Square Roots: Free Response

Take the first one by hunting for perfect-square factors. The radicand is odd, so $4$ and $100$ are no use, but its digits add to $9$, so it is divisible by $9$, and $441 \div 9 = 49$. Both factors are perfect squares, so the product rule finishes the job:

$\sqrt{441} = \sqrt{9 \times 49} = \sqrt{9}\times\sqrt{49} = 3 \times 7 = 21.$

The check is to square the answer and see whether the radicand comes back, and it does: $21^2 = 441$.

The other two need no arithmetic at all, because the square and the radical are inverse operations and cancel whenever the number underneath is not negative:

$\sqrt{23^2} = 23 \qquad\text{and}\qquad \left(\sqrt{38}\right)^2 = 38.$

Working out $23^2 = 529$ and then hunting for the root of $529$ would reach the same place after far more work, and the second one cannot be done that way at all, since $\sqrt{38}$ cannot be evaluated as a whole number or as a decimal that stops. The radical form records it exactly, but there is nothing to compute before squaring it. That the middle step can be skipped is what calling the two operations inverse actually buys you.

First find the pair. Since $484 = 4 \times 121$ and both factors are perfect squares,

$\sqrt{484} = \sqrt{4}\times\sqrt{121} = 2 \times 11 = 22.$

The classmate is right that $22$ is not alone in squaring to $484$, because a negative times a negative is positive:

$22^2 = 484 \qquad\text{and}\qquad (-22)^2 = (-22)\times(-22) = 484.$

So the question "what squares to $484$?" genuinely has two answers. What cannot have two answers is the symbol. If $\sqrt{484}$ named both numbers, then $\sqrt{484} + 1$ would be $23$ and $-21$ at the same time, and no line of working containing a radical could be checked, or even read. A piece of notation that names two numbers is not a number, and only a number can be carried through a calculation.

The agreement that fixes this is to hand back the non-negative member of the pair, the principal square root, so $\sqrt{484} = 22$ and nothing else. The negative solution has not been lost, only moved out of the symbol and onto the page in front of it:

$-\sqrt{484} = -22.$

Both facts then live side by side without contradicting each other: two numbers square to $484$, and the radical names one of them.

Test the rule where nobody checked it. The lesson states this cancelling for a number that is zero or positive, and Riya has quietly dropped that condition, so try a negative value. Take $x = -17$:

$(-17)^2 = (-17)\times(-17) = 289, \qquad \sqrt{289} = 17.$

Her left side is $17$ and her right side is $-17$, so the rule is false as written. One counterexample is enough to bring down a claim about every number, and this one is not a freak case: the same thing happens for every negative $x$, because squaring destroys the sign and the radical is required to come back on the non-negative side.

There are two honest repairs.

The narrow one puts the missing condition back: $\sqrt{x^2} = x$ whenever $x$ is zero or positive. That is the form the lesson states, and it is true.

The wider one keeps every number in play by asking what $\sqrt{x^2}$ actually returns. It returns the size of $x$ with the sign discarded, and that is exactly what absolute value means:

$\sqrt{x^2} = |x|.$

Check it on both signs. For the counterexample, $\sqrt{(-17)^2} = 17 = |-17|$, and going the other way $\sqrt{17^2} = 17 = |17|$. Riya's version is the special case of this one in which $x$ was already non-negative, which is why it looked right on every example she had tried.

Simplifying a radical means finding the largest perfect square that divides the radicand.

For $63$, run through the perfect squares. It is odd, so $4$ is out; $9$ divides it, since $63 = 9 \times 7$; and nothing larger fits. Split it there:

$\sqrt{63} = \sqrt{9 \times 7} = \sqrt{9}\times\sqrt{7} = 3\sqrt{7}.$

The $7$ stays under the radical, because $7$ is prime and so has no perfect-square factor except $1$.

For $180$, several perfect squares divide it: $4$, $9$ and $36$. The largest is $36$, since $180 = 36 \times 5$:

$\sqrt{180} = \sqrt{36 \times 5} = \sqrt{36}\times\sqrt{5} = 6\sqrt{5}.$

Pulling out $4$ instead is not wrong, only unfinished:

$\sqrt{180} = \sqrt{4 \times 45} = 2\sqrt{45}.$

The radicand $45$ still holds the perfect square $9$, so the same rule has to be applied a second time: $2\sqrt{45} = 2 \times 3\sqrt{5} = 6\sqrt{5}$. Taking the largest perfect square first arrives in one step instead of two, and it is the only way to be sure at a glance that nothing is left inside.

