Radicals and Rational Exponents: practice solutions, Grade 11

Rate this post

Practice solutions Grade 11 : Radicals and Rational Exponents — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

Sommaire

1 Evaluating roots ★★★

a) \(12^2 = 144\), so \(\sqrt{144} = 12\).

b) \(5^3 = 125\), so \(\sqrt[3]{125} = 5\).

c) \(3^4 = 81\), so \(\sqrt[4]{81} = 3\).

d) \((-4)^3 = -64\), so \(\sqrt[3]{-64} = -4\).

e) \(\sqrt{\dfrac{49}{25}} = \dfrac{\sqrt{49}}{\sqrt{25}} = \dfrac{7}{5}\).

2 Radical form to exponent form ★★★

a) \(7^{1/2} = \sqrt{7}\).

b) \(5^{2/3} = \sqrt[3]{5^2} = \sqrt[3]{25}\).

c) \(x^{3/4} = \sqrt[4]{x^3}\).

d) \((2y)^{1/5} = \sqrt[5]{2y}\).

3 Exponent form from radicals ★★★

a) \(\sqrt{11} = 11^{1/2}\).

b) \(\sqrt[3]{a^5} = a^{5/3}\).

c) \(\sqrt[5]{b^2} = b^{2/5}\).

d) \(\dfrac{1}{\sqrt{z}} = \dfrac{1}{z^{1/2}} = z^{-1/2}\).

4 Evaluating rational exponents ★★★

a) \(9^{3/2} = (\sqrt{9})^3 = 3^3 = 27\).

b) \(8^{4/3} = (\sqrt[3]{8})^4 = 2^4 = 16\).

c) \(32^{2/5} = (\sqrt[5]{32})^2 = 2^2 = 4\).

d) \(100^{-1/2} = \dfrac{1}{\sqrt{100}} = \dfrac{1}{10}\).

5 Pulling out perfect squares ★★★

a) \(\sqrt{50} = \sqrt{25\cdot 2} = 5\sqrt{2}\).

b) \(\sqrt{48} = \sqrt{16\cdot 3} = 4\sqrt{3}\).

c) \(\sqrt{200} = \sqrt{100\cdot 2} = 10\sqrt{2}\).

d) \(\sqrt{75} = \sqrt{25\cdot 3} = 5\sqrt{3}\).

6 True or false? ★★★

a) False. For \(x = -3\), \(\sqrt{9} = 3\), not \(-3\). In general \(\sqrt{x^2} = |x|\).

b) True. \((-2)^3 = -8\).

c) False. The left side is \(3 + 4 = 7\) and the right side is \(5\).

d) False. \(16^{1/2} = \sqrt{16} = 4\); the exponent \(\tfrac{1}{2}\) means a square root, not half.

7 Where is it defined? ★★★

a) We need \(x - 5 \ge 0\), so \(x \ge 5\).

b) We need \(7 - 2x \ge 0\), so \(x \le \dfrac{7}{2}\).

c) The index is odd, so every real number works.

d) We need \(3x + 9 \ge 0\), so \(x \ge -3\).

8 How far is the horizon? ★★★

a) \(d = 1.22\sqrt{36} = 1.22 \times 6 = 7.32\) miles.

b) \(d = 1.22\sqrt{144} = 1.22 \times 12 = 14.64\) miles.

c) \(1.22\sqrt{h} = 12.2\), so \(\sqrt{h} = 10\) and \(h = 100\) ft (about 30 m).

9 Simplifying with variables ★★★

a) \(\sqrt{18x^5} = \sqrt{9\cdot 2\cdot x^4\cdot x} = 3x^2\sqrt{2x}\).

b) \(\sqrt[3]{40a^4} = \sqrt[3]{8\cdot 5\cdot a^3\cdot a} = 2a\sqrt[3]{5a}\).

c) \(\sqrt{12}\cdot\sqrt{27} = \sqrt{324} = 18\).

d) \(\sqrt{5}\cdot\sqrt{20} = \sqrt{100} = 10\).

10 Rationalizing denominators ★★★

a) \(\sqrt{8} = 2\sqrt{2}\), so \(\dfrac{6}{2\sqrt{2}} = \dfrac{3}{\sqrt{2}} = \dfrac{3\sqrt{2}}{2}\).

b) Multiply by the conjugate: \(\dfrac{4(\sqrt{7}+\sqrt{3})}{7-3} = \sqrt{7}+\sqrt{3}\).

c) Multiply top and bottom by \(\sqrt[3]{4}\), since \(\sqrt[3]{2}\cdot\sqrt[3]{4} = \sqrt[3]{8} = 2\): \(\dfrac{3\sqrt[3]{4}}{2}\).

