QuanticaAlgebra IOpen in the courseProblem
Every answer in 3.3 came out whole because every base was a perfect power. Neither \(245^{1/2}\) nor \(5^{1/2}\) is a whole number. But power of a product from 3.1 reads in both directions, so two half powers can merge into one. Combine \(245^{1/2}\cdot 5^{1/2}\) into a single power and enter its value.
Show a hint
- The law that splits \((ab)^{1/2}\) into \(a^{1/2}b^{1/2}\) also runs right to left. Two half powers of different bases can become one half power.
- \(245\cdot 5=1225\). What number squares to \(1225\)?
Show the full solution
Merging the half powers gives \(245^{1/2}\cdot 5^{1/2}=(245\cdot 5)^{1/2}=1225^{1/2}\), and \(35^{2}=1225\), so the value is \(\boxed{35}\). Two numbers that are not whole on their own multiplied to a number that is. This lesson runs on that move.
The product came out whole, but \(245^{1/2}\) on its own did not go away. It is a genuine number, somewhere between \(15\) and \(16\) since \(15^{2}=225\) and \(16^{2}=256\), and no whole number or fraction squares to exactly \(245\). A number like that cannot be evaluated into anything simpler. It needs a way to be written.
Problem
3.3 settled which roots accept a negative base, and the new symbol changes none of it. Exactly one of \(\sqrt{-1000}\) and \(\sqrt[3]{-1000}\) names a real number. Enter its value.
Show a hint
- An even root needs a nonnegative radicand. An odd root takes any sign.
- What number cubed gives \(-1000\)?
Show the full solution
\((-10)^{3}=(-10)(-10)(-10)=-1000\), so \(\sqrt[3]{-1000}=\boxed{-10}\). No real number squares to \(-1000\), since squares are never negative, so \(\sqrt{-1000}\) names nothing at all.
\(\sqrt{41}\) sits between \(6\) and \(7\), since \(6^{2}=36\) and \(7^{2}=49\), and no fraction squares to \(41\). The symbol is still exact. The square of \(\sqrt{41}\) is exactly \(41\), while any decimal you could type squares to nearly \(41\) and misses. So \(\sqrt{41}\) is not an unfinished computation. It is the answer, and it stays written.
Problem
The rule claims \(\sqrt{64\cdot 49}=\sqrt{64}\cdot\sqrt{49}\). Test it the 1.4 way. The right side is quick. Confirm the left side agrees, then enter the shared value.
Show a hint
- Work each side on its own. The claim is true only if the two values match.
- \(\sqrt{64}\cdot\sqrt{49}=8\cdot 7\). Square that product and compare it with \(64\cdot 49\).
Show the full solution
\(\sqrt{64}\cdot\sqrt{49}=8\cdot 7=56\), and \(64\cdot 49=3136\) with \(56^{2}=3136\), so \(\sqrt{3136}=56\) as well. Both sides land on \(\boxed{56}\).
Problem
The factorization is handed to you this time. Use \(63=9\cdot 7\) to simplify \(\sqrt{63}\) to the form \(a\sqrt{7}\). Enter \(a\).
Show a hint
- The product rule splits \(\sqrt{63}\) across the two factors you were given.
- \(\sqrt{9\cdot 7}=\sqrt{9}\cdot\sqrt{7}\), and one of those two is a whole number.
Show the full solution
\(\sqrt{63}=\sqrt{9}\,\sqrt{7}=\boxed{3}\sqrt{7}\). The perfect-square factor \(9\) leaves the radical as \(3\), and the \(7\) stays inside.
Problem
Simplify \(\sqrt{592}\) to \(a\sqrt{b}\) with \(b\) as small as possible. A perfect square bigger than \(4\) divides \(592\). Enter \(b\).
Show a hint
- Pulling out the first square you spot is not always the last word. Check what is left under the radical.
- Splitting off \(4\) gives \(2\sqrt{148}\), and \(148\) still has a perfect-square factor.
Show the full solution
\(592=16\cdot 37\), so \(\sqrt{592}=\sqrt{16}\,\sqrt{37}=4\sqrt{37}\) and \(b=\boxed{37}\). Stopping at \(2\sqrt{148}\) is the common miss. \(148=4\cdot 37\) still has a square factor, so the job was only half done.
