Simplifying Radical Expressions - Math 2

When $\sqrt[3]{54x^7y^{10}}$ is simplified, which form correctly represents the result?
Answer: $3x^2y^3\sqrt[3]{2xy}$ $\sqrt[3]{54x^7y^{10}} = \sqrt[3]{27 \cdot 2 \cdot x^6 \cdot x \cdot y^9 \cdot y} = \sqrt[3]{27} \cdot \sqrt[3]{x^6} \cdot \sqrt[3]{y^9} \cdot \sqrt[3]{2xy} = 3x^2y^3\sqrt[3]{2xy}$ . Choice A fails to factor out the perfect cube 27. Choice C incorrectly distributes the exponents. Choice D incorrectly factors 54 as having a factor of 6 cubed.
If $a = \sqrt{7 + 4\sqrt{3}}$ and $b = \sqrt{7 - 4\sqrt{3}}$ , what is the value of $a^2 + b^2$ ?
Answer: $14$ By definition, $a^2 = 7 + 4\sqrt{3}$ and $b^2 = 7 - 4\sqrt{3}$ . Therefore: $a^2 + b^2 = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 7 + 7 + 4\sqrt{3} - 4\sqrt{3} = 14$ . Choice A might result from incorrectly computing $7 + 7 = 10$ instead of $14$ . Choice B could come from errors in handling the $\sqrt{3}$ terms. Choice D might result from computing $a^2 + b^2 + 2ab$ instead of just $a^2 + b^2$ .
Which of the following expressions is NOT equivalent to $2\sqrt{50} - \sqrt{98} + \sqrt{18}$ ?
Answer: $2\sqrt{32} - \sqrt{50}$ First, simplify the original expression: $2\sqrt{50} - \sqrt{98} + \sqrt{18} = 2(5\sqrt{2}) - 7\sqrt{2} + 3\sqrt{2} = 10\sqrt{2} - 7\sqrt{2} + 3\sqrt{2} = 6\sqrt{2}$ . Choice A shows the intermediate step. Choice B is the simplified form. Choice C: $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$ , which is equivalent. Choice D: $2\sqrt{32} - \sqrt{50} = 2(4\sqrt{2}) - 5\sqrt{2} = 8\sqrt{2} - 5\sqrt{2} = 3\sqrt{2} \neq 6\sqrt{2}$ .
If $\sqrt{x^2 + 6x + 9} = x + 3$ , what constraint must be placed on $x$ ?
Answer: $x \geq -3$ Since $x^2 + 6x + 9 = (x + 3)^2$ , we have $\sqrt{(x + 3)^2} = |x + 3|$ . For this to equal $x + 3$ (without absolute value), we need $x + 3 \geq 0$ , which means $x \geq -3$ . Choice A is incorrect because when $x < -3$ , $\sqrt{(x + 3)^2} = -(x + 3) \neq x + 3$ . Choice C is too restrictive since $x = -3$ works. Choice D is incorrect because $x = -3$ actually satisfies the equation.
If $\sqrt{48x^3y^5} = 4xy^2\sqrt{3xy}$ , what must be true about the values of $x$ and $y$ ?
Answer: $x \geq 0$ and $y \geq 0$ When simplifying $\sqrt{48x^3y^5}$ , we get $\sqrt{16 \cdot 3 \cdot x^2 \cdot x \cdot y^4 \cdot y} = 4xy^2\sqrt{3xy}$ . For this to be valid as written (without absolute value signs), we need $x \geq 0$ so that $\sqrt{x^2} = x$ and $y \geq 0$ so that $\sqrt{y^4} = y^2$ (not $|y|^2$ ). Choice B is incorrect because $y$ cannot be negative without absolute value notation. Choice C is incorrect because $x$ cannot be negative. Choice D is incorrect because $y$ cannot be negative.
Which of the following expressions is equivalent to $\sqrt[6]{64x^{15}y^{18}}$ ?
Answer: $2x^2y^3\sqrt[6]{x^3}$ When you encounter radical expressions with variables, you need to systematically break down each component using the properties of exponents and radicals. The key insight is that $\sqrt[6]{a} = a^{\frac{1}{6}}$ , so you can rewrite the expression using fractional exponents and then simplify. Starting with $\sqrt[6]{64x^{15}y^{18}}$ , you can separate this into $\sqrt[6]{64} \cdot \sqrt[6]{x^{15}} \cdot \sqrt[6]{y^{18}}$ . For the numerical part: $\sqrt[6]{64} = \sqrt[6]{2^6} = 2$ For the y-term: $\sqrt[6]{y^{18}} = y^{\frac{18}{6}} = y^3$ For the x-term: $\sqrt[6]{x^{15}} = x^{\frac{15}{6}} = x^{\frac{5}{2}}$ . Since $\frac{15}{6} = 2\frac{3}{6} = 2\frac{1}{2}$ , you can write this as $x^{2 + \frac{1}{2}} = x^2 \cdot x^{\frac{1}{2}} = x^2 \cdot \sqrt[6]{x^3}$ (since $x^{\frac{1}{2}} = x^{\frac{3}{6}} = \sqrt[6]{x^3}$ ). This gives you $2x^2y^3\sqrt[6]{x^3}$ , which is choice C. Choice A incorrectly leaves the x-exponent as $\frac{5}{2}$ instead of separating it properly. Choice B has the wrong coefficient (4 instead of 2) and the wrong remaining radical. Choice D fails to simplify the x-exponent at all, leaving $x^{15}$ under the radical. Study tip: When simplifying radicals with large exponents, always divide the exponent by the radical's index, then express any remainder as a radical to ensure complete simplification.
