Roots, rational exponents and rationalising
Tonight's idea
You know the exponent laws for whole numbers: \(a^m a^n = a^{m+n}\) and \((a^m)^n = a^{mn}\). Nobody defined \(x^{1/2}\); it was forced. If the laws are to keep working, \(\left(x^{1/2}\right)^2 = x^{1/2 \cdot 2} = x\). So \(x^{1/2}\) must be a number whose square is \(x\): a square root, and we pick the non-negative one. The same argument gives \(x^{1/n} = \sqrt[n]{x}\), \(x^{m/n} = \left(\sqrt[n]{x}\right)^m\), \(x^0 = 1\) and \(x^{-n} = 1/x^n\).
Simplifying. \(\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt2\): pull out the biggest square. \(\sqrt{a}\sqrt{b} = \sqrt{ab}\), but \(\sqrt{a} + \sqrt{b}\) does not simplify.
Rationalising. For \(\dfrac{1}{\sqrt5 - \sqrt3}\), multiply top and bottom by the conjugate \(\sqrt5 + \sqrt3\). The bottom becomes \(5 - 3 = 2\) by the difference of squares, so the answer is \(\dfrac{\sqrt5 + \sqrt3}{2}\).
Nested radicals. Sometimes \(\sqrt{a + 2\sqrt b}\) is secretly \(\sqrt m + \sqrt n\). Squaring the guess gives \(m + n + 2\sqrt{mn}\), so you need \(m + n = a\) and \(mn = b\): two numbers with a given sum and product.
Watch one: three kinds of tidy-up
\(8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4\). Take the root first; the numbers stay small. Likewise \(16^{-3/4} = 1/\left(\sqrt[4]{16}\right)^3 = \tfrac18\).
\(\dfrac{6}{\sqrt7 - 1} = \dfrac{6(\sqrt7 + 1)}{(\sqrt7 - 1)(\sqrt7 + 1)} = \dfrac{6(\sqrt7 + 1)}{6} = \sqrt7 + 1\).
\(\sqrt{7 - 4\sqrt3}\): write \(4\sqrt3\) as \(2\sqrt{12}\) so it matches the form \(a + 2\sqrt b\), then look for \(m + n = 7\), \(mn = 12\): that is 4 and 3. Guess \(2 - \sqrt3\) (minus, because of the minus inside). Check: \((2 - \sqrt3)^2 = 4 - 4\sqrt3 + 3\). Yes, and it is positive, as a square root must be.
Your turn
- Evaluate without a calculator: \(27^{2/3}\), \(32^{-2/5}\), \(\left(\tfrac{1}{16}\right)^{-3/4}\).
- Rationalise the denominators: \(\dfrac{2}{3 - \sqrt7}\) and \(\dfrac{\sqrt3 + 1}{\sqrt3 - 1}\).
- Solve: \(x^{3/2} = 27\) (raise both sides to the power \(\tfrac23\)), and \(\sqrt{x + 7} = x - 5\) (this one produces two candidates; check both).
- Un-nest: \(\sqrt{11 + 6\sqrt2}\) and \(\sqrt{5 - 2\sqrt6}\). Check each answer by squaring.
Stretch: a radical that goes on forever
Let \(x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}\), going on forever. Explain why \(x = \sqrt{2 + x}\), solve for \(x\), and say why one root of your quadratic must be rejected.
Answers
- \(27^{2/3} = 3^2 = 9\). \(32^{-2/5} = 1/2^2 = \tfrac14\). \(\left(\tfrac1{16}\right)^{-3/4} = 16^{3/4} = 2^3 = 8\).
- \(\dfrac{2(3 + \sqrt7)}{9 - 7} = 3 + \sqrt7\). \(\dfrac{(\sqrt3 + 1)^2}{3 - 1} = \dfrac{4 + 2\sqrt3}{2} = 2 + \sqrt3\).
- \(x = 27^{2/3} = 9\). Squaring the second: \(x + 7 = x^2 - 10x + 25\), so \(x^2 - 11x + 18 = 0\), \(x = 2\) or \(9\). But \(x = 2\) gives \(\sqrt9 = 3\) against \(2 - 5 = -3\): false (squaring hid the sign). \(x = 9\) gives \(\sqrt{16} = 4 = 9 - 5\). Only \(x = 9\).
- \(\sqrt{11 + 6\sqrt2} = \sqrt{11 + 2\sqrt{18}}\): \(m + n = 11\), \(mn = 18\), so 9 and 2: \(3 + \sqrt2\). Check: \(9 + 6\sqrt2 + 2\). \(\sqrt{5 - 2\sqrt6}\): 3 and 2: \(\sqrt3 - \sqrt2\). Check: \(3 - 2\sqrt6 + 2\).
Stretch. The expression under the outermost root is \(2\) plus a copy of the whole thing, so \(x = \sqrt{2 + x}\). Squaring: \(x^2 - x - 2 = (x - 2)(x + 1) = 0\). A square root is never negative, so \(x = -1\) is rejected and \(x = 2\).