The product first, where the rule applies. Both factors are perfect squares, so the radical comes apart cleanly:

$\sqrt{9 \times 144} = \sqrt{9}\times\sqrt{144} = 3 \times 12 = 36.$

A check confirms it: $9 \times 144 = 1296$, and $36^2 = 1296$.

Now the sum, where it does not apply. The radicand is $9 + 144 = 153$, which is not a perfect square, so trap it between the perfect squares on either side. Counting up, $144 = 12^2$ sits just below it and $169 = 13^2$ is the next one above:

$144 < 153 < 169, \qquad\text{so}\qquad 12 < \sqrt{153} < 13.$

Splitting the radical across the sum instead produces a number that is not even inside that range:

$\sqrt{9} + \sqrt{144} = 3 + 12 = 15.$

The true root is below $13$ and the split answer is $15$, so the split overshoots by more than $2$. That is not a rounding slip. The two expressions are simply different quantities, and part C says what the difference is made of.

Everything here follows from one description of the symbol: $\sqrt{n}$ is the non-negative number whose square is $n$. So to test whether some expression is $\sqrt{n}$, square it, and check as well that it is not negative.

Take the product. Squaring $\sqrt{a}\times\sqrt{b}$ means multiplying it by itself, and multiplication may be reordered freely:

$\left(\sqrt{a}\times\sqrt{b}\right)^2 = \sqrt{a}\times\sqrt{a}\times\sqrt{b}\times\sqrt{b} = a \times b.$

It is also non-negative, being a product of two non-negative numbers. It squares to $a \times b$ and it is not negative, which is the entire description of $\sqrt{a \times b}$, so the two are the same number.

Now run the identical test on the sum. Squaring $\sqrt{a} + \sqrt{b}$ means multiplying that sum by itself, and each term of the first bracket multiplies each term of the second:

$\left(\sqrt{a} + \sqrt{b}\right)^2 = \sqrt{a}\sqrt{a} + \sqrt{a}\sqrt{b} + \sqrt{b}\sqrt{a} + \sqrt{b}\sqrt{b}.$

The first and last of those are $a$ and $b$, and the middle two are equal to each other, so

$\left(\sqrt{a} + \sqrt{b}\right)^2 = a + b + 2\sqrt{a}\sqrt{b}.$

There is the spoiler: two cross terms that the product version never produced. Squaring $\sqrt{a} + \sqrt{b}$ does not give $a + b$, it gives $a + b$ plus something more, and unless one of $a$ and $b$ is zero that extra piece is genuinely positive, so $\sqrt{a} + \sqrt{b}$ is the root of something larger than $a + b$ rather than of $a + b$ itself.

That extra piece also says exactly when the sum version would be safe. Both roots are non-negative, so $2\sqrt{a}\sqrt{b}$ is zero exactly when one of them is zero, which happens exactly when $a$ or $b$ is zero. That is the only case, and it is a useless one: with $b = 0$ the claim reads $\sqrt{a + 0} = \sqrt{a} + 0$. For every other pair of non-negative numbers the cross terms are genuinely present, and part B already measured them. With $a = 9$ and $b = 144$ they add $2 \times 3 \times 12 = 72$, and $153 + 72 = 225$, which is why the split answer of $15$ is the root of $225$ rather than of $153$.

The area of a square is its side multiplied by itself, so the side is the square root of the area. That makes the first side $\sqrt{1225}$ feet, and $1225$ is too large to recognize from the table of small squares, so factor it instead.

It ends in a $5$, so it is divisible by $5$, and the quotient ends in a $5$ as well: $1225 \div 5 = 245$ and $245 \div 5 = 49$. So $1225 = 5 \times 5 \times 49 = 25 \times 49$, a product of two perfect squares, which is exactly the situation the product rule handles:

$\sqrt{1225} = \sqrt{25 \times 49} = \sqrt{25}\times\sqrt{49} = 5 \times 7 = 35.$

The first plot has sides of $35$ feet. Check by squaring: $35^2 = 35 \times 35 = 1225$, which is the area given.

Only the non-negative root is of any use here, and the radical hands back only that one anyway. The number $-35$ also squares to $1225$, but a side of $-35$ feet is not something anyone can lay out on the ground.

