QuanticaAlgebra IOpen in the courseProblem
3.4 ended by pointing at the division that does not come out clean. Divide \(6\) by \(\sqrt{3}\). One student stops at \(\frac{6}{\sqrt{3}}\), another writes \(2\sqrt{3}\). The forms look nothing alike, so test them the 1.4 way. Square each exactly, and both land on the same whole number. Enter it.
Show a hint
- Squaring removes each radical exactly, so both forms square to plain numbers you can compare.
- Use \(\left(\frac{a}{\sqrt{b}}\right)^{2}=\frac{a^{2}}{b}\) on the first form and \((c\sqrt{b})^{2}=c^{2}\cdot b\) on the second.
Show the full solution
\(\left(\frac{6}{\sqrt{3}}\right)^{2}=\frac{36}{3}=12\) and \((2\sqrt{3})^{2}=4\cdot 3=\boxed{12}\). Two positive numbers with the same square must be equal, so both forms name the same number and both answers to the division are correct. This lesson decides which form counts as finished.
Both forms are right, and they are equal. Fractions had the same problem. \(\frac{69}{92}\) reduces to \(\frac{3}{4}\), one number with two spellings, and lowest terms settled which spelling counts as finished. Radicals need a standard form of their own. Before choosing it, name the kind of number sitting in that denominator.
Problem
Of the four numbers \(\sqrt{225}\), \(\sqrt{33}\), \(-\frac{7}{20}\), and \(0.6\), how many are rational? Enter the count.
Show a hint
- Test each one against the definition, writable as one integer over another. A radical can hide a whole number.
- \(15^{2}=225\). Is \(33\) a perfect square?
Show the full solution
\(\sqrt{225}=15\), \(-\frac{7}{20}\) is already a fraction of integers, and \(0.6=\frac{3}{5}\), while \(33\) is no perfect square, so \(\sqrt{33}\) is irrational and the count is \(\boxed{3}\). Judge each number by its value, not by how it happens to be written.
Problem
\(\frac{\sqrt{13}}{\sqrt{13}}\) equals \(1\), and multiplying by \(1\) changes nothing. Multiply \(\frac{9}{\sqrt{13}}\) by it, then use the product rule from 3.4 on the bottom. Enter the denominator of the result.
Show a hint
- The whole question is what \(\sqrt{13}\cdot\sqrt{13}\) equals.
- By the product rule, \(\sqrt{13}\cdot\sqrt{13}=\sqrt{13\cdot 13}=\sqrt{169}\).
Show the full solution
\(\frac{9}{\sqrt{13}}\cdot\frac{\sqrt{13}}{\sqrt{13}}=\frac{9\sqrt{13}}{\sqrt{169}}=\frac{9\sqrt{13}}{13}\), so the denominator is \(\boxed{13}\). The radical did not vanish, it moved to the numerator, and the denominator became an integer.
Problem
Rationalize \(\frac{4}{\sqrt{11}}\). Write the result as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(a+b+c\).
Show a hint
- Multiply by \(1\) written as \(\frac{\sqrt{11}}{\sqrt{11}}\).
- The result is \(\frac{4\sqrt{11}}{11}\). Read off the three numbers.
Show the full solution
\(\frac{4}{\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}}=\frac{4\sqrt{11}}{11}\), so \(a+b+c=4+11+11=\boxed{26}\). Nothing cancels here, since \(4\) and \(11\) share no factor.
Problem
Rationalize \(\frac{508}{\sqrt{127}}\) and reduce. The result is a whole multiple of the radical, \(a\sqrt{127}\). Enter \(a\).
Show a hint
- Rationalizing is not the last step here. A fraction appears, and it reduces.
- After the move you have \(\frac{508\sqrt{127}}{127}\). That fraction is not in lowest terms yet.
Show the full solution
\(\frac{508}{\sqrt{127}}=\frac{508\sqrt{127}}{127}=4\sqrt{127}\), since \(508=4\cdot 127\), so \(a=\boxed{4}\). Stopping at \(\frac{508\sqrt{127}}{127}\) is the common miss. That answer is correct but not in lowest terms.
