9}$. It is not 0.3, because $0.3^2$ = 0.09. With one decimal place, write 0.9 as 0.90: $\sqrt{0.90} = \frac{\sqrt{90}}{10} \approx 0.949$. In the same way $\sqrt{0.4} \approx 0.632$, while $\sqrt{0.04}$ = 0.2.

Moving the decimal point. If you know one root, you know a whole family. Moving the point two places in the number moves it one place in the square root. Since $\sqrt{5776}$ = 76:

• $\sqrt{57.76}$ = 7.6, $\sqrt{0.5776}$ = 0.76, $\sqrt{0.005776}$ = 0.076 and $\sqrt{577600}$ = 760.
• But $\sqrt{577.6}$ is not 76 with the point moved. The point moved an odd number of places, so the digits change: $\sqrt{577.6} \approx 24.03$.

For cube roots, three places in the number make one place in the root. Since $\sqrt[3]{4913}$ = 17, $\sqrt[3]{4.913}$ = 1.7 and $\sqrt[3]{0.004913}$ = 0.17.

Fractions. Take the root of the top and the bottom separately: $\sqrt{\frac{169}{625}} = \frac{13}{25}$. A mixed number must become an improper fraction first: $\sqrt{5\frac{4}{9}} = \sqrt{\frac{49}{9}} = \frac{7}{3} = 2\frac{1}{3}$. Taking the root of $\frac{4}{9}$ alone and keeping the 5 (to get $5\frac{2}{3}$) is the usual mistake.

When the bottom is not a perfect square, multiply the top and bottom to make it one: $\sqrt{\frac{5}{3}} = \sqrt{\frac{15}{9}} = \frac{\sqrt{15}}{3}$. So if $\sqrt{15}$ = 3.873, then $\sqrt{\frac{5}{3}}$ = 1.291.

Recurring decimals. Change them to fractions first (see Chapter 1): $\sqrt{0.\overline{4}} = \sqrt{\frac{4}{9}} = \frac{2}{3} = 0.\overline{6}$ and $\sqrt{1.\overline{7}} = \sqrt{\frac{16}{9}} = \frac{4}{3} = 1.\overline{3}$.

Between 0 and 1, roots are bigger. For a number between 0 and 1, the square root is bigger than the number and the square is smaller: $0.25^2$ = 0.0625, which is less than 0.25, which is less than $\sqrt{0.25}$ = 0.5. "Arrange in order" questions lean on this.

Cube Roots: Prime Factors, Unit Digit and Decimals

Prime factors. Group equal primes in threes and take one prime from each group.

• $9261 = 3^3 \times 7^3$, so $\sqrt[3]{9261} = 3 \times 7$ = 21.
• $13824 = 2^9 \times 3^3$, so $\sqrt[3]{13824} = 2^3 \times 3$ = 24.

Cube roots of products split up and join up freely: $\sqrt[3]{54} \times \sqrt[3]{32} = \sqrt[3]{1728}$ = 12, and $\sqrt[3]{16} \times \sqrt[3]{4} = \sqrt[3]{64}$ = 4.

The unit-digit method for perfect cubes. This settles any perfect cube up to six digits in about ten seconds.

Unlike squares, each ending gives exactly one possible last digit. Only 2 and 8 swap, and 3 and 7 swap.

1. The last digit of the cube fixes the last digit of the root.
2. Drop the last three digits. Find the largest cube that fits into what is left; its root is the first part of the answer.

• $\sqrt[3]{185193}$: it ends in 3, so the root ends in 7. Dropping 193 leaves 185, and $5^3$ = 125 fits but $6^3$ = 216 does not. The root is 57.
• $\sqrt[3]{778688}$: it ends in 8, so the root ends in 2. Dropping 688 leaves 778, and $9^3$ = 729 fits but $10^3$ = 1000 does not. The root is 92.

This only works when you know the number is a perfect cube, which SSC questions either state or make clear from the options.