The second side is $\sqrt{700}$ feet. Put it in exact form first, by pulling out the largest perfect-square factor. Since $700 = 100 \times 7$ and $100$ is a perfect square,

$\sqrt{700} = \sqrt{100 \times 7} = \sqrt{100}\times\sqrt{7} = 10\sqrt{7}.$

That is the exact side length, $10\sqrt{7}$ feet, with the $7$ staying underneath because it is prime.

Exact is not the same as useful to someone holding a tape measure, so trap it as well. Hunt for the perfect squares either side of $700$. Squares of sides in roughly the right range give $26^2 = 676$ just below and $27^2 = 729$ just above:

$676 < 700 < 729.$

Taking roots keeps the order, because a larger number has a larger root:

$26 < \sqrt{700} < 27.$

So each side of the second plot is between $26$ and $27$ feet. It leans toward $26$, because $700$ is only $24$ above $676$ while it is $29$ below $729$, so the area sits a little nearer the lower perfect square.

The trap settles the first half on its own. The side is longer than $26$ feet and shorter than $27$ feet:

$26 < \sqrt{700} < 27.$

A $26$ foot cut is therefore definitely too short and would leave a gap on every side of the plot. A $27$ foot cut is definitely long enough, and it is the shortest whole number that is, because $26$ and $27$ are consecutive: there is no whole number in between to try. The crew pays for that certainty with roughly half a foot of waste per side, which is the price of ordering in whole feet.

The second half is about the number itself rather than the order form. The area $700$ is not a perfect square: it sits strictly between $26^2$ and $27^2$, so no whole number squares to it. The square root of a whole number that is not a perfect square is irrational, which means its decimal expansion runs on forever without ever settling into a repeating block. Every decimal anyone can write down stops, and a decimal that stops can always be written as a fraction, so no such decimal can be this root.

You can watch the approximations improve without ever arriving. Squaring $26.4$ gives $696.96$, which is under $700$, and squaring $26.5$ gives $702.25$, which is over, so the side is between $26.4$ and $26.5$ feet. Another digit narrows the interval again, and so on with no end. That is why the honest way to record the length is the exact form $10\sqrt{7}$ feet, with a rounded value written beside it only where something actually has to be cut.

Turn the condition on the root into a condition on $n$ by squaring. That is safe here, because everything in sight is positive and a larger positive number has a larger square:

$8 < \sqrt{n} < 9 \quad\text{becomes}\quad 64 < n < 81.$

Both ends stay strict, and that is exactly right: $\sqrt{64} = 8$ is not strictly greater than $8$, and $\sqrt{81} = 9$ is not strictly less than $9$, so neither $64$ nor $81$ belongs in the list. The whole numbers that do belong start at $65$ and end at $80$.

Count them by subtracting the ends and adding one, since both of those are themselves in the list:

$80 - 65 + 1 = 16.$

So $16$ whole numbers have a square root strictly between $8$ and $9$. Not one of them is a perfect square, because the nearest perfect squares are the $64$ and $81$ that were just ruled out.

Count each run the same way, by subtracting the ends of the list of numbers actually inside and adding one.

Between $3^2 = 9$ and $4^2 = 16$, the numbers strictly inside run from $10$ to $15$:

$15 - 10 + 1 = 6.$

Between $12^2 = 144$ and $13^2 = 169$, they run from $145$ to $168$:

$168 - 145 + 1 = 24.$

A run of six near the start of the number line has become a run of twenty four by the time you reach the squares of $12$ and $13$, and it keeps going. Each new square is built from the one before by adding a strip along two sides and a corner, and that strip gets longer every time, so the step from one perfect square to the next is always larger than the step before it.

The consequence is about how thinly the perfect squares are spread. Up to $169$ there are only $13$ of them, so fewer than one in ten of the first $169$ whole numbers has a whole-number root, and the proportion keeps falling as the runs stretch. That is the sense in which a whole-number root is the exception rather than the rule: the perfect squares are scattered thinly and get thinner, and every whole number in the gaps has a square root that no whole number can name.

Start by granting what is true, because most of it is. The estimates are in the right region, since $64 = 8^2$ and $81 = 9^2$ give

$64 < 68 < 81, \qquad\text{so}\qquad 8 < \sqrt{68} < 9.$

The refinements are genuine too. Squaring each estimate shows it closing in from below:

$8.2^2 = 67.24, \qquad 8.24^2 = 67.8976, \qquad 8.246^2 = 67.996516.$

Every one of those is nearer to $68$ than the one before it, exactly as the student says.

