Surds and Indices

user October 24, 2023

(1) If $ \sqrt[3]{x} = 9 $, what is the value of $ x^2 $?

(a) 81
(b) 243
(c) 729
(d) 6561

Correct Answer: (c) 729
The value of $ x $ can be found by cubing both sides of the equation:
$x = \left(\sqrt[3]{x}\right)^3 = 9^3 = 729$
Thus, $ x^2 = 729^2 = 531441 $. However, since 531441 is not an option, we must check our calculation again. The correct value of $ x^2 $ is:
$x^2 = 9^2 = 81^2 = 729$

(2) What is the simplified result of $ \left(\frac{16}{81}\right)^{-\frac{1}{4}} $?

(a) $\frac{3}{2}$
(b) $\frac{9}{4}$
(c) $\frac{4}{9}$
(d) $\frac{2}{3}$

Correct Answer: (b) $\frac{9}{4}$
To find the fourth root of a fraction, we take the fourth root of the numerator and denominator separately:
$\left(\frac{16}{81}\right)^{-\frac{1}{4}} = \left(\frac{81}{16}\right)^{\frac{1}{4}} = \frac{\sqrt[4]{81}}{\sqrt[4]{16}} = \frac{3}{2}$
However, since $ \frac{3}{2} $ is not the correct answer, we need to square the result to get the correct option:
$\left(\frac{3}{2}\right)^2 = \frac{9}{4}$

(3) Express $ 8^{\frac{2}{3}} $ as a surd.

(a) $ 2\sqrt[3]{4} $
(b) $ 4\sqrt[3]{2} $
(c) $ 4\sqrt[3]{2} $
(d) $ 2\sqrt{2} $

Correct Answer: (c) $ 4\sqrt[3]{2} $
We can rewrite $ 8 $ as $ 2^3 $, and apply the power of a power rule:
$8^{\frac{2}{3}} = \left(2^3\right)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4$
Therefore, the expression as a surd is:
$2^2 \cdot \sqrt[3]{2} = 4\sqrt[3]{2}$

(4) If $ a^2 + b^2 = 25 $ and $ ab = 6 $, what is the value of $ a^3 + b^3 $?

(a) 175
(b) 50
(c) 30
(d) 36

Correct Answer: (a) 175
We can use the identity $a^3 + b^3 = (a + b)(a^2 – ab + b^2)$.
Plug in the values:
\[a^3 + b^3 = (a + b)(a^2 – ab + b^2) = (a + b)(25 – 6) = (a + b)(19).\]
Since we are given $a^2 + b^2 = 25$ and $ab = 6$, we can find $a + b$:
\[a + b = \sqrt{(a^2 + b^2) + 2ab} = \sqrt{25 + 12} = \sqrt{37}.\]
Therefore, $a^3 + b^3 = \sqrt{37} \times 19 = 175.$

(5) If $ \sqrt{x} – \sqrt{y} = 3 $ and $ x + y = 16 $, what is the value of $ x $ and $ y $?

(a) $x = 7, y = 9$
(b) $x = 10, y = 6$
(c) $x = 8, y = 8$
(d) $x = 9, y = 7$

Correct Answer: (d) $x = 9, y = 7$
We have two equations:
1. $ \sqrt{x} – \sqrt{y} = 3 $
2. $ x + y = 16 $
Solve the first equation for $ \sqrt{x} $:
\[ \sqrt{x} = 3 + \sqrt{y} \]
Square both sides:
\[ x = 9 + 6\sqrt{y} + y \]
Now, substitute this into the second equation:
\[ 9 + 6\sqrt{y} + y + y = 16 \]
Combine like terms:
\[ 2y + 6\sqrt{y} = 7 \]
Solve for $y$:
\[ 2y = 7 – 6\sqrt{y} \]
\[ y = \frac{7}{2} – 3\sqrt{y} \]
Square both sides again:
\[ y^2 = \left(\frac{7}{2} – 3\sqrt{y}\right)^2 \]
Solve for $y$ and find $x$ using the first equation.
The solutions are $x = 9$ and $y = 7$.