Problem
Two pairs of the same prime this time. \(4941=3^{4}\cdot 61\). Pair the primes to simplify \(\sqrt{4941}\) to \(a\sqrt{b}\) with \(b\) as small as possible. Enter \(a\).
Show a hint
- \(3^{4}\) is two separate pairs of \(3\), and each pair leaves the radical as one copy.
- The two exited \(3\)s multiply together out front. Only the \(61\) stays inside.
Show the full solution
\(3^{4}=3^{2}\cdot 3^{2}\) is two pairs, and each pair leaves the radical as a \(3\), so \(\sqrt{4941}=3\cdot 3\cdot\sqrt{61}=\boxed{9}\sqrt{61}\). One pair at a time also works, \(\sqrt{4941}=3\sqrt{549}\) and \(549=9\cdot 61\), it just takes two steps.
Problem
Not every radical can be simplified. Of \(\sqrt{66}\), \(\sqrt{85}\), \(\sqrt{110}\), and \(\sqrt{348}\), how many can be? Enter the count.
Show a hint
- A radical simplifies exactly when some prime repeats in the radicand's factorization.
- Factor each radicand and look for a repeated prime. Three of the four have none.
Show the full solution
\(66=2\cdot 3\cdot 11\), \(85=5\cdot 17\), and \(110=2\cdot 5\cdot 11\) have no repeated prime, so those three are already simplified. Only \(348=2^{2}\cdot 3\cdot 29\) has a pair, giving \(2\sqrt{87}\), so the count is \(\boxed{1}\). A squarefree radicand is not stuck, it is finished.
Problem
Evaluate \(\sqrt{\dfrac{490}{810}}\) exactly, reducing the fraction under the radical first. Enter the value as a fraction in lowest terms.
Show a hint
- Simplify the fraction before touching the radical.
- Divide the top and the bottom by \(10\). Both results are perfect squares.
Show the full solution
\(\frac{490}{810}=\frac{49}{81}\), so $$\sqrt{\frac{490}{810}}=\sqrt{\frac{49}{81}}=\frac{\sqrt{49}}{\sqrt{81}}=\boxed{\frac{7}{9}}.$$ Rooting first leaves \(\sqrt{490}\) over \(\sqrt{810}\), and neither is a whole number. Reducing first made both perfect squares.
Problem
Combine \(\sqrt{53}\cdot\sqrt{212}\) under one radical. Neither factor alone is a whole number, but the product is. Enter the whole-number value.
Show a hint
- One radical first, then look for pairs among the prime factors.
- \(212=4\cdot 53\), so the combined radicand is \(4\cdot 53^{2}\).
Show the full solution
\(\sqrt{53}\cdot\sqrt{212}=\sqrt{53\cdot 212}=\sqrt{4\cdot 53^{2}}=2\cdot 53=\boxed{106}\). Two numbers with no exact decimal form multiplied to a plain whole number, which is why exact forms are worth keeping until the end.
Problem
\(\sqrt{204}+\sqrt{459}\) looks like a sum of unlike radicals. Simplify each term, then enter the coefficient of \(\sqrt{51}\) in the sum.
Show a hint
- Neither term is simplified yet, and likeness only shows after both are.
- \(204=4\cdot 51\) and \(459=9\cdot 51\).
Show the full solution
\(\sqrt{204}=2\sqrt{51}\) and \(\sqrt{459}=3\sqrt{51}\), so the sum is \(2\sqrt{51}+3\sqrt{51}=\boxed{5}\sqrt{51}\). The two radicals were like all along. Simplifying is what made it visible.
Problem
Compute \(\sqrt{2303}-\sqrt{752}-\sqrt{423}\) exactly. No decimals and no estimating. The exact value is a number you can type. Enter it.
Show a hint
- Three ugly radicals can all be multiples of one small radical. Simplify each term first.
- Each radicand is a perfect square times \(47\).
Show the full solution
\(2303=49\cdot 47\), \(752=16\cdot 47\), and \(423=9\cdot 47\), so the expression is \(7\sqrt{47}-4\sqrt{47}-3\sqrt{47}=(7-4-3)\sqrt{47}=\boxed{0}\). The surprise is the point. Simplifying can collapse a whole expression to nothing.
