Rationalizing and Radical Conjugates: Free Response

Multiply top and bottom by the missing copy of $\sqrt6$, which is exactly $1$ in disguise.

$\frac{7}{\sqrt6}\cdot\frac{\sqrt6}{\sqrt6}=\frac{7\sqrt6}{6}.$

The denominator is $(\sqrt6)^2=6$, a whole number, so the radical is gone from downstairs.

Write the radicand as a power of its prime base: $4=2^2$, so $\sqrt[3]{4}=2^{2/3}$. The exponent still needed to reach a whole power of $2$ is $\tfrac13$, which is $\sqrt[3]{2}$, not $\sqrt[3]{16}$.

$\frac{3}{\sqrt[3]{4}}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{3\sqrt[3]{2}}{\sqrt[3]{8}}=\frac{3\sqrt[3]{2}}{2}.$

The denominator became the whole number $2$, and the multiplier was the smallest one that would do it.

Rewrite each radical as a rational power of $a$: $\sqrt[n]{a}=a^{1/n}$ and $\sqrt[n]{a^{\,n-1}}=a^{(n-1)/n}$. The exponent law for multiplying like bases says the exponents add:

$a^{1/n}\cdot a^{(n-1)/n}=a^{\frac1n+\frac{n-1}{n}}=a^{\frac{n}{n}}=a^1=a.$

The two fractional exponents were built to add to $1$ no matter what $n$ is, so the product simplifies to $a$ itself, whatever the index. That step alone works for any positive real $a$, but restricting $a$ to a positive integer is what makes the result actually radical-free: a whole number carries no radical at all, whereas if $a$ were itself irrational the product would just be $a$ again, an irrational number, so nothing would have been cleared.

Multiply top and bottom by the radical conjugate, $4-\sqrt{13}$.

$\frac{1}{4+\sqrt{13}}\cdot\frac{4-\sqrt{13}}{4-\sqrt{13}}=\frac{4-\sqrt{13}}{4^2-13}=\frac{4-\sqrt{13}}{3}.$

The denominator became the whole number $3$, since the cross terms in $(4+\sqrt{13})(4-\sqrt{13})$ cancel and only $16-13$ survives.

Multiply the two exact expressions directly, without ever approximating $\sqrt{13}$.

$\frac{4-\sqrt{13}}{3}\cdot(4+\sqrt{13})=\frac{(4-\sqrt{13})(4+\sqrt{13})}{3}=\frac{16-13}{3}=\frac{3}{3}=1.$

The cross terms cancel again exactly as they did in part A, and the surviving numerator is once more a whole number, so the product is exactly $1$ with no rounding anywhere in the check. A check built from two rounded decimals might round to something that merely looks like $1$, but rounded arithmetic can never PROVE the product is exactly $1$, the way this exact symbolic computation just did.

Write the answer from part A in the $p+q\sqrt{13}$ form directly.

$\frac{4-\sqrt{13}}{3}=\frac43-\frac13\sqrt{13},$

so $p=\tfrac43$ and $q=-\tfrac13$, both rational.

Because that is again a number of the exact same form the lesson proved is closed under reciprocals, adding it to, multiplying it by, or dividing it into any other nonzero exact length of the form $p'+q'\sqrt{13}$ produces another number of that same exact form, with no error introduced. A rounded decimal has no such guarantee: it is only ever close to the true value, so every further computation built on it compounds that first rounding rather than staying exact.

Expand the product the same way a square-root conjugate pair is expanded.

$(2+\sqrt[3]{11})(2-\sqrt[3]{11})=2^2-\left(\sqrt[3]{11}\right)^2=4-\sqrt[3]{121}.$

Since $121$ is not a perfect cube, $\sqrt[3]{121}$ is irrational, so the result $4-\sqrt[3]{121}$ still carries a radical. The sign flip only cancelled the first power of $\sqrt[3]{11}$; the squared power, $\left(\sqrt[3]{11}\right)^2$, was never touched.

Use the sum-of-cubes multiplier $x^2-xy+y^2$ with $x=2$ and $y=\sqrt[3]{11}$.

$(2+\sqrt[3]{11})\left(4-2\sqrt[3]{11}+\sqrt[3]{121}\right)=2^3+\left(\sqrt[3]{11}\right)^3=8+11=19.$

So the denominator becomes the whole number $19$, and

$\frac{1}{2+\sqrt[3]{11}}=\frac{4-2\sqrt[3]{11}+\sqrt[3]{121}}{19}.$

Since $19$ shares no common factor with $4$, $2$, or $1$, this fraction is already fully reduced.

For a square root, the sequence is just $\sqrt d,\ (\sqrt d)^2=d$: two terms, with nothing strictly between them. A single sign flip cancels the one irrational term, $\sqrt d$, directly, which is why it always works there.

