Radical Operations - Accuplacer Aaf

1. Which expression is equivalent to $\sqrt{x^2 + 6x + 9}$ for $x < -3$?
   Answer: $-x-3$ The expression under the radical, $x^2 + 6x + 9$, is a perfect square trinomial, which factors to $(x+3)^2$. Therefore, $\sqrt{x^2 + 6x + 9} = \sqrt{(x+3)^2}$. The square root of a squared expression is its absolute value: $|x+3|$. Since the problem states that $x < -3$, the value of $x+3$ will be negative. The absolute value of a negative number is its opposite. Therefore, $|x+3| = -(x+3) = -x-3$.
2. The legs of a right triangle have lengths $\sqrt{5}$ inches and $2\sqrt{3}$ inches. What is the perimeter of the triangle?
   Answer: $\sqrt{5} + 2\sqrt{3} + \sqrt{17}$ inches First, find the length of the hypotenuse, c, using the Pythagorean theorem, $a^2 + b^2 = c^2$. Let $a = \sqrt{5}$ and $b = 2\sqrt{3}$. Then $c^2 = (\sqrt{5})^2 + (2\sqrt{3})^2 = 5 + (4 \cdot 3) = 5 + 12 = 17$. So, $c = \sqrt{17}$. The perimeter is the sum of the lengths of the three sides: $\sqrt{5} + 2\sqrt{3} + \sqrt{17}$. These radicals cannot be simplified or combined further.
3. If the area of a rectangle is 7 and its width is $3 - \sqrt{2}$, what is its length?
   Answer: $3 + \sqrt{2}$ Area of a rectangle is length times width (A = l * w). To find the length, divide the area by the width: $l = \frac{A}{w} = \frac{7}{3 - \sqrt{2}}$. To simplify, rationalize the denominator by multiplying by the conjugate, $3 + \sqrt{2}$: $\frac{7}{3 - \sqrt{2}} \cdot \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{7(3 + \sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{21 + 7\sqrt{2}}{9 - 2} = \frac{21 + 7\sqrt{2}}{7}$. Dividing each term in the numerator by 7 gives $3 + \sqrt{2}$.
4. For $x > 0$, which expression is equivalent to $x^{3/2} (x^{1/2} + 2x^{-1/2})$?
   Answer: $x^2 + 2x$ Distribute $x^{3/2}$ to both terms inside the parentheses. The first term becomes $x^{3/2} \cdot x^{1/2} = x^{3/2 + 1/2} = x^{4/2} = x^2$. The second term becomes $x^{3/2} \cdot 2x^{-1/2} = 2x^{3/2 - 1/2} = 2x^{2/2} = 2x^1 = 2x$. The sum is $x^2 + 2x$.
5. For $x > 0$ and $y > 0$, what is the product of $\sqrt{18xy^2}$ and $\sqrt{6x^3y}$ in simplest form?
   Answer: $6x^2y\sqrt{3y}$ First, multiply the expressions under the radical: $\sqrt{(18xy^2)(6x^3y)} = \sqrt{108x^4y^3}$. Now, simplify the resulting radical. The number 108 can be factored as $36 \cdot 3$. The variable parts are $x^4$ and $y^3 = y^2 \cdot y$. So, $\sqrt{36 \cdot 3 \cdot x^4 \cdot y^2 \cdot y} = \sqrt{36}\sqrt{x^4}\sqrt{y^2}\sqrt{3y} = 6x^2y\sqrt{3y}$.
6. Which of the following integers is the value of $(\sqrt{50} - \sqrt{2})^2$?
   Answer: 32 First, simplify the expression inside the parentheses. $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$. The expression becomes $(5\sqrt{2} - \sqrt{2})^2$. Combining the like terms gives $(4\sqrt{2})^2$. Squaring this expression results in $4^2 \cdot (\sqrt{2})^2 = 16 \cdot 2 = 32$.
7. Which of the following expressions is equivalent to $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$?
   Answer: $5 - 2\sqrt{6}$ To simplify, rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - \sqrt{2}$. Numerator: $(\sqrt{3} - \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}$. Denominator: $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$. The expression simplifies to $\frac{5 - 2\sqrt{6}}{1} = 5 - 2\sqrt{6}$.
8. Which expression is equivalent to $\sqrt{50} + \sqrt{8} - \sqrt{18}$?
   Answer: $4\sqrt{2}$ First, simplify each radical by factoring out perfect squares. $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$. $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$. $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$. Now substitute: $\sqrt{50} + \sqrt{8} - \sqrt{18} = 5\sqrt{2} + 2\sqrt{2} - 3\sqrt{2} = (5 + 2 - 3)\sqrt{2} = 4\sqrt{2}$.
