Rationalizing Denominators: Free Response

Multiply the numerator and denominator by $\sqrt{10}$, since $10$ has no perfect-square factor to pull out first.

$\frac{6}{\sqrt{10}} = \frac{6}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}} = \frac{6\sqrt{10}}{10}.$

The $6$ and the $10$ share a factor of $2$, so reduce the fraction:

$\frac{6\sqrt{10}}{10} = \frac{3\sqrt{10}}{5}.$

Here $18$ does have a perfect-square factor, so pull it out first: $\sqrt{18} = \sqrt{9\cdot 2} = 3\sqrt{2}$.

$\frac{9}{\sqrt{18}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}.$

Now rationalize the smaller fraction that is left:

$\frac{3}{\sqrt{2}} = \frac{3}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.$

Compare what happened in the two parts. In part A, $10 = 2\times 5$ has no perfect-square factor greater than $1$, so there was nothing to simplify before rationalizing; the work was rationalize, then reduce.

In part B, $18 = 9\times 2$ carries the perfect-square factor $9$. Pulling it out first shrank the radical itself:

$\sqrt{18} = \sqrt{9\cdot 2} = 3\sqrt{2},$

which turned the fraction into $\frac{3}{\sqrt{2}}$ before any rationalizing began, keeping every later number small.

So the test is on the radicand alone, before any multiplying: does it have a perfect-square factor greater than $1$? If yes, pull it out first. If no, there is nothing to gain by looking for one, so rationalize directly and reduce whatever fraction is left at the end.

The conjugate keeps both terms the same and flips only the sign between them.

$3+\sqrt{7}\ \longrightarrow\ 3-\sqrt{7}.$

$\sqrt{11}-\sqrt{6}\ \longrightarrow\ \sqrt{11}+\sqrt{6}.$

The second pair works exactly like the first: it makes no difference that both terms happen to carry a radical.

Each product is a difference of two squares, where the middle terms cancel.

$(3+\sqrt{7})(3-\sqrt{7}) = 3^2-(\sqrt{7})^2 = 9-7 = 2.$

$(\sqrt{11}-\sqrt{6})(\sqrt{11}+\sqrt{6}) = (\sqrt{11})^2-(\sqrt{6})^2 = 11-6 = 5.$

Both products come out as whole numbers with no radical remaining, which is exactly what a conjugate is for.

Expand the repeated product the same way you would any binomial squared.

$(4+\sqrt{3})(4+\sqrt{3}) = 16+4\sqrt{3}+4\sqrt{3}+3 = 19+8\sqrt{3}.$

Both middle terms carry the SAME sign, so they add together instead of cancelling, and the radical $\sqrt{3}$ is still there in the result.

Compare the conjugate product:

$(4+\sqrt{3})(4-\sqrt{3}) = 16-4\sqrt{3}+4\sqrt{3}-3 = 13.$

Here the middle terms carry OPPOSITE signs, $-4\sqrt3$ and $+4\sqrt3$, so they cancel to zero and only the two squared terms remain. That cancellation is the one property a repeated same-sign product never has, no matter what number replaces $\sqrt3$.

The move that clears a two-term radical denominator is multiplying by the CONJUGATE, the same two terms with the sign between them flipped. The student instead multiplied $3-\sqrt2$ by an identical copy of itself, which squares the expression rather than pairing it with its sign-flipped twin.

$(3-\sqrt{2})(3-\sqrt{2}) = 9-3\sqrt2-3\sqrt2+2 = 11-6\sqrt2.$

Both middle terms are $-3\sqrt2$, so they add together instead of cancelling, and $\sqrt2$ is still there. The correct multiplier is the conjugate of the original denominator, whose opposite-signed middle terms would cancel instead.

Multiply the numerator and denominator by the conjugate $3+\sqrt2$.

$\frac{5}{3-\sqrt2}\cdot\frac{3+\sqrt2}{3+\sqrt2} = \frac{5(3+\sqrt2)}{3^2-(\sqrt2)^2}.$

The denominator is $9-2=7$, and the numerator distributes to $15+5\sqrt2$:

$\frac{5(3+\sqrt2)}{9-2} = \frac{15+5\sqrt2}{7}.$

No common factor divides $15$, $5$, and $7$ together, so this is fully simplified.