Decimals and fractions. The number of decimal places must be a multiple of 3, and the root has a third as many: $\sqrt[3]{0.000343}$ = 0.07 and $\sqrt[3]{0.001728}$ = 0.12. For a fraction take the cube root of the top and the bottom: $\sqrt[3]{\frac{343}{1000}} = \frac{7}{10}$. A mixed number or an expression must become a single fraction first:

$$\sqrt[3]{3\frac{3}{8}} = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} \qquad \sqrt[3]{1 - \frac{189}{216}} = \sqrt[3]{\frac{27}{216}} = \frac{3}{6} = \frac{1}{2}$$

Negative numbers. A cube root keeps the sign: $\sqrt[3]{-2197} = -13$ and $\sqrt[3]{-0.008} = -0.2$.

Mixed expressions. Take each root on its own and then combine. $\sqrt[3]{0.000064} \div \sqrt{0.0016}$ = 0.04 ÷ 0.04 = 1. The cube root of 0.000064 is 0.04 (six places, so two), and the square root of 0.0016 is also 0.04 (four places, so two).

Approximate Roots

When the number under the root is not a perfect square, the options usually say "approximately" or "closest to". Two steps handle almost all of them.

Step 1: bracket it. Find the perfect squares on either side. For $\sqrt{150}$: $12^2$ = 144 and $13^2$ = 169, so the root is between 12 and 13, and much closer to 12.

Step 2: correct it. If $N = a^2 + d$ with $d$ small compared with $a^2$, then

$$\sqrt{N} \approx a + \frac{d}{2a}$$

So $\sqrt{150} \approx 12 + \frac{6}{24}$ = 12.25 (the true value is 12.247). For a number just below a square, $d$ is negative: $\sqrt{80} = \sqrt{81 - 1} \approx 9 - \frac{1}{18} \approx 8.944$.

Why it works: $(a + e)^2 = a^2 + 2ae + e^2$. When $e$ is small, $e^2$ is tiny, so $2ae \approx d$ and $e \approx \frac{d}{2a}$.

The same idea works for cube roots: $\sqrt[3]{a^3 + d} \approx a + \frac{d}{3a^2}$. So $\sqrt[3]{130} \approx 5 + \frac{5}{75} \approx 5.067$ (the true value is 5.066).

Using the roots you know. Pull out a perfect square, then use the table of roots at the start of this section.

• $\sqrt{300} = 10\sqrt{3} \approx 17.32$
• $\sqrt{2000} = 20\sqrt{5} \approx 44.72$
• $\sqrt{0.5} = \frac{\sqrt{50}}{10} = \frac{5\sqrt{2}}{10} \approx 0.707$
• $\sqrt{0.08} = \frac{\sqrt{8}}{10} \approx 0.283$

Expressions. Round each root, but keep track of which way you rounded. $\sqrt{626} + \sqrt{399} \approx 25 + 20$ = 45. The first root is a touch above 25 and the second a touch below 20, so the errors nearly cancel (the true value is 44.995).

Options in "closest to" questions are usually far apart, so bracketing alone often decides it. Use the $\frac{d}{2a}$ correction only when two options are close together.

Expressions and Equations with Roots

Nested roots: work from the inside out. In a question like $\sqrt{218 + \sqrt{37 + \sqrt{144}}}$, the innermost root comes out first, and in a set question each step lands on a perfect square.

$$\sqrt{144} = 12, \qquad \sqrt{37 + 12} = \sqrt{49} = 7, \qquad \sqrt{218 + 7} = \sqrt{225} = 15$$

Decimal versions work the same way: $\sqrt{0.16 + \sqrt{0.0081}} = \sqrt{0.16 + 0.09} = \sqrt{0.25}$ = 0.5. If a step does not land on a perfect square, recheck the step before it.

Join roots before you work them out. For positive numbers, $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ and $\sqrt{a} \div \sqrt{b} = \sqrt{\frac{a}{b}}$.

• $\sqrt{12} \times \sqrt{27} = \sqrt{324}$ = 18
• $\sqrt{0.2} \times \sqrt{0.8} = \sqrt{0.16}$ = 0.4
• $\sqrt{98} \div \sqrt{2} = \sqrt{49}$ = 7

Adding is a different story: $\sqrt{a + b}$ is not $\sqrt{a} + \sqrt{b}$. $\sqrt{16 + 9} = \sqrt{25}$ = 5, while $\sqrt{16} + \sqrt{9}$ = 7.

"What comes in place of ?" Get the root on its own, then square (or cube) both sides.