$\sqrt d \;\longrightarrow\; (\sqrt d)^2=d \qquad\text{versus}\qquad \sqrt[3]{d} \;\longrightarrow\; (\sqrt[3]{d})^2 \;\longrightarrow\; (\sqrt[3]{d})^3=d.$

For a cube root, the sequence has three terms, and $\left(\sqrt[3]{d}\right)^2$ sits strictly between the first term $\sqrt[3]{d}$ and the last term $d$, itself irrational since $d$ is not a perfect cube. A sign flip still cancels the first term, $\sqrt[3]{d}$, but has no effect on the middle term, since squaring is blind to sign regardless of which sign was flipped. One cancellation cannot clear two independent irrational terms, so the three-term multiplier is needed instead, built to cancel both at once.

Multiply top and bottom by $\sqrt{50}+7$, the partner that clears the numerator instead of the denominator.

$\sqrt{50}-7=\frac{\left(\sqrt{50}-7\right)\left(\sqrt{50}+7\right)}{\sqrt{50}+7}=\frac{50-49}{\sqrt{50}+7}=\frac{1}{\sqrt{50}+7}.$

The delicate subtraction of two nearly equal numbers has become a plain division.

Since $7^2=49<50$ and $7.1^2=50.41>50$, we know $7<\sqrt{50}<7.1$. Add $7$ throughout:

$14<\sqrt{50}+7<14.1.$

A larger denominator gives a smaller fraction, so taking reciprocals flips the inequality:

$\frac{1}{14.1}<\frac{1}{\sqrt{50}+7}<\frac1{14},$

which, by part A, traps $\sqrt{50}-7$ in that same narrow interval, roughly between $0.0709$ and $0.0714$.

Both $\sqrt{50}-7$ and $\sqrt{197}-14$ fit the same pattern.

$\sqrt{n^2+1}-n=\frac{1}{\sqrt{n^2+1}+n},$

with $n=7$ in the first case and $n=14$ in the second. The numerator is fixed at $1$ in both, but the denominator $\sqrt{n^2+1}+n$ grows as $n$ grows: it is roughly $14$ when $n=7$ and roughly $28$ when $n=14$.

A fixed numerator over a larger denominator is a smaller number, so $\sqrt{197}-14$ is itself smaller than $\sqrt{50}-7$, and any interval built from similarly tight decimal bounds on the larger $n$ will be correspondingly narrower too. This is the same crowding effect the lesson noted for consecutive square roots: the bigger the numbers involved, the closer together their roots sit.

Multiply by the conjugate $\sqrt{n+2}+\sqrt n$ over itself, the same move used for a gap of one.

$\sqrt{n+2}-\sqrt n=\frac{\left(\sqrt{n+2}-\sqrt n\right)\left(\sqrt{n+2}+\sqrt n\right)}{\sqrt{n+2}+\sqrt n}=\frac{(n+2)-n}{\sqrt{n+2}+\sqrt n}=\frac{2}{\sqrt{n+2}+\sqrt n}.$

The surviving numerator is the gap between the two radicands, $2$, not the $1$ from the lesson's version. Nothing in that computation picked out a particular value of $n$: it only used $(\sqrt{n+2})^2=n+2$ and $(\sqrt n)^2=n$, which hold for every $n\ge0$, so the formula is valid for the whole family at once, not just for one instance of it.

Set $n=25$ in the formula from part A. Since $\sqrt{25}=5$,

$\sqrt{27}-\sqrt{25}=\frac{2}{\sqrt{27}+5}.$

Given $5.19<\sqrt{27}<5.2$, add $5$ throughout to get $10.19<\sqrt{27}+5<10.2$. Taking reciprocals flips the inequality, and multiplying through by the numerator $2$ gives

$\frac{2}{10.2}<\sqrt{27}-\sqrt{25}<\frac{2}{10.19},$

which is roughly between $0.1961$ and $0.1963$.

In every case of this kind, multiplying $\sqrt{n+k}-\sqrt n$ by its conjugate over itself leaves the same shape of numerator.

$\sqrt{n+k}-\sqrt n=\frac{(n+k)-n}{\sqrt{n+k}+\sqrt n}=\frac{k}{\sqrt{n+k}+\sqrt n},$

and the $n$'s cancel regardless of what $k$ is, leaving exactly $k$. The denominator, $\sqrt{n+k}+\sqrt n$, keeps the same algebraic form, a sum of two square roots, for every $k$; only its numeric value changes with $k$, through the radicand $n+k$.

So the numerator of the rationalized form is always just the gap $k$ between the two integers under the roots, nothing more. Doubling $k$ from $1$ to $2$ therefore doubles the numerator directly, from $1$ to $2$, because that numerator was never anything but $k$ in the first place.