11 Exponent rules ★★★

a) \(\dfrac{2}{3}+\dfrac{5}{6} = \dfrac{4}{6}+\dfrac{5}{6} = \dfrac{9}{6} = \dfrac{3}{2}\), so the result is \(x^{3/2}\).

b) Multiply each exponent by 6: \(x^{3}y^{-2} = \dfrac{x^3}{y^2}\).

c) \(\dfrac{3}{4}-\dfrac{1}{4} = \dfrac{1}{2}\), so \(x^{1/2}\).

d) \(\left(-\dfrac{2}{3}\right)\left(-\dfrac{3}{2}\right) = 1\), so the result is \(x\).

12 Negative rational exponents ★★★

a) \(\left(\dfrac{27}{8}\right)^{-2/3} = \left(\dfrac{8}{27}\right)^{2/3} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}\).

b) \(81^{-3/4} = \dfrac{1}{(\sqrt[4]{81})^3} = \dfrac{1}{27}\).

c) \(64^{5/6} = (\sqrt[6]{64})^5 = 2^5 = 32\).

13 A square root equation ★★★

Square both sides: \(3x - 5 = 16\), so \(3x = 21\) and \(x = 7\).

Check: \(\sqrt{3\cdot 7 - 5} = \sqrt{16} = 4\). The solution is \(x = 7\).

14 A cube root equation ★★★

Cube both sides: \(x + 4 = -8\), so \(x = -12\).

Check: \(\sqrt[3]{-12+4} = \sqrt[3]{-8} = -2\). The solution is \(x = -12\).

15 Isolate first ★★★

Isolate the radical: \(\sqrt{x-3} = 4\). Square: \(x - 3 = 16\), so \(x = 19\).

Check: \(\sqrt{16} + 5 = 9\). The solution is \(x = 19\).

16 Domain, range and table ★★★

a) We need \(x + 2 \ge 0\), so the domain is \(x \ge -2\). Since \(\sqrt{x+2} \ge 0\), the range is \(y \ge -3\).

b) \(f(-2) = 0 - 3 = -3\); \(f(-1) = 1 - 3 = -2\); \(f(2) = 2 - 3 = -1\); \(f(7) = 3 - 3 = 0\).

c) Shift the parent graph 2 units left and 3 units down. The starting point moves from \((0,0)\) to \((-2,-3)\).

17 An extraneous solution ★★★

Square both sides: \(2x + 15 = x^2\), so \(x^2 - 2x - 15 = 0\), i.e. \((x-5)(x+3) = 0\). Candidates: \(x = 5\) and \(x = -3\).

Check \(x = 5\): \(\sqrt{25} = 5\). True.

Check \(x = -3\): \(\sqrt{9} = 3 \ne -3\). False, so \(-3\) is extraneous.

The only solution is \(x = 5\).

18 Both candidates survive ★★★

Square: \(5x + 1 = x^2 + 2x + 1\), so \(x^2 - 3x = 0\), i.e. \(x(x-3) = 0\). Candidates: \(0\) and \(3\).

\(x = 0\): \(\sqrt{1} = 1\) and \(0 + 1 = 1\). True.

\(x = 3\): \(\sqrt{16} = 4\) and \(3 + 1 = 4\). True.

Both are solutions: there is no extraneous solution here, which is why the check is needed to know for sure.

19 Two radicals ★★★

Isolate one radical: \(\sqrt{x+6} = 1 + \sqrt{x-1}\).

Square both sides: \(x + 6 = 1 + 2\sqrt{x-1} + (x - 1) = x + 2\sqrt{x-1}\).

So \(6 = 2\sqrt{x-1}\), then \(\sqrt{x-1} = 3\) and \(x - 1 = 9\), giving \(x = 10\).

Check: \(\sqrt{16} - \sqrt{9} = 4 - 3 = 1\). The solution is \(x = 10\).

20 Equations with rational exponents ★★★

a) Raise both sides to the power \(\tfrac{3}{2}\): \(|x| = 25^{3/2} = 125\). So \(x = 125\) or \(x = -125\). Check: \((\sqrt[3]{-125})^2 = (-5)^2 = 25\). Both work.

b) Raise both sides to the power \(\tfrac{2}{3}\): \(x + 2 = 64^{2/3} = 16\), so \(x = 14\). Check: \(16^{3/2} = 4^3 = 64\).

21 Find the equation from the graph ★★★

a) The curve starts at \(P(-4,-3)\), so \(h = -4\) and \(k = -3\). Then \(R(0,-1)\) gives \(a\sqrt{4} - 3 = -1\), so \(2a = 2\) and \(a = 1\). Thus \(f(x) = \sqrt{x+4} - 3\). Check with \(Q\): \(\sqrt{1} - 3 = -2\); with \(S\): \(\sqrt{9} - 3 = 0\).

b) Domain \(x \ge -4\); range \(y \ge -3\).

c) \(\sqrt{x+4} - 3 = 1\), so \(\sqrt{x+4} = 4\), \(x + 4 = 16\) and \(x = 12\).