Operations with Radicals: Free Response

Split each radicand into a perfect square times whatever is left over, choosing the largest perfect square available.

Since $54 = 9 \cdot 6$ and $9$ is a perfect square:

$\sqrt{54} = \sqrt9\,\sqrt6 = 3\sqrt6.$

Since $96 = 16 \cdot 6$ and $16$ is a perfect square:

$\sqrt{96} = \sqrt{16}\,\sqrt6 = 4\sqrt6.$

Both leftover radicands are $6$, which has no square factor beyond $1$, so neither simplifies further.

All three terms are now multiples of the same radical, so the distributive law collects them by their coefficients alone.

$\sqrt{54}+\sqrt{96}-\sqrt{24} = 3\sqrt6 + 4\sqrt6 - 2\sqrt6 = (3+4-2)\sqrt6 = 5\sqrt6.$

As a check, $\sqrt{54}+\sqrt{96}-\sqrt{24} \approx 7.348 + 9.798 - 4.899 = 12.247$, and $5\sqrt6 \approx 12.247$.

The classmate's error is not in the rule quoted. Two radical terms in simplest form really do need a matching index and radicand to combine, and $54$, $96$, and $24$ really are three different numbers.

The error is in when that comparison was run. A radicand as first written does not yet show what number the radical actually is; only the simplified form does. Here every one of the three terms is secretly a multiple of $\sqrt6$:

$54 = 9\cdot6,\qquad 96=16\cdot6,\qquad 24=4\cdot6.$

Once that common factor is exposed, the like-radical test has something to compare, and it says the three terms do combine. Running the test on the raw radicands instead of the simplified ones is what produced the wrong conclusion.

Multiply the coefficients together and the radicands together, then simplify.

$(3\sqrt2)(5\sqrt6) = 15\sqrt{12} = 15\cdot2\sqrt3 = 30\sqrt3.$

The radicand $12=4\cdot3$ hid a perfect square, so the simplifying step is not optional even though the original product looked finished.

Distribute every term of the first factor against every term of the second.

$3\sqrt2\cdot2\sqrt2 = 12, \qquad 3\sqrt2\cdot(-\sqrt7) = -3\sqrt{14}, \qquad \sqrt7\cdot2\sqrt2 = 2\sqrt{14}, \qquad \sqrt7\cdot(-\sqrt7) = -7.$

Add the four pieces. The two rational terms combine to $12-7=5$, and the two radical terms combine to $-3\sqrt{14}+2\sqrt{14}=-\sqrt{14}$:

$(3\sqrt2+\sqrt7)(2\sqrt2-\sqrt7) = 5 - \sqrt{14}.$

In part B the two factors carried different coefficients on $\sqrt2$ (a $3$ and a $2$), so the two cross terms, $-3\sqrt{14}$ and $+2\sqrt{14}$, were unequal in size and only partly cancelled, leaving $-\sqrt{14}$ behind.

Here both factors carry the same coefficient, $3\sqrt2$. The two cross terms are now

$3\sqrt2\cdot(-\sqrt7) = -3\sqrt{14}, \qquad \sqrt7\cdot3\sqrt2 = 3\sqrt{14},$

which are exact opposites and cancel completely. What remains is $(3\sqrt2)^2-(\sqrt7)^2 = 18-7=11$, with no radical term at all. Matching the coefficient is what turns a partial cancellation into a total one.

The area of a rectangle is length times width, so distribute the two binomials.

$4\sqrt3\cdot\sqrt3 = 12, \qquad 4\sqrt3\cdot2\sqrt2 = 8\sqrt6, \qquad \sqrt2\cdot\sqrt3 = \sqrt6, \qquad \sqrt2\cdot2\sqrt2 = 4.$

Add the four pieces, combining the rational terms and the radical terms separately:

$(4\sqrt3+\sqrt2)(\sqrt3+2\sqrt2) = (12+4) + (8\sqrt6+\sqrt6) = 16 + 9\sqrt6 \text{ cm}^2.$

The width repeats the length's two radicals with only the connecting sign flipped, so this is the conjugate product $(a+b)(a-b)=a^2-b^2$ with $a=3\sqrt5$ and $b=\sqrt{11}$.

$(3\sqrt5+\sqrt{11})(3\sqrt5-\sqrt{11}) = (3\sqrt5)^2 - (\sqrt{11})^2 = 45 - 11 = 34 \text{ cm}^2.$

Both edge lengths are positive, since $3\sqrt5\approx6.71$ is larger than $\sqrt{11}\approx3.32$.

Compare how each width was built from its length.