Expand the two possibilities in letters, with $b$ standing for a radical term. Squaring the same expression gives

$(a\pm b)^2 = a^2\pm 2ab+b^2,$

which always keeps a middle term $\pm 2ab$. If $b$ carries a radical, that middle term still does, so the denominator is never rational. Multiplying by the conjugate instead gives

$(a+b)(a-b) = a^2-ab+ab-b^2 = a^2-b^2,$

where the two middle terms are exact opposites, $-ab$ and $+ab$, forced to cancel by the fact that one factor has the opposite sign from the other. A repeated same-sign product can never reproduce that cancellation, since both of its middle terms are forced to carry the SAME sign as each other, whatever binomial it started from.

Multiply the numerator and denominator by the conjugate of $\sqrt7+\sqrt3$.

$\frac{8}{\sqrt7+\sqrt3}\cdot\frac{\sqrt7-\sqrt3}{\sqrt7-\sqrt3} = \frac{8(\sqrt7-\sqrt3)}{7-3}.$

The denominator is $7-3=4$, and the numerator distributes to $8\sqrt7-8\sqrt3$:

$\frac{8\sqrt7-8\sqrt3}{4} = 2\sqrt7-2\sqrt3.$

Every term shares a factor of $4$, so the fraction reduces all the way down.

Before reaching for the conjugate, check the two radicals: $\sqrt8=\sqrt{4\cdot2}=2\sqrt2$, so the denominator simplifies to

$\sqrt8-\sqrt2 = 2\sqrt2-\sqrt2 = \sqrt2,$

a SINGLE root instead of two. The problem is now $\frac{6}{\sqrt2}$, which clears the ordinary way:

$\frac{6}{\sqrt2} = \frac{6}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2} = \frac{6\sqrt2}{2} = 3\sqrt2.$

The conjugate would have reached the same value, but simplifying the radicals first made the whole conjugate step unnecessary.

Simplify each radical in both denominators on its own, before comparing anything.

In $\sqrt7+\sqrt3$, neither $7$ nor $3$ has a perfect-square factor, so neither radical simplifies, and the two roots share no common radicand. There is nothing to combine, so a true two-term conjugate is needed.

In $\sqrt8-\sqrt2$, the radicand $8$ does have a perfect-square factor, so pull it out:

$\sqrt8 = \sqrt{4\cdot 2} = 2\sqrt2.$

Once that is done, both terms are multiples of the SAME radical $\sqrt2$, so they combine into one term by ordinary subtraction, the way $2x-x=x$ does for any matching variable.

The test: simplify each radical first. If the two simplified radicals share the same number under the root, they combine into a single root and no conjugate is needed. If they do not, the denominator is a genuine two-term expression, and the conjugate is the only way to clear it.

Multiply the numerator and denominator by the conjugate of $\sqrt2-\sqrt6$.

$\frac{1}{\sqrt2-\sqrt6}\cdot\frac{\sqrt2+\sqrt6}{\sqrt2+\sqrt6} = \frac{\sqrt2+\sqrt6}{2-6}.$

The denominator is $2-6=-4$, a NEGATIVE rational number:

$\frac{\sqrt2+\sqrt6}{-4} = -\frac{\sqrt2+\sqrt6}{4}.$

The radical never went away improperly; the difference of squares is simply negative here because $2<6$.

Multiply by the conjugate $\sqrt p+\sqrt q$, exactly as for any two-root denominator.

$\frac{1}{\sqrt p-\sqrt q}\cdot\frac{\sqrt p+\sqrt q}{\sqrt p+\sqrt q} = \frac{\sqrt p+\sqrt q}{p-q}.$

Since $p<q$, the difference $p-q$ is negative, so this result has a negative denominator for EVERY such $p$ and $q$, not just for one chosen pair. To write it with a positive denominator, multiply the top and bottom by $-1$, itself a form of $1$:

$\frac{\sqrt p+\sqrt q}{p-q} = \frac{-1(\sqrt p+\sqrt q)}{-1(p-q)} = \frac{-(\sqrt p+\sqrt q)}{q-p} = -\frac{\sqrt p+\sqrt q}{q-p}.$

Because $p<q$ was assumed and never dropped, $q-p$ is positive for every valid choice, so this form always has a positive denominator.

Substitute $p=2$ and $q=6$ into the general formula from part B.

$\frac{\sqrt p+\sqrt q}{p-q}\ \Big|_{p=2,\,q=6} = \frac{\sqrt2+\sqrt6}{2-6} = \frac{\sqrt2+\sqrt6}{-4},$

which matches part A exactly, confirming the general result on a case already worked by hand.

A negative denominator is not a warning sign here. The conjugate's product is DEFINED to be $p-q$, and whenever the first radicand is smaller than the second, that difference is negative by ordinary arithmetic, exactly as $2-6=-4$ is negative. The fraction is still fully rationalized: no radical remains in the denominator, which is the only thing the word rationalized ever promised.