9. Which of the following is equivalent to $\frac{4}{2 - \sqrt{6}}$?
   Answer: $-4 - 2\sqrt{6}$ To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $2 + \sqrt{6}$. This gives $\frac{4}{2 - \sqrt{6}} \cdot \frac{2 + \sqrt{6}}{2 + \sqrt{6}} = \frac{4(2 + \sqrt{6})}{(2)^2 - (\sqrt{6})^2} = \frac{8 + 4\sqrt{6}}{4 - 6} = \frac{8 + 4\sqrt{6}}{-2}$. Dividing each term in the numerator by -2 gives $\frac{8}{-2} + \frac{4\sqrt{6}}{-2} = -4 - 2\sqrt{6}$.
10. Which of the following values is the greatest?
    Answer: $5\sqrt{2}$ To compare the values, it is helpful to square each one to remove the radical. $(2\sqrt{11})^2 = 4 \cdot 11 = 44$. $(3\sqrt{5})^2 = 9 \cdot 5 = 45$. $(4\sqrt{3})^2 = 16 \cdot 3 = 48$. $(5\sqrt{2})^2 = 25 \cdot 2 = 50$. Since 50 is the largest squared value, $5\sqrt{2}$ is the greatest of the original values.
11. Which of the following is equivalent to $(2\sqrt{3} - \sqrt{5})(\sqrt{3} + 3\sqrt{5})$?
    Answer: $-9 + 5\sqrt{15}$ Use the FOIL method to multiply the binomials. First: $(2\sqrt{3})(\sqrt{3}) = 2 \cdot 3 = 6$. Outer: $(2\sqrt{3})(3\sqrt{5}) = 6\sqrt{15}$. Inner: $(-\sqrt{5})(\sqrt{3}) = -\sqrt{15}$. Last: $(-\sqrt{5})(3\sqrt{5}) = -3 \cdot 5 = -15$. Combine the terms: $6 + 6\sqrt{15} - \sqrt{15} - 15 = (6 - 15) + (6\sqrt{15} - \sqrt{15}) = -9 + 5\sqrt{15}$.
12. If $\sqrt{x + 12} = \sqrt{3x - 4}$, what is the value of $x$?
    Answer: $8$ Since both sides are equal and both are square roots, we can square both sides: $(\sqrt{x + 12})^2 = (\sqrt{3x - 4})^2$, which gives $x + 12 = 3x - 4$. Solving for x: $x + 12 = 3x - 4$, so $12 + 4 = 3x - x$, which gives $16 = 2x$, so $x = 8$. We must check this solution in the original equation: $\sqrt{8 + 12} = \sqrt{20}$ and $\sqrt{3(8) - 4} = \sqrt{24 - 4} = \sqrt{20}$. Both sides equal $\sqrt{20}$, so $x = 8$ is valid. Choice A gives $\sqrt{16} = 4$ and $\sqrt{8} = 2\sqrt{2} \approx 2.83$, which are not equal. Choice B gives $\sqrt{18} = 3\sqrt{2} \approx 4.24$ and $\sqrt{14} \approx 3.74$, which are not equal. Choice D gives $\sqrt{24} = 2\sqrt{6} \approx 4.90$ and $\sqrt{32} = 4\sqrt{2} \approx 5.66$, which are not equal.
13. The expression $\sqrt{54 + \sqrt{121}}$ simplifies to which of the following?
    Answer: $\sqrt{65}$ According to the order of operations, simplify the innermost radical first. $\sqrt{121} = 11$. The expression becomes $\sqrt{54 + 11}$. Adding the numbers inside the radical gives $\sqrt{65}$. Since 65 has no perfect square factors (its factors are 5 and 13), the radical cannot be simplified further.
14. Which of the following is equivalent to $\frac{\sqrt{75}}{3} - \sqrt{48}$? Give your answer in simplest form.
    Answer: $-\frac{7\sqrt{3}}{3}$ First, simplify both radical terms. $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$. So the first term is $\frac{5\sqrt{3}}{3}$. The second term is $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$. The expression is now $\frac{5\sqrt{3}}{3} - 4\sqrt{3}$. To subtract, find a common denominator: $\frac{5\sqrt{3}}{3} - \frac{12\sqrt{3}}{3} = \frac{5\sqrt{3} - 12\sqrt{3}}{3} = \frac{-7\sqrt{3}}{3}$.