Problem
Rationalize \(\frac{10}{\sqrt{15}}\). Write the result as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(a+b+c\).
Show a hint
- After the move, the plain fraction part is not yet in lowest terms.
- \(\gcd(10,15)=5\). Cancel it between the coefficient and the denominator.
Show the full solution
\(\frac{10}{\sqrt{15}}=\frac{10\sqrt{15}}{15}=\frac{2\sqrt{15}}{3}\), so \(a+b+c=2+15+3=\boxed{20}\). The \(5\) cancels between \(10\) and \(15\). The radicand \(15\) stays put, it is not part of the plain fraction.
Problem
Simplify the radical first this time, the 3.4 way, then rationalize \(\frac{226}{\sqrt{452}}\). Everything collapses to a single radical \(\sqrt{b}\). Enter \(b\).
Show a hint
- \(452\) has a perfect-square factor. Take it out before touching the fraction.
- \(\sqrt{452}=2\sqrt{113}\), so the fraction is \(\frac{226}{2\sqrt{113}}=\frac{113}{\sqrt{113}}\).
Show the full solution
\(\sqrt{452}=2\sqrt{113}\), so \(\frac{226}{\sqrt{452}}=\frac{226}{2\sqrt{113}}=\frac{113}{\sqrt{113}}=\frac{113\sqrt{113}}{113}=\sqrt{113}\), and \(b=\boxed{113}\). The brute road \(\frac{226\sqrt{452}}{452}\) lands in the same place, just with uglier numbers along the way.
Problem
Rationalize \(\frac{214}{5\sqrt{107}}\) and reduce fully. Write the result as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(a+b+c\).
Show a hint
- Multiply the top and bottom by \(\sqrt{107}\). The result is not yet in lowest terms.
- That gives \(\frac{214\sqrt{107}}{535}\), and \(535=5\cdot 107\).
Show the full solution
\(\frac{214}{5\sqrt{107}}=\frac{214\sqrt{107}}{5\cdot 107}=\frac{214\sqrt{107}}{535}=\frac{2\sqrt{107}}{5}\), so \(a+b+c=2+107+5=\boxed{114}\). The common factor is easy to miss because it is the radicand itself. \(107\) divides both \(214\) and \(535\).
Problem
Write \(\sqrt{\dfrac{19}{27}}\) as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Here \(a\) may be \(1\). Enter \(a+b+c\).
Show a hint
- The fraction is already in lowest terms, so split it with the quotient rule.
- \(\sqrt{27}=3\sqrt{3}\), so the job is to rationalize \(\frac{\sqrt{19}}{3\sqrt{3}}\).
Show the full solution
\(\sqrt{\frac{19}{27}}=\frac{\sqrt{19}}{\sqrt{27}}=\frac{\sqrt{19}}{3\sqrt{3}}=\frac{\sqrt{19}\cdot\sqrt{3}}{3\cdot 3}=\frac{\sqrt{57}}{9}\), so \(a+b+c=1+57+9=\boxed{67}\). The numerator radicals merged into \(\sqrt{57}\) by the product rule.
Problem
Write \(\sqrt{\dfrac{105}{48}}\) as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive, reducing the fraction under the radical before anything else. Here \(a\) may be \(1\). Enter \(a+b+c\).
Show a hint
- \(105\) and \(48\) share a factor, and the reduced denominator is special.
- \(\frac{105}{48}=\frac{35}{16}\), and \(16\) is a perfect square.
Show the full solution
\(\frac{105}{48}=\frac{35}{16}\), so \(\sqrt{\dfrac{105}{48}}=\sqrt{\dfrac{35}{16}}=\frac{\sqrt{35}}{4}\) and \(a+b+c=1+35+4=\boxed{40}\). Reducing first left nothing to rationalize. The denominator came out whole on its own.
