Arithmetic with Roots

Both numbers slide under one radical. $$\sqrt{3}\times \sqrt{12}=\sqrt{3\times 12}=\sqrt{36}=\boxed{{\displaystyle 6}}$$ Two irrational roots can multiply to a whole number, and that happens exactly when the product underneath is a perfect square.

The product rule joins the two roots. $$\sqrt{6}\times \sqrt{10}=\sqrt{6\times 10}=\sqrt{60}$$ Since $60=4\times 15$ and $4$ is the largest perfect square dividing $60$, we get $\sqrt{60}=\sqrt{4}\times \sqrt{15}=2\sqrt{15}$, so the number in front is $\boxed{{\displaystyle 2}}$. Joining the roots is only half the job. Whatever lands under the radical still needs simplifying the 8.3 way.

The coefficients multiply and the radicands multiply, so $4\times 2=8$ and $\sqrt{3}\times \sqrt{6}=\sqrt{18}$, giving $8\sqrt{18}$. Since $18=9\times 2$, $\sqrt{18}=3\sqrt{2}$, and that $3$ folds into the $8$ for $8\times 3=24$. The simplest form is $24\sqrt{2}$, so the whole number in front is $\boxed{{\displaystyle 24}}$. Simplify last. Stopping at $8\sqrt{18}$ gives the right value in an unfinished form.

Squaring means multiplying the expression by itself, so group the coefficients and the roots. The coefficients give $5\times 5=25$, and $\sqrt{2}\times \sqrt{2}=\sqrt{4}=2$. $${\left(5\sqrt{2}\right)}^2=25\times 2=\boxed{{\displaystyle 50}}$$ A root times itself returns the number underneath, so $(k\sqrt{n}{)}^2=k^2\times n$ is a whole number whenever $k$ and $n$ are.

Both terms carry the same root, so add the coefficients and keep $\sqrt{7}$. $$6\sqrt{7}+2\sqrt{7}=(6+2)\sqrt{7}=8\sqrt{7}$$ The whole number in front is $\boxed{{\displaystyle 8}}$. This is $6x+2x=8x$ from 6.1 with $\sqrt{7}$ standing in for the variable.

Take each root on its own. Since $6\times 6=36$ and $8\times 8=64$, $$\sqrt{36}+\sqrt{64}=6+8=\boxed{{\displaystyle 14}}$$ The tempting shortcut gives $\sqrt{36+64}=\sqrt{100}=10$, a different number, which shows a root does not carry across a plus sign.

Simplify each radical first. Since $\sqrt{8}=2\sqrt{2}$, the first term is $3\times 2\sqrt{2}=6\sqrt{2}$, and since $72=36\times 2$, the second is $\sqrt{72}=6\sqrt{2}$. $$3\sqrt{8}-\sqrt{72}=6\sqrt{2}-6\sqrt{2}=\boxed{{\displaystyle 0}}$$ Two radicals that look different can turn out equal, which is why every term gets simplified before you combine.

The quotient rule puts the division under one radical. $$\frac{\sqrt{252}}{\sqrt{7}}=\sqrt{\frac{252}{7}}=\sqrt{36}=\boxed{{\displaystyle 6}}$$ Neither $\sqrt{252}$ nor $\sqrt{7}$ is a whole number by itself, yet the quotient is, because $252÷7$ lands on a perfect square.

The root splits across the fraction, and both pieces are perfect squares. $$\sqrt{\frac{25}{49}}=\frac{\sqrt{25}}{\sqrt{49}}=\frac{5}{7}$$ Since $5$ and $7$ share no common factor, that is already lowest terms, $\boxed{{\displaystyle 5\mathrm{/}7}}$. Squaring back checks it, since ${\left(\frac{5}{7}\right)}^2=\frac{25}{49}$.

Two digits after the point means hundredths, so $0.64=\frac{64}{100}$. Now split the root across the fraction. $$\sqrt{0.64}=\frac{\sqrt{64}}{\sqrt{100}}=\frac{8}{10}=\boxed{{\displaystyle 0.8}}$$ Squaring checks it, since $0.8\times 0.8=0.64$.

Pull the largest perfect square out of each radicand. That gives $\sqrt{12}=2\sqrt{3}$, $\sqrt{50}=5\sqrt{2}$, $\sqrt{63}=3\sqrt{7}$, and $\sqrt{8}=2\sqrt{2}$. The two $\sqrt{2}$ terms combine into $7\sqrt{2}$, while $2\sqrt{3}$ and $3\sqrt{7}$ have no partner, so the sum is $$7\sqrt{2}+2\sqrt{3}+3\sqrt{7},$$ which is $\boxed{{\displaystyle 3}}$ distinct radical terms. The count is just how many different square-free numbers are left under the roots, and you can only see that after simplifying.