15. If $\sqrt{2x - 1} + 3 = 7$, what is the value of $x$?
    Answer: $\frac{17}{2}$ When you encounter an equation with a square root, your goal is to isolate the radical term first, then eliminate it by squaring both sides. Starting with $\sqrt{2x - 1} + 3 = 7$, subtract 3 from both sides to isolate the square root: $\sqrt{2x - 1} = 4$. Now square both sides to eliminate the radical: $(\sqrt{2x - 1})^2 = 4^2$, which gives you $2x - 1 = 16$. Add 1 to both sides: $2x = 17$. Finally, divide by 2: $x = \frac{17}{2}$. Let's verify: substituting $x = \frac{17}{2}$ back into the original equation gives $\sqrt{2 \cdot \frac{17}{2} - 1} + 3 = \sqrt{17 - 1} + 3 = \sqrt{16} + 3 = 4 + 3 = 7$ ✓ Choice A ($9$) comes from solving $2x - 1 = 17$ instead of $2x - 1 = 16$—a calculation error when squaring 4. Choice B ($\frac{15}{2}$) results from forgetting to add 1 back when solving $2x = 16$, giving $x = 8$, then mistakenly writing it as a fraction. Choice C ($8$) comes from correctly getting $2x = 16$ but then dividing incorrectly or from solving $2x - 1 = 15$. Always check your solution in radical equations by substituting back into the original equation. This catches both arithmetic errors and extraneous solutions that can arise when squaring both sides of an equation.
16. Which of the following is equivalent to $\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{2+\sqrt{3}}$?
    Answer: 1 This is a telescoping sum. Rationalize each term separately. $\frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2} = \frac{\sqrt{2}-1}{1} = \sqrt{2}-1$. $\frac{1}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} = \frac{\sqrt{3}-\sqrt{2}}{1} = \sqrt{3}-\sqrt{2}$. $\frac{1}{2+\sqrt{3}} = \frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3}$. Now add the simplified terms: $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (2-\sqrt{3})$. The terms $\sqrt{2}$ and $-\sqrt{2}$ cancel, and the terms $\sqrt{3}$ and $-\sqrt{3}$ cancel, leaving $-1 + 2 = 1$.
17. For $x > 0$, which of the following is equivalent to $\sqrt{48x^3} + x\sqrt{12x}$?
    Answer: $6x\sqrt{3x}$ First, simplify each radical term. For the first term, $\sqrt{48x^3} = \sqrt{16 \cdot 3 \cdot x^2 \cdot x} = 4x\sqrt{3x}$. For the second term, $x\sqrt{12x} = x\sqrt{4 \cdot 3x} = x \cdot 2\sqrt{3x} = 2x\sqrt{3x}$. Now, add the like terms: $4x\sqrt{3x} + 2x\sqrt{3x} = 6x\sqrt{3x}$.
18. Simplify $\sqrt{12x^2y} \cdot \sqrt{18xy^3}$ where $x \geq 0$ and $y \geq 0$.
    Answer: $6xy^2\sqrt{6x}$ Use the property $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$: $\sqrt{12x^2y} \cdot \sqrt{18xy^3} = \sqrt{12x^2y \cdot 18xy^3} = \sqrt{216x^3y^4}$. Now factor out perfect squares: $216 = 36 \cdot 6$, $x^3 = x^2 \cdot x$, and $y^4 = (y^2)^2$. So $\sqrt{216x^3y^4} = \sqrt{36x^2y^4 \cdot 6x} = \sqrt{36x^2y^4} \cdot \sqrt{6x} = 6xy^2\sqrt{6x}$. Choice B incorrectly has $x^2$ outside the radical, which would require $x^4$ under the radical originally. Choice C has the wrong coefficient (18 instead of 6). Choice D has an even larger wrong coefficient (36 instead of 6).
19. Which of the following is the simplified form of $(3\sqrt{2} - 2\sqrt{3})^2$?
    Answer: $30 - 12\sqrt{6}$ To square the binomial, use the formula $(a-b)^2 = a^2 - 2ab + b^2$. Here, $a = 3\sqrt{2}$ and $b = 2\sqrt{3}$. So, $a^2 = (3\sqrt{2})^2 = 9 \cdot 2 = 18$. $b^2 = (2\sqrt{3})^2 = 4 \cdot 3 = 12$. The middle term is $-2ab = -2(3\sqrt{2})(2\sqrt{3}) = -12\sqrt{6}$. Combining these gives $18 - 12\sqrt{6} + 12 = 30 - 12\sqrt{6}$.
20. Which of the following is equivalent to $\frac{4}{\sqrt{8}}$?
    Answer: $\sqrt{2}$ First, simplify the denominator: $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$. So $\frac{4}{\sqrt{8}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}}$. To rationalize the denominator, multiply both numerator and denominator by $\sqrt{2}$: $\frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$. Choice B would result from stopping at $\frac{2\sqrt{2}}{2}$ without simplifying to $\sqrt{2}$. Choice C would result from incorrectly rationalizing $\frac{1}{\sqrt{2}}$ instead of $\frac{2}{\sqrt{2}}$. Choice D appears to involve an incorrect denominator manipulation.

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