Problem
Rationalize \(\frac{84}{\sqrt[3]{131}}\), a cube root in the denominator this time. The result has the form \(\frac{84\sqrt[3]{d}}{131}\). Enter \(d\).
Show a hint
- How many copies of \(131\) must sit under a cube root before it comes out whole?
- You need \(131^{3}\) in the denominator, so multiply by \(\frac{\sqrt[3]{131^{2}}}{\sqrt[3]{131^{2}}}\).
Show the full solution
\(\frac{84}{\sqrt[3]{131}}\cdot\frac{\sqrt[3]{131^{2}}}{\sqrt[3]{131^{2}}}=\frac{84\sqrt[3]{131^{2}}}{\sqrt[3]{131^{3}}}=\frac{84\sqrt[3]{17161}}{131}\), so \(d=131^{2}=\boxed{17161}\). One extra copy would have left \(\sqrt[3]{131^{2}}\) in the denominator, still irrational.
Problem
One expression, every tool. Write \(\frac{18}{\sqrt{14}}+2\sqrt{14}\) as a single fraction \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Rationalize first, then combine like radicals as in 3.4 over a common denominator as in 2.5. Enter \(a+b+c\).
Show a hint
- Rationalize and reduce the first term. Both terms then become multiples of \(\sqrt{14}\).
- \(\frac{18}{\sqrt{14}}=\frac{9\sqrt{14}}{7}\) and \(2\sqrt{14}=\frac{14\sqrt{14}}{7}\).
Show the full solution
\(\frac{18}{\sqrt{14}}=\frac{18\sqrt{14}}{14}=\frac{9\sqrt{14}}{7}\), and \(2\sqrt{14}=\frac{14\sqrt{14}}{7}\), so the sum is \(\frac{9\sqrt{14}+14\sqrt{14}}{7}=\frac{23\sqrt{14}}{7}\) and \(a+b+c=23+14+7=\boxed{44}\). Standard form is what made the two terms combinable at all.
Chapter 3 stretched the exponent laws to zero, negatives, and fractions, turned fractional powers into radicals, and closed with a standard form for answers. All of it was rewriting, one expression traded for an equal one. Chapter 4 turns to solving, where a claim of equality can be true or false. 4.1 starts with what an equation says.
Practice these ideas
Practice
Exactly one of \(\sqrt{81}\) and \(\sqrt{87}\) is rational. Decide which one, and enter the value of the rational one as a plain integer.
Show the solution
\(81=9^{2}\), so \(\sqrt{81}=\boxed{9}\), which is rational. \(87\) sits between \(9^{2}=81\) and \(10^{2}=100\), so no integer squares to \(87\), and 3.4 said no fraction does either, so \(\sqrt{87}\) is irrational.
Practice
To rationalize a fraction whose denominator is \(\sqrt{95}\), you multiply by \(\frac{\sqrt{95}}{\sqrt{95}}\). Enter the value of that multiplier as a single number.
Show the solution
\(\sqrt{95}\) is a nonzero number, and any nonzero number over itself equals \(\boxed{1}\). Multiplying by \(1\) changes how a fraction looks, never its value, and that is what makes rationalizing legal.
Practice
Rationalizing works because of one product. Enter the exact value of \(\sqrt{103}\cdot\sqrt{103}\).
Show the solution
\(\sqrt{103}\cdot\sqrt{103}=\sqrt{103^{2}}=\boxed{103}\). This product is what every rationalization uses. One more copy of the radical turns it into an integer.
Practice
Rationalize \(\frac{12}{\sqrt{89}}\) and write it as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(c\).
Show the solution
\(\frac{12}{\sqrt{89}}\cdot\frac{\sqrt{89}}{\sqrt{89}}=\frac{12\sqrt{89}}{89}\), already in lowest terms, so \(c=\boxed{89}\).
Practice
Rationalize \(\frac{16}{\sqrt{55}}\) and write it as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(a+b+c\), a plus b plus c, as a single number.