• $\sqrt{?} + \sqrt{529} = 41$: $\sqrt{529}$ = 23, so $\sqrt{?}$ = 18 and ? = 324.
• $\sqrt[3]{?} = 0.4$: cubing both sides, ? = 0.064.
• $\sqrt{?} \div \sqrt{441} = 0.02$: $\sqrt{441}$ = 21, so $\sqrt{?}$ = 0.42 and ? = 0.1764.

Equations with a root. Same idea: get the root alone, then square.

$$\sqrt{1 + \frac{x}{121}} = \frac{12}{11} \;\Rightarrow\; 1 + \frac{x}{121} = \frac{144}{121} \;\Rightarrow\; x = 144 - 121 = 23$$

The side without the root must not be negative, because a root sign never gives a negative value. $\sqrt{x + 5} = -3$ has no solution, even though squaring would suggest $x = 4$.

[SSC Tip] In $\sqrt{1 + \frac{x}{k^2}} = \frac{m}{k}$, the answer is always $x = m^2 - k^2$. So $\sqrt{1 + \frac{x}{169}} = \frac{14}{13}$ gives $x$ = 196 − 169 = 27 straight away.

Nested and Infinite Roots

The dots in $\sqrt{20 + \sqrt{20 + \sqrt{20 + \cdots}}}$ mean the pattern never stops. That is exactly what makes these easy: the part under the first root is the whole expression again.

The plus form. Call the value $y$. Then $y = \sqrt{20 + y}$, so $y^2 = 20 + y$, which gives $y^2 - y - 20 = 0$, or $(y - 5)(y + 4) = 0$. The value is positive, so $y$ = 5.

The shortcut: write the number as a product of two consecutive numbers, $x = n(n + 1)$. Then

$$\sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} = n + 1 \qquad \sqrt{x - \sqrt{x - \sqrt{x - \cdots}}} = n$$

The plus form gives the bigger factor and the minus form gives the smaller one. With 20 = 4 × 5, the plus form is 5 and the minus form is 4. With 12 = 3 × 4, they are 4 and 3. With 2 = 1 × 2, $\sqrt{2 + \sqrt{2 + \cdots}}$ = 2.

[Tier 2] If the number is not of the form $n(n + 1)$, solve $y^2 - y - x = 0$ and keep the positive root. For $x$ = 1 that gives $\frac{1 + \sqrt{5}}{2}$.

The product form. If $y = \sqrt{x\sqrt{x\sqrt{x \cdots}}}$, then $y = \sqrt{xy}$, so $y^2 = xy$ and $y = x$. So $\sqrt{6\sqrt{6\sqrt{6 \cdots}}}$ = 6.

The product form with a fixed number of roots. Use indices (the laws are in the Indices and Powers section). Reading from the outside, the first $x$ is under one root, the next under two, and so on, so the powers are $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, …

$$\sqrt{x\sqrt{x}} = x^{\frac{3}{4}} \qquad \sqrt{x\sqrt{x\sqrt{x}}} = x^{\frac{7}{8}} \qquad \sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}} = x^{\frac{15}{16}}$$

With $n$ roots the power is $\frac{2^n - 1}{2^n}$. So $\sqrt{2\sqrt{2\sqrt{2}}} = 2^{\frac{7}{8}}$.

The division form. If $y = \sqrt{x \div \sqrt{x \div \sqrt{x \div \cdots}}}$, then $y^2 = \frac{x}{y}$, so $y^3 = x$ and $y = \sqrt[3]{x}$. So $\sqrt{27 \div \sqrt{27 \div \cdots}}$ = 3.

Cube-root versions [Tier 2]. For $y = \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x \cdots}}}$, $y^3 = xy$, so $y = \sqrt{x}$. And $\sqrt[3]{x + \sqrt[3]{x + \cdots}} = n$ when $x = n^3 - n = (n - 1)n(n + 1)$. So $\sqrt[3]{24 + \sqrt[3]{24 + \cdots}}$ = 3, because 24 = 2 × 3 × 4.

Solved Examples

Question 1

Find the value of $96^2 + 54^2$.

Answer. Both numbers sit close to a base. 96 is 4 below 100, so its square is 96 − 4 = 92 followed by $4^2$ = 16, which is 9216. 54 is 4 above 50, so its square is 25 + 4 = 29 followed by 16, which is 2916. Adding them, 9216 + 2916 = 12,132.