Take the three pieces one at a time. Squaring. ${\left(2\sqrt{5}\right)}^2=2^2\times {\left(\sqrt{5}\right)}^2=4\times 5=20$. Multiplying. $\sqrt{72}\times \sqrt{2}=\sqrt{144}=12$. Dividing. ${\displaystyle \frac{\sqrt{180}}{\sqrt{5}}}=\sqrt{{\displaystyle \frac{180}{5}}}=\sqrt{36}=6$. Now it is plain arithmetic. $$20+12-6=\boxed{{\displaystyle 26}}$$ Each rule turned a radical into a whole number, so nothing irrational survived to the end.

The product rule merges the two roots. $$\sqrt{5}\times \sqrt{45}=\sqrt{5\times 45}=\sqrt{225}=\boxed{{\displaystyle 15}}$$ Two roots that are irrational on their own can multiply to a whole number when the product inside is a perfect square.

Both numbers go under one root, so $\sqrt{2}\times \sqrt{32}=\sqrt{64}$, and $8\times 8=64$, giving $\boxed{{\displaystyle 8}}$. Neither root is a whole number by itself, but the product inside is a perfect square, so the radical clears.

The radicands multiply, so $\sqrt{6}\times \sqrt{14}=\sqrt{84}$. Since $84=4\times 21$ and $21$ has no perfect-square factor left, $\sqrt{84}=\sqrt{4}\times \sqrt{21}=2\sqrt{21}$, so the whole number in front is $\boxed{{\displaystyle 2}}$. Combining is not the last step. The new radicand usually needs simplifying too.

The coefficients multiply and the roots multiply. That gives $3\times 5=15$ and $\sqrt{2}\times \sqrt{2}=\sqrt{4}=2$, so the value is $15\times 2=\boxed{{\displaystyle 30}}$. A root times itself always returns the number underneath, which is what clears the radical here.

The coefficients give $2\times 5=10$ and the radicands give $\sqrt{3}\times \sqrt{6}=\sqrt{18}$, so the product is $10\sqrt{18}$. Since $18=9\times 2$, $\sqrt{18}=3\sqrt{2}$, and $10\times 3=30$. The answer is $30\sqrt{2}$, so the whole number in front is $\boxed{{\displaystyle 30}}$. Multiply first, simplify last.

Squaring hits both parts, so $4\times 4=16$ and $\sqrt{5}\times \sqrt{5}=5$. That leaves $16\times 5=\boxed{{\displaystyle 80}}$. Squaring $k\sqrt{n}$ always gives $k^2\times n$, with the radical gone.

Squaring hits both parts. The coefficients give $6\times 6=36$, and $\sqrt{3}\times \sqrt{3}=3$. $${\left(6\sqrt{3}\right)}^2=36\times 3=\boxed{{\displaystyle 108}}$$ A root times itself returns the number underneath, which is why the irrational part disappears completely.

All three terms carry $\sqrt{3}$, so only the coefficients change. The bare $\sqrt{3}$ counts as $1\sqrt{3}$, and $10-4+1=7$, giving $7\sqrt{3}$, so $k=\boxed{{\displaystyle 7}}$. Forgetting that a bare root has a coefficient of $1$ is the usual slip here.

Take each root on its own. Since $12\times 12=144$ and $5\times 5=25$, the sum is $12+5=\boxed{{\displaystyle 17}}$. Adding under one root instead would give $\sqrt{169}=13$, a different number, which is why a root never crosses a plus sign.

Since $27=9\times 3$, $\sqrt{27}=3\sqrt{3}$, and since $48=16\times 3$, $\sqrt{48}=4\sqrt{3}$. Both terms carry $\sqrt{3}$ now, so add the coefficients, $3+4=7$, giving $7\sqrt{3}$ and $k=\boxed{{\displaystyle 7}}$. Radicals that look unlike often match once each one is simplified.

Simplify the second radical first. Since $50=25\times 2$, $\sqrt{50}=\sqrt{25}\times \sqrt{2}=5\sqrt{2}$, so $$5\sqrt{2}-\sqrt{50}=5\sqrt{2}-5\sqrt{2}=\boxed{{\displaystyle 0}}$$ Five of something minus five of the same thing is nothing, and the two terms only look different before you simplify.

The quotient rule puts the division under one root. $$\frac{\sqrt{300}}{\sqrt{3}}=\sqrt{\frac{300}{3}}=\sqrt{100}=\boxed{{\displaystyle 10}}$$ Neither root is a whole number alone, but the quotient inside is a perfect square, so the radical clears.

The root splits across the fraction. $$\sqrt{\frac{9}{64}}=\frac{\sqrt{9}}{\sqrt{64}}=\frac{3}{8}$$ Since $3$ and $8$ share no common factor, that is already lowest terms, $\boxed{{\displaystyle 3\mathrm{/}8}}$. Squaring back checks it, since ${\left(\frac{3}{8}\right)}^2=\frac{9}{64}$.

The 81 sits in the hundredths place, so $0.81={\displaystyle \frac{81}{100}}$. Now split the root across the fraction. $$\sqrt{0.81}=\frac{\sqrt{81}}{\sqrt{100}}=\frac{9}{10}=\boxed{{\displaystyle 0.9}}$$ Squaring checks it, since $0.9\times 0.9=0.81$.