Lesson 12 · from §95 of the 1910 text
Rationalizing Factors
Rationalizing factors
Division by a radical is easiest when the radical is first removed from the divisor. A rationalizing factor of an expression is a factor that makes the product rational, free of radicals.
- \sqrt3 is a rationalizing factor of \sqrt3, since \sqrt3 \cdot \sqrt3 = 3; also of 2\sqrt3 and of \sqrt{12}, since \sqrt{12} \cdot \sqrt3 = \sqrt{36} = 6.
- \sqrt[3]{9} is a rationalizing factor of \sqrt[3]{3}, since \sqrt[3]{3} \cdot \sqrt[3]{9} = \sqrt[3]{27} = 3. For a cube root, choose the factor that makes the radicand a perfect cube.
- For a binomial with square roots, change the sign between the terms: \sqrt5 + \sqrt2 has the rationalizing factor \sqrt5 - \sqrt2, because \left(\sqrt5 + \sqrt2\right)\left(\sqrt5 - \sqrt2\right) = 5 - 2 = 3. The same works for 2 + \sqrt7 and 2 - \sqrt7, and for \sqrt{a} + \sqrt{b} and \sqrt a - \sqrt b.
Every expression has many rationalizing factors (\sqrt3, 2\sqrt3, \sqrt{12} all work for \sqrt{12}); any one that makes the product rational is accepted. The product you give must match the factor you chose.
Typing answers: type the factor in the first box, like
sqrt(3),cbrt(9)orsqrt(5) - sqrt(2), and the product in the second, like3orx - a.
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Determine a rationalizing factor for each of the following expressions and find the product of the expression and the factor:
\sqrt5
3\sqrt6
2\sqrt7
\sqrt8
\sqrt{32}
\sqrt{27}
\sqrt[3]2
\sqrt[3]3
2\sqrt[3]5
\sqrt[3]{16}
\sqrt[3]{25}
\sqrt[3]{36}
\sqrt[3]{49}
\sqrt2 + 3
\sqrt3 - \sqrt2
3 + \sqrt7
3\sqrt2 - 5
4\sqrt3 - \sqrt2
2\sqrt5 + 7\sqrt6
\sqrt x - \sqrt a
\sqrt{3a} + \sqrt x
3\sqrt x - a\sqrt2
\sqrt{x + a} + \sqrt x
\sqrt x - \sqrt{a - x}
\sqrt{2ax - x^2} - \sqrt{ax}