Question 2

Find $27^3$ without long multiplication.

Answer. 27 is 3 below 30, so I use $(30 - 3)^3 = 30^3 - 3 \times 30^2 \times 3 + 3 \times 30 \times 3^2 - 3^3$. That is 27000 − 8100 + 810 − 27 = 19,683. A quick check: 27 = $3^3$, so $27^3 = 3^9$, and $3^9$ is also 19683.

Question 3

If $1^2 + 2^2 + 3^2 + \cdots + 15^2 = 1240$, find $2^2 + 4^2 + 6^2 + \cdots + 30^2$.

Answer. Every number in the second sum is double a number in the first, so every square is 4 times as big: $2^2 + 4^2 + \cdots + 30^2 = 2^2\left(1^2 + 2^2 + \cdots + 15^2\right)$. The answer is 4 × 1240 = 4960. Multiplying by 2 instead of 4 is the trap here.

Question 4

Find $\sqrt{5184}$ using prime factors.

Answer. I keep halving: 5184 → 2592 → 1296 → 648 → 324 → 162 → 81. That is six 2s, and 81 = $3^4$. So $5184 = 2^6 \times 3^4$. Halving each power gives $2^3 \times 3^2$ = 8 × 9 = 72.

Question 5

Find $\sqrt{54756}$ by long division.

Answer. Pairs from the right: 5 | 47 | 56.

• $2^2$ = 4 fits into 5, so the root starts with 2, and 1 is left.
• Bringing down 47 gives 147. Double 2 is 4, and 43 × 3 = 129 fits (44 × 4 = 176 is too big). The root is now 23, and 18 is left.
• Bringing down 56 gives 1856. Double 23 is 46, and 464 × 4 = 1856 exactly.

So $\sqrt{54756}$ = 234.

Question 6

Find $\sqrt{12 \times 75 \times 49}$.

Answer. I don't multiply it out. 12 = $2^2 \times 3$, 75 = $3 \times 5^2$ and 49 = $7^2$. Together that is $2^2 \times 3^2 \times 5^2 \times 7^2$, and every power is even. The root is 2 × 3 × 5 × 7 = 210.

Question 7

Find $\sqrt{0.0169} + \sqrt{1.69} + \sqrt{169}$.

Answer. $\sqrt{169}$ = 13. 1.69 has two decimal places, so its root has one: 1.3. 0.0169 has four places, so its root has two: 0.13. Adding, 0.13 + 1.3 + 13 = 14.43.

Question 8

Given that $\sqrt{5329} = 73$, find $\sqrt{53.29} + \sqrt{0.5329} + \sqrt{0.005329} + \sqrt{0.00005329}$.

Answer. Each number has the digits of 5329 with the point moved an even number of places, so each root is 73 with the point moved half as far: 7.3, 0.73, 0.073 and 0.0073. Their sum is 8.1103.

Question 9

Simplify $\sqrt{6\frac{1}{4}} + \sqrt{1\frac{9}{16}}$.

Answer. First I change both into improper fractions: $6\frac{1}{4} = \frac{25}{4}$ and $1\frac{9}{16} = \frac{25}{16}$. Their roots are $\frac{5}{2}$ and $\frac{5}{4}$. Adding, $\frac{10}{4} + \frac{5}{4} = \frac{15}{4}$, which is $3\frac{3}{4}$.

Question 10

Find $\sqrt{0.\overline{4}}$. (a) 0.2 (b) $0.\overline{2}$ (c) 0.6 (d) $0.\overline{6}$

Answer. A recurring decimal has to become a fraction first: $0.\overline{4} = \frac{4}{9}$. Its root is $\frac{2}{3}$, which is 0.666… So the answer is (d) $0.\overline{6}$. Taking the root of the digit 4 and keeping the bar gives option (b), which is wrong.

Question 11

If $\sqrt{15}$ = 3.873, find $\sqrt{\frac{5}{3}}$ and $\sqrt{\frac{3}{5}}$.

