Arithmetic with Roots

Both numbers slide under one radical. $$\sqrt{3}\times\sqrt{12}=\sqrt{3\times12}=\sqrt{36}=\boxed{6}$$ 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. $$\sqrt{6}\times\sqrt{10}=\sqrt{6\times10}=\sqrt{60}$$ Since \(60=4\times15\) 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 \(\boxed{2}\). 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 \(\boxed{24}\). 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 \(\sqrt{2}\times\sqrt{2}=\sqrt{4}=2\). $$\left(5\sqrt{2}\right)^2=25\times 2=\boxed{50}$$ 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 \(\boxed{8}\). 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 \(8\times 8=64\), $$\sqrt{36}+\sqrt{64}=6+8=\boxed{14}$$ 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 \(\sqrt{72}=6\sqrt{2}\). $$3\sqrt{8}-\sqrt{72}=6\sqrt{2}-6\sqrt{2}=\boxed{0}$$ 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. $$\frac{\sqrt{252}}{\sqrt{7}}=\sqrt{\frac{252}{7}}=\sqrt{36}=\boxed{6}$$ Neither \(\sqrt{252}\) nor \(\sqrt{7}\) is a whole number by itself, yet the quotient is, because \(252\div 7\) 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, \(\boxed{5/7}\). 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. $$\sqrt{0.64}=\frac{\sqrt{64}}{\sqrt{100}}=\frac{8}{10}=\boxed{0.8}$$ 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 \(\boxed{3}\) 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. \(\dfrac{\sqrt{180}}{\sqrt{5}}=\sqrt{\dfrac{180}{5}}=\sqrt{36}=6\). Now it is plain arithmetic. $$20+12-6=\boxed{26}$$ Each rule turned a radical into a whole number, so nothing irrational survived to the end.

The product rule merges the two roots. $$\sqrt{5}\times\sqrt{45}=\sqrt{5\times 45}=\sqrt{225}=\boxed{15}$$ 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 \(\boxed{8}\). 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 \(\boxed{2}\). 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 \(15\times 2=\boxed{30}\). 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 \(\boxed{30}\). Multiply first, simplify last.

Squaring hits both parts, so \(4\times 4=16\) and \(\sqrt{5}\times\sqrt{5}=5\). That leaves \(16\times 5=\boxed{80}\). 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 \(\sqrt{3}\times\sqrt{3}=3\). $$\left(6\sqrt{3}\right)^2=36\times 3=\boxed{108}$$ 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 \(k=\boxed{7}\). 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 \(12+5=\boxed{17}\). 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 \(k=\boxed{7}\). 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 $$5\sqrt{2}-\sqrt{50}=5\sqrt{2}-5\sqrt{2}=\boxed{0}$$ 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. $$\frac{\sqrt{300}}{\sqrt{3}}=\sqrt{\frac{300}{3}}=\sqrt{100}=\boxed{10}$$ 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, \(\boxed{3/8}\). Squaring back checks it, since \(\left(\tfrac{3}{8}\right)^2=\tfrac{9}{64}\).

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