Radicals and Rational Exponents: math test solutions, Grade 11 – download the PDF

1 Roots and exponents / 3 pts

a) \(13\) (0.5 pt). b) \(-6\) (0.5 pt). c) \(\sqrt[4]{625} = 5\) (0.5 pt). d) \(7^3 = 343\) (0.5 pt). e) \(\dfrac{1}{2^2} = \dfrac{1}{4}\) (0.5 pt). f) \(\left(\dfrac{2}{3}\right)^3 = \dfrac{8}{27}\) (0.5 pt).

2 Simplifying radicals / 4 pts

a) \(\sqrt{49\cdot 2} = 7\sqrt{2}\) (1 pt).

b) \(\sqrt[3]{27\cdot 2\cdot x^3\cdot x} = 3x\sqrt[3]{2x}\) (1.5 pt: 0.5 factoring, 1 result).

c) \(\dfrac{10\sqrt{5}}{5} = 2\sqrt{5}\) (0.5 pt).

d) \(\dfrac{5(3+\sqrt{2})}{9-2} = \dfrac{5(3+\sqrt{2})}{7}\) (1 pt).

3 Exponent rules / 3 pts

a) \(\dfrac{5}{6}+\dfrac{2}{6} = \dfrac{7}{6}\), so \(x^{7/6}\) (1 pt).

b) \(8^{2/3}\cdot a^{4} = 4a^4\) (1 pt).

c) \(x^{-2}y^{3} = \dfrac{y^3}{x^2}\) (1 pt).

4 Solving equations / 4 pts

a) \(4x + 1 = 49\), so \(x = 12\); check \(\sqrt{49} = 7\) (1.5 pt).

b) \(\sqrt[3]{2x-5} = 3\), so \(2x - 5 = 27\) and \(x = 16\); check \(\sqrt[3]{27}+1 = 4\) (1.5 pt).

c) \(x - 3 = 125^{2/3} = 25\), so \(x = 28\); check \(25^{3/2} = 125\) (1 pt).

5 Extraneous solution / 3 pts

Square both sides: \(3x + 7 = x^2 + 2x + 1\) (1 pt). Then \(x^2 - x - 6 = 0\), so \((x-3)(x+2) = 0\) and the candidates are \(3\) and \(-2\) (1 pt).

Check \(x = 3\): \(\sqrt{16} = 4 = 3+1\), valid. Check \(x = -2\): \(\sqrt{1} = 1 \ne -1\), extraneous. The solution is \(x = 3\) (1 pt).

6 A radical function / 3 pts

a) \(x + 3 \ge 0\) gives domain \(x \ge -3\) (0.5 pt); range \(y \ge -4\) (0.5 pt).

b) \(f(-3) = -4\), \(f(-2) = 2-4 = -2\), \(f(1) = 4-4 = 0\), \(f(6) = 6-4 = 2\) (1 pt).

c) \(f(x) = 0\) gives \(\sqrt{x+3} = 2\), so \(x = 1\): the \(x\)-intercept is \((1, 0)\) (0.5 pt). \(f(0) = 2\sqrt{3} - 4 \approx -0.54\): the \(y\)-intercept is \((0, 2\sqrt{3}-4)\) (0.5 pt).