Answer. I make each bottom a perfect square. $\sqrt{\frac{5}{3}} = \sqrt{\frac{15}{9}} = \frac{\sqrt{15}}{3}$ = 3.873 ÷ 3 = 1.291. And $\sqrt{\frac{3}{5}} = \sqrt{\frac{15}{25}} = \frac{\sqrt{15}}{5}$ = 3.873 ÷ 5 = 0.7746.

Question 12

Find the cube root of 438976.

Answer. It ends in 6, so the root ends in 6. Dropping the last three digits leaves 438. $7^3$ = 343 fits into 438 but $8^3$ = 512 does not, so the first digit is 7. The cube root is 76.

Question 13

Find $\sqrt[3]{1 - \frac{61}{125}}$.

Answer. First the subtraction: $1 - \frac{61}{125} = \frac{125 - 61}{125} = \frac{64}{125}$. Both parts are cubes, so the cube root is $\frac{4}{5}$, which is 0.8.

Question 14

Find $\sqrt[3]{0.003375} \times \sqrt[3]{-0.008}$.

Answer. 3375 = $15^3$, and 0.003375 has six decimal places, so its cube root is 0.15. The cube root of −0.008 is −0.2, because $0.2^3$ = 0.008 and a cube root keeps the sign. The product is 0.15 × (−0.2) = −0.03.

Question 15

$\sqrt{85}$ is closest to: (a) 9.12 (b) 9.22 (c) 9.32 (d) 9.42

Answer. $9^2$ = 81, so 85 = 81 + 4. Using $a + \frac{d}{2a}$, the root is about 9 + $\frac{4}{18}$ = 9 + 0.22 = 9.22. The answer is (b) 9.22. (The true value is 9.2195.)

Question 16

$\sqrt{2000}$ is closest to: (a) 4.47 (b) 44.4 (c) 44.7 (d) 45.2

Answer. 2000 = 400 × 5, so $\sqrt{2000} = 20\sqrt{5}$, which is about 20 × 2.236 = 44.72. The answer is (c) 44.7. Option (a) is $\sqrt{20}$, a slip with the decimal point.

Question 17

Find $\sqrt{353 + \sqrt{50 + \sqrt{196}}}$.

Answer. From the inside: $\sqrt{196}$ = 14. Then 50 + 14 = 64, and $\sqrt{64}$ = 8. Then 353 + 8 = 361, and $\sqrt{361}$ = 19.

Question 18

Find $\sqrt{0.04 + \sqrt{0.0025}}$.

Answer. $\sqrt{0.0025}$ = 0.05 (four places, so two). Then 0.04 + 0.05 = 0.09, and $\sqrt{0.09}$ = 0.3.

Question 19

If $\sqrt{1 + \frac{x}{196}} = \frac{15}{14}$, find $x$.

Answer. Squaring both sides, $1 + \frac{x}{196} = \frac{225}{196}$. So $\frac{x}{196} = \frac{29}{196}$ and $x$ = 29. (The shortcut gives it at once: 225 − 196 = 29.)

Question 20

What comes in place of ? in $\sqrt{0.0169 \times ?} = 5.2$?

Answer. $\sqrt{0.0169}$ = 0.13, so $\sqrt{?}$ = 5.2 ÷ 0.13 = 40, and ? = $40^2$ = 1600. Squaring first works too: 0.0169 × ? = 27.04, and 27.04 ÷ 0.0169 = 1600.

Question 21

Find $\sqrt{42 + \sqrt{42 + \sqrt{42 + \cdots}}}$ and $\sqrt{42 - \sqrt{42 - \sqrt{42 - \cdots}}}$.

Answer. 42 = 6 × 7. The plus form gives the bigger factor and the minus form gives the smaller one, so the values are 7 and 6. Checking the first: if the value is 7, then $\sqrt{42 + 7} = \sqrt{49}$ = 7, which fits.

Question 22

Write $\sqrt{7\sqrt{7\sqrt{7}}}$ as a power of 7. What would the value be if the roots went on forever?

Answer. The outer 7 is under one root (power $\frac{1}{2}$), the next is under two ($\frac{1}{4}$) and the last is under three ($\frac{1}{8}$). Adding, $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$, so the value is $7^{\frac{7}{8}}$. This matches $\frac{2^3 - 1}{2^3}$ for three roots. If the roots never stopped, the value would be 7 itself.