Show the solution
\(\frac{16}{\sqrt{55}}=\frac{16\sqrt{55}}{55}\). Since \(16\) and \(55\) share no factor, this is finished, and \(a+b+c=16+55+55=\boxed{126}\).
Practice
Rationalizing never changes a number's value, and squaring can prove it. Square \(\frac{50}{\sqrt{8}}\) exactly and enter the result as a fraction in lowest terms.
Show the solution
\(\left(\frac{50}{\sqrt{8}}\right)^{2}=\frac{50^{2}}{(\sqrt{8})^{2}}=\frac{2500}{8}=\boxed{\frac{625}{2}}\). Rationalized or not, \(\frac{50}{\sqrt{8}}\) squares to this same number, which is what calling the two forms equal means.
Practice
Rationalize \(\frac{60}{\sqrt{30}}\) and reduce. The result is a whole multiple of the radical, \(a\sqrt{30}\). Enter \(a\).
Show the solution
\(\frac{60}{\sqrt{30}}=\frac{60\sqrt{30}}{30}=2\sqrt{30}\), so \(a=\boxed{2}\). Stopping at \(\frac{60\sqrt{30}}{30}\) leaves the fraction unreduced, so it is not standard form yet.
Practice
A student rationalizes \(\frac{58}{\sqrt{58}}\) by multiplying only the denominator by \(\sqrt{58}\), gets \(\frac{58}{58}\), and answers \(1\). The real value is \(\sqrt{b}\). Enter \(b\).
Show the solution
\(\frac{58}{\sqrt{58}}\cdot\frac{\sqrt{58}}{\sqrt{58}}=\frac{58\sqrt{58}}{58}=\sqrt{58}\), so \(b=\boxed{58}\). The student's one-sided move divided the value by \(\sqrt{58}\), which no rewrite is allowed to do.
Practice
Rationalize \(\frac{25}{2\sqrt{139}}\) and write it as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(c\).
Show the solution
\(\frac{25}{2\sqrt{139}}\cdot\frac{\sqrt{139}}{\sqrt{139}}=\frac{25\sqrt{139}}{2\cdot 139}=\frac{25\sqrt{139}}{278}\). Since \(25\) and \(278\) share no factor, \(c=\boxed{278}\).
Practice
Rationalize \(\frac{93}{4\sqrt{62}}\) and reduce fully. Write it as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(a+b+c\), a plus b plus c, as a single number.
Show the solution
\(\frac{93}{4\sqrt{62}}=\frac{93\sqrt{62}}{4\cdot 62}=\frac{93\sqrt{62}}{248}\). Both \(93=3\cdot 31\) and \(248=8\cdot 31\) carry a \(31\), so the fraction reduces to \(\frac{3\sqrt{62}}{8}\), and \(a+b+c=3+62+8=\boxed{73}\).
Practice
Write \(\sqrt{\dfrac{21}{40}}\) as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive, where \(a\) may be \(1\). Enter \(b+c\), b plus c, as a single number.
Show the solution
\(\sqrt{\frac{21}{40}}=\frac{\sqrt{21}}{\sqrt{40}}=\frac{\sqrt{21}}{2\sqrt{10}}=\frac{\sqrt{21}\cdot\sqrt{10}}{2\cdot 10}=\frac{\sqrt{210}}{20}\), so \(b+c=210+20=\boxed{230}\). Simplifying \(\sqrt{40}\) first kept every number small.
Practice
Write \(\sqrt{\dfrac{258}{75}}\) as \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive, where \(a\) may be \(1\). Reduce the fraction under the radical first. Enter \(b+c\), b plus c, as a single number.
Show the solution
\(\frac{258}{75}=\frac{86}{25}\), so \(\sqrt{\frac{258}{75}}=\frac{\sqrt{86}}{\sqrt{25}}=\frac{\sqrt{86}}{5}\) and \(b+c=86+5=\boxed{91}\). Reducing first left a perfect square downstairs, so there was nothing to rationalize.