The step that does not follow is the last one. "Each estimate is nearer than the last" does not deliver "some estimate is exact". Getting closer forever and arriving are different things, and this is a case where they come apart. The radicand $68$ is not a perfect square, since it sits strictly between $8^2$ and $9^2$ and no whole number squares to it, and the square root of a whole number that is not a perfect square is irrational: its decimal never ends and never falls into a repeating block. A decimal that stops can always be written as a fraction, so no decimal that stops can be this number, no matter how many digits are written.

What the sequence does achieve is worth naming, because it is not nothing. It produces better and better approximations, and it pins the exact value inside a narrower and narrower trap. The last estimate above already gives

$8.246 < \sqrt{68} < 8.247,$

since $8.247^2 = 68.013009$ is above $68$, and that is finer than any measurement a garden or a workshop is likely to need. The exact value is written $\sqrt{68}$, or $2\sqrt{17}$ once the perfect-square factor $4$ is pulled out, and that symbol is the only thing that records it exactly.

Take the three roots first.

For $324$, split the radicand into perfect-square factors. It is even, and $324 = 4 \times 81$, so

$\sqrt{324} = \sqrt{4}\times\sqrt{81} = 2 \times 9 = 18.$

For $1$, the root is $1$, since $1^2 = 1$.

For the fraction, take the root of the top and the root of the bottom, which is the quotient form of the same splitting rule:

$\sqrt{\frac{1}{25}} = \frac{\sqrt{1}}{\sqrt{25}} = \frac{1}{5}.$

Now set each root beside its radicand. The root $18$ is far below $324$, so here rooting made the number smaller. The root $1$ is neither above nor below $1$, so here it changed nothing. And $\tfrac{1}{5}$ is above $\tfrac{1}{25}$: over the common denominator $\tfrac{1}{5} = \tfrac{5}{25}$, which is five times as large as $\tfrac{1}{25}$, so here rooting made the number bigger.

All three outcomes occur, which already settles that "a root is smaller than the number it came from" is not a rule.

Give the root a name. Let $r$ stand for $\sqrt{n}$, where $n$ is zero or positive. Squaring undoes rooting, so the radicand is that root multiplied by itself:

$n = r \times r.$

The question asks how $n$ compares with $r$, and that line turns it into a comparison between $r \times r$ and $r$. So everything depends on what multiplying $r$ by $r$ does, and multiplication answers that: multiplying a positive number by something greater than $1$ makes it bigger, multiplying it by $1$ leaves it alone, and multiplying it by something between $0$ and $1$ makes it smaller.

If $r > 1$, then $r \times r > r$, so $n > r$ and the radicand is larger than its root. And $r > 1$ happens exactly when $n > 1$, since $n$ is $r$ multiplied by itself.

If $0 < r < 1$, then $r \times r < r$, so $n < r$ and the radicand is smaller than its root. That is the case that broke the rule in part A, and again $0 < r < 1$ happens exactly when $0 < n < 1$.

If $r$ is $0$ or $1$, then $r \times r = r$ and the two agree.

No further examples are needed, because the argument never looked at any particular number. It is worth seeing why the fraction case feels so strange. Multiplying by a proper fraction takes only a part of what you had, so a fifth of a fifth is a small piece of an already small quantity, which lands it below the number it came from. Nothing grew when the root was taken: the squaring shrank the number, and the root only undid that.

A counterexample has to come from below $1$ and above $0$, since that is the region where part B says the comparison reverses. Take $0.36$:

$\sqrt{0.36} = 0.6, \qquad\text{and}\qquad 0.6 > 0.36.$

Check the choice the honest way, by squaring the root back: $0.6 \times 0.6 = 0.36$, so $0.6$ really is the root, and it really is larger than the number it came from. The claim is false, and one number is enough to say so.

The corrected claim states the condition that was assumed all along but never written down. For a radicand greater than $1$, the square root is smaller than the radicand. For a radicand strictly between $0$ and $1$, the square root is larger. At $0$ and at $1$ the root equals the radicand, since $\sqrt{0} = 0$ and $\sqrt{1} = 1$, and those two are the turning points rather than stray exceptions: they are the only numbers a root can equal, because they are the only non-negative numbers that are unchanged when multiplied by themselves.

Now apply it with no arithmetic at all. The number $0.49$ lies between $0$ and $1$, so its root is above it. The number $90$ is greater than $1$, so its root is below it. Neither decision needed a root evaluated; the position of the radicand relative to $1$ settled both, which is what a rule with its condition attached is worth.