In Plate 2, the width $(3\sqrt5-\sqrt{11})$ uses the exact same two radicals as the length $(3\sqrt5+\sqrt{11})$, with only the connecting sign reversed. That is precisely the conjugate pattern, and part B showed why it matters: the two cross terms in the expansion are negatives of each other and cancel, leaving only the rational terms:

$3\sqrt5\cdot(-\sqrt{11}) + \sqrt{11}\cdot3\sqrt5 = -3\sqrt{55}+3\sqrt{55} = 0.$

In Plate 1, the width $(\sqrt3+2\sqrt2)$ is not the length $(4\sqrt3+\sqrt2)$ with a sign flipped; the coefficients on $\sqrt3$ and $\sqrt2$ do not match between the two factors at all. The cross terms in that product are not opposites, so instead of cancelling they combine:

$8\sqrt6 + \sqrt6 = 9\sqrt6 \neq 0,$

and a radical remains in the area. The difference is entirely in whether the width is the length's conjugate, not in anything about the numbers being irrational.

Rewrite both factors as powers of the same base. Since $125=5^3$,

$\sqrt[4]{125} = 5^{3/4}, \qquad \sqrt5 = 5^{1/2}.$

Multiplying powers of the same base adds the exponents, over their common denominator $4$:

$5^{3/4}\cdot5^{1/2} = 5^{3/4+2/4} = 5^{5/4}.$

An exponent past $1$ means a whole factor of $5$ comes out, since $5/4 = 1+1/4$:

$5^{5/4} = 5^1\cdot5^{1/4} = 5\sqrt[4]5.$

Write both factors as powers of $2$ and add the exponents over the common denominator $15$:

$2^{1/3}\cdot2^{1/5} = 2^{5/15+3/15} = 2^{8/15}.$

Since $8/15$ is already less than $1$, no whole power comes out this time, and the exponent converts back to a radical of index $15$:

$2^{8/15} = \sqrt[15]{2^8} = \sqrt[15]{256}.$

Both parts A and B start by rewriting a radical as a power of its base, $5^{3/4}$ or $2^{1/3}$, and that step only makes sense if the radical already names a real number in the first place; the exponent notation is just a new name for something that already exists.

Whether a radical names a real number depends on the index's parity together with the radicand's sign, not on the sign of the base by itself. Take $\sqrt[4]{-16}$. If it named a real number $r$, then $r^4=-16$:

$r^4 = -16.$

But a fourth power of a real number is never negative, so no real $r$ satisfies that equation. $\sqrt[4]{-16}$ is not a real number, and there is nothing to relabel as $(-16)^{1/4}$.

The odd-index factor in the very same product behaves completely differently. An odd power keeps the sign of its base, so every real number, negative ones included, is some real number's cube:

$\left(-\sqrt[3]{16}\right)^3 = -16.$

So $\sqrt[3]{-16}=-\sqrt[3]{16}$ is a perfectly good real number, even though its radicand is the same $-16$ that broke the fourth root.

The product $\sqrt[4]{-16}\cdot\sqrt[3]{-16}$ therefore fails at its very first factor, before any exponent has been added or any base examined for its sign. The requirement is not that a base be nonnegative in general; it is that each radical being converted must already be real, which for an even index forces a nonnegative radicand and for an odd index requires nothing at all.

The quotient rule combines the two roots into the root of a single quotient.

$\frac{\sqrt{162}}{\sqrt6} = \sqrt{\frac{162}{6}} = \sqrt{27} = \sqrt9\,\sqrt3 = 3\sqrt3.$

Divide the coefficients and the radicands separately, then apply the quotient rule to what is left under the roots.

$\frac{18\sqrt{200}}{3\sqrt8} = \frac{18}{3}\sqrt{\frac{200}{8}} = 6\sqrt{25} = 6\cdot5 = 30.$

The quotient rule always applies the same way, regardless of what the resulting radicand turns out to be:

$\frac{\sqrt5}{\sqrt3} = \sqrt{\frac53}.$

The three parts differ only in what that radicand is. In part B the radicand became $25$, a perfect square outright, so the root evaluated to a whole number with nothing left over. In part A the radicand became $27$, which is not itself a perfect square but does have a perfect-square factor, $27=9\cdot3$, so the root only partly resolved, leaving $3\sqrt3$.

Here the radicand is $5/3$, which is not even a whole number, let alone a perfect square. There is no perfect-square factor to look for in a fraction that is not an integer, so the question that resolved parts A and B does not even apply here, and the radical stays exactly where the quotient rule left it.