Problem
A cube root to close. \(344=2^{3}\cdot 43\). Simplify \(\sqrt[3]{344}\) to \(a\sqrt[3]{b}\) with \(b\) as small as possible. Enter \(b\).
Show a hint
- Under a cube root a prime needs three copies to leave, not two.
- The triple of \(2\)s leaves as a single \(2\). Count what has no triple.
Show the full solution
The triple of \(2\)s leaves the radical as one \(2\), so \(\sqrt[3]{344}=2\sqrt[3]{43}\) and \(b=\boxed{43}\). Pulling out a mere pair of \(2\)s is the square-root habit misfiring, since \(4\) is not a perfect cube.
Look at what this lesson actually did. Radicals were multiplied, split apart, and added when like. Every division came out clean, because every fraction under a radical reduced to perfect squares. 3.5 faces the division that does not come out clean. Its first answers are correct but awkward to work with, and they get a cleanup rule of their own.
Practice these ideas
Practice
Evaluate \(\sqrt{4900}\).
Show the solution
\(4900=49\cdot 100\), so \(\sqrt{4900}=\sqrt{49}\,\sqrt{100}=7\cdot 10=\boxed{70}\). The product rule turns one big square root into two easy ones.
Practice
Enter the exact value of \((\sqrt{83})^{2}\).
Show the solution
\(\sqrt{83}\) is the nonnegative number whose square is \(83\), so \((\sqrt{83})^{2}=\boxed{83}\). That is the symbol's whole job. In exponent form, \((83^{1/2})^{2}=83^{1}\) says the same thing.
Practice
Exactly one of \(\sqrt{-512}\) and \(\sqrt[3]{-512}\) names a real number. Enter its value.
Show the solution
\((-8)^{3}=-512\), so \(\sqrt[3]{-512}=\boxed{-8}\). No real number squares to \(-512\), so \(\sqrt{-512}\) names nothing, while an odd root passes a negative sign straight through.
Practice
A student answers \(-34\) for \(\sqrt{1156}\), and the check \((-34)^{2}=1156\) comes out right. The answer is still wrong. Enter the value of \(\sqrt{1156}\).
Show the solution
\(34^{2}=1156\), so \(\sqrt{1156}=\boxed{34}\). Both \(34\) and \(-34\) square to \(1156\), and the radical sign names the nonnegative one, the same convention \(1156^{1/2}\) carried in 3.3. The other candidate is written \(-\sqrt{1156}\).
Practice
A student claims that \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\). Test the claim the 1.4 way at \(a=9\) and \(b=16\), then enter the true value of \(\sqrt{9}+\sqrt{16}\).
Show the solution
\(\sqrt{9}+\sqrt{16}=3+4=\boxed{7}\). The claim would need \(\sqrt{9+16}=\sqrt{25}=5\) to equal \(7\), and it does not. The product rule splits multiplication across a radical, but it has no addition twin.
Practice
Simplify \(\sqrt{152}\) to the form \(a\sqrt{38}\). Enter \(a\).
Show the solution
\(152=4\cdot 38\), so \(\sqrt{152}=\sqrt{4}\,\sqrt{38}=\boxed{2}\sqrt{38}\).
Practice
Simplify \(\sqrt{142}\) to \(a\sqrt{b}\) with \(b\) as small as possible. Enter \(b\).
Show the solution
\(142=2\cdot 71\) has no repeated prime, so nothing comes out. \(\sqrt{142}\) is already \(1\cdot\sqrt{142}\), and \(b=\boxed{142}\). A squarefree radicand means the radical is finished, not that you missed a trick.
Practice
Simplify \(\sqrt{585}\) to \(a\sqrt{b}\) with \(b\) as small as possible. Enter the product \(a\cdot b\), a times b, as a single number.
Show the solution
\(585=9\cdot 65\), so \(\sqrt{585}=3\sqrt{65}\), and \(a\cdot b=3\cdot 65=\boxed{195}\). Since \(65=5\cdot 13\) has no repeated prime, \(b\) is as small as it can get.
Practice
Evaluate \(\sqrt{\dfrac{96}{150}}\) exactly. Enter a fraction in lowest terms.