Practice
Evaluate \(\sqrt{\dfrac{284}{639}}\) exactly, reducing the fraction under the radical first. Enter your answer as a fraction in lowest terms.
Show the solution
\(284=4\cdot 71\) and \(639=9\cdot 71\), so \(\frac{284}{639}=\frac{4}{9}\) and \(\sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}=\boxed{\frac{2}{3}}\). No radical remains, since both parts of the reduced fraction are perfect squares.
Practice
Rationalize the cube-root denominator in \(\frac{1}{\sqrt[3]{109}}\). The result has the form \(\frac{\sqrt[3]{n}}{109}\). Enter \(n\).
Show the solution
\(\frac{1}{\sqrt[3]{109}}\cdot\frac{\sqrt[3]{109^{2}}}{\sqrt[3]{109^{2}}}=\frac{\sqrt[3]{109^{2}}}{\sqrt[3]{109^{3}}}=\frac{\sqrt[3]{11881}}{109}\), so \(n=\boxed{11881}\). Multiplying by just one more copy would leave \(\sqrt[3]{109^{2}}\) downstairs, still irrational.
Practice
\(22201=149^{2}\), so \(\sqrt[3]{22201}\) already holds two copies of \(149\). To rationalize \(\frac{1}{\sqrt[3]{22201}}\), multiply by \(\frac{\sqrt[3]{n}}{\sqrt[3]{n}}\) with \(n\) as small as possible. Enter \(n\).
Show the solution
\(\sqrt[3]{149^{2}}\cdot\sqrt[3]{149}=\sqrt[3]{149^{3}}=149\), so \(n=\boxed{149}\) and the fraction becomes \(\frac{\sqrt[3]{149}}{149}\). Multiplying by \(\frac{\sqrt[3]{22201^{2}}}{\sqrt[3]{22201^{2}}}\) also works but brings in far bigger numbers for the same answer.
Practice
Compute \(\frac{39}{\sqrt{78}}+\frac{\sqrt{78}}{2}\). Once the first term is in standard form, the sum collapses to a single radical \(\sqrt{n}\). Enter \(n\).
Show the solution
\(\frac{39}{\sqrt{78}}=\frac{39\sqrt{78}}{78}=\frac{\sqrt{78}}{2}\), so the sum is \(\frac{\sqrt{78}}{2}+\frac{\sqrt{78}}{2}=\frac{2\sqrt{78}}{2}=\sqrt{78}\) and \(n=\boxed{78}\). The two terms were equal all along, and standard form is what made that visible.
Practice
Write \(\frac{47}{\sqrt{94}}+\frac{\sqrt{94}}{4}\) as one fraction \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive. Enter \(a+b+c\), a plus b plus c, as a single number.
Show the solution
\(\frac{47}{\sqrt{94}}=\frac{47\sqrt{94}}{94}=\frac{\sqrt{94}}{2}=\frac{2\sqrt{94}}{4}\), so the sum is \(\frac{2\sqrt{94}}{4}+\frac{\sqrt{94}}{4}=\frac{3\sqrt{94}}{4}\) and \(a+b+c=3+94+4=\boxed{101}\). Neither term could combine with the other until the radical left the denominator.
Practice
Write \(\frac{201}{\sqrt{134}}-\sqrt{134}\) as one fraction \(\frac{a\sqrt{b}}{c}\) with \(b\) squarefree, the fraction in lowest terms, and \(c\) positive, where \(a\) may be \(1\). Enter \(b+c\), b plus c, as a single number.
Show the solution
\(\frac{201}{\sqrt{134}}=\frac{201\sqrt{134}}{134}=\frac{3\sqrt{134}}{2}\), and \(\sqrt{134}=\frac{2\sqrt{134}}{2}\), so the difference is \(\frac{3\sqrt{134}}{2}-\frac{2\sqrt{134}}{2}=\frac{\sqrt{134}}{2}\) and \(b+c=134+2=\boxed{136}\). The whole coefficient collapses to \(1\), which is why the ask allows \(a=1\).