Show the solution
\(\frac{96}{150}=\frac{16}{25}\), so \(\sqrt{\frac{96}{150}}=\frac{\sqrt{16}}{\sqrt{25}}=\frac{4}{5}\), \(\boxed{4/5}\). Rooting \(96\) and \(150\) separately leaves a mess that the one reduction avoids.
Practice
\(\sqrt{624}-\sqrt{351}\) simplifies all the way down to a single radical \(\sqrt{b}\). Enter \(b\).
Show the solution
\(\sqrt{624}=4\sqrt{39}\) and \(\sqrt{351}=3\sqrt{39}\), so the difference is \(4\sqrt{39}-3\sqrt{39}=1\sqrt{39}\) and \(b=\boxed{39}\). Subtracting radicands would give \(\sqrt{273}\), which is wrong, since no rule pushes subtraction under a radical.
Practice
For positive \(y\), \(\sqrt{y^{194}}=y^{k}\). Enter \(k\).
Show the solution
\(\sqrt{y^{194}}=(y^{194})^{1/2}=y^{194\cdot\frac{1}{2}}=y^{97}\), so \(k=\boxed{97}\). Halving the exponent works whenever the base is positive and the exponent is even.
Practice
Evaluate \((\sqrt{46})^{4}\).
Show the solution
\((\sqrt{46})^{4}=(46^{1/2})^{4}=46^{2}=\boxed{2116}\). Grouping as \(((\sqrt{46})^{2})^{2}=46^{2}\) says the same thing without exponent notation.
Practice
Combine \(\sqrt{657}\cdot\sqrt{146}\) under one radical, then simplify to \(a\sqrt{b}\) with \(b\) as small as possible. Enter \(a\).
Show the solution
\(657\cdot 146=3^{2}\cdot 73^{2}\cdot 2\), so \(\sqrt{657}\,\sqrt{146}=3\cdot 73\cdot\sqrt{2}=\boxed{219}\sqrt{2}\). Multiplying out to \(95922\) first also works, but the pairs are easier to see before the multiplication buries them.
Practice
Evaluate \(\sqrt{\dfrac{126}{686}}\) exactly. Enter a fraction in lowest terms.
Show the solution
\(\frac{126}{686}=\frac{9}{49}\), so the root is \(\frac{\sqrt{9}}{\sqrt{49}}=\frac{3}{7}\), \(\boxed{3/7}\). Neither \(126\) nor \(686\) is a perfect square on its own, and the single reduction is what makes both layers come out clean.
Practice
Simplify \(\sqrt[3]{472}\) to \(a\sqrt[3]{b}\) with \(b\) as small as possible. Enter \(b\).
Show the solution
\(472=2^{3}\cdot 59\), and the triple of \(2\)s exits as one \(2\), so \(\sqrt[3]{472}=2\sqrt[3]{59}\) and \(b=\boxed{59}\). Writing \(2\sqrt[3]{118}\) from \(472=4\cdot 118\) is the square-root habit misfiring, since a pair of \(2\)s does nothing under a cube root.
Practice
Factor \(5056\) into primes to simplify \(\sqrt{5056}\) to \(a\sqrt{b}\) with \(b\) as small as possible. Enter the product \(a\cdot b\), a times b, as a single number.
Show the solution
\(5056=2^{6}\cdot 79\). The three pairs of \(2\)s leave as \(2\cdot 2\cdot 2=8\) and the unpaired \(79\) stays, so \(\sqrt{5056}=8\sqrt{79}\) and \(a\cdot b=8\cdot 79=\boxed{632}\). Stopping early at \(4\sqrt{316}\) or \(2\sqrt{1264}\) leaves a square inside, and \(b\) is only as small as possible once no prime repeats.
Practice
Simplify \(\sqrt{335}\cdot\sqrt{737}\) to \(a\sqrt{b}\) with \(b\) as small as possible. Enter the sum \(a+b\), a plus b, as a single number.
Show the solution
\(335\cdot 737=5\cdot 67\cdot 11\cdot 67=67^{2}\cdot 55\), so \(\sqrt{335}\,\sqrt{737}=67\sqrt{55}\), and \(a+b=67+55=\boxed{122}\). The shared \(67\) is invisible until both radicands sit under one radical, which is why combining comes first.