If $x = \sqrt{2 + \sqrt{3}}$ and $y = \sqrt{2 - \sqrt{3}}$ , what is the value of $x^2y^2$ ?
Answer: $1$ When you encounter nested radicals like these, the key insight is to look for opportunities to simplify through multiplication, since the expressions often have complementary structures. Let's calculate $x^2y^2$ by first finding $x^2$ and $y^2$ separately: $x^2 = (\sqrt{2 + \sqrt{3}})^2 = 2 + \sqrt{3}$ $y^2 = (\sqrt{2 - \sqrt{3}})^2 = 2 - \sqrt{3}$ Now we can find $x^2y^2$ : $x^2y^2 = (2 + \sqrt{3})(2 - \sqrt{3})$ This is a difference of squares pattern: $(a + b)(a - b) = a^2 - b^2$ , where $a = 2$ and $b = \sqrt{3}$ : $x^2y^2 = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$ Looking at the wrong answers: Choice (A) gives 4, which you'd get if you mistakenly calculated just $2^2$ and forgot to subtract $(\sqrt{3})^2$ . Choice (B) gives 2, which might result from incorrectly simplifying the original expressions or making arithmetic errors. Choice (D) gives $\sqrt{3}$ , which could come from confusing this with $xy$ or making errors in the radical manipulation. The correct answer is (C) 1. Study tip: When you see complementary nested radicals (expressions that differ only in the sign between terms), immediately think about multiplying them together. The radical terms will often cancel out cleanly, leaving you with rational numbers that are much easier to work with.
What is the simplified form of $\sqrt{\frac{32x^7y^4}{8x^3y^2}}$ where $x > 0$ and $y > 0$ ?
Answer: $2x^2y$ First simplify the fraction under the radical: $\frac{32x^7y^4}{8x^3y^2} = \frac{32}{8} \cdot \frac{x^7}{x^3} \cdot \frac{y^4}{y^2} = 4x^4y^2$ . Then $\sqrt{4x^4y^2} = \sqrt{4} \cdot \sqrt{x^4} \cdot \sqrt{y^2} = 2x^2y$ . Choice B results from incorrectly simplifying $x^4$ as $x \cdot \sqrt{x}$ instead of $x^2$ . Choice C comes from not simplifying $\sqrt{4} = 2$ . Choice D incorrectly includes an extra $\sqrt{x}$ factor.
If $\sqrt{x + 12} + \sqrt{x} = 6$ and $x \geq 0$ , what is the value of $\sqrt{x + 12} - \sqrt{x}$ ?
Answer: $2$ Let $a = \sqrt{x + 12}$ and $b = \sqrt{x}$ . We're given that $a + b = 6$ and want to find $a - b$ . We can use the identity $(a + b)(a - b) = a^2 - b^2$ . Here, $a^2 - b^2 = (x + 12) - x = 12$ . So $(a + b)(a - b) = 12$ , which gives us $6(a - b) = 12$ , therefore $a - b = 2$ . Choice A (1) might result from incorrectly thinking $a - b = \frac{a^2 - b^2}{a + b} - 1$ . Choice C (3) could come from assuming $a - b = \frac{a + b}{2}$ . Choice D (4) might result from confusing this with $\frac{2(a^2 - b^2)}{a + b}$ .
If $\sqrt{32m^4n^6} - 2m^2n^3\sqrt{2} = km^2n^3\sqrt{2}$ , what is the value of $k$ ?
Answer: $k = 2$ First simplify $\sqrt{32m^4n^6} = \sqrt{16 \cdot 2 \cdot m^4 \cdot n^6} = 4m^2n^3\sqrt{2}$ . So the equation becomes $4m^2n^3\sqrt{2} - 2m^2n^3\sqrt{2} = km^2n^3\sqrt{2}$ . This simplifies to $2m^2n^3\sqrt{2} = km^2n^3\sqrt{2}$ , so $k = 2$ . Choice B would result from incorrectly not subtracting. Choice C would result from adding instead of subtracting. Choice D would result from incorrectly assuming the terms cancel completely.
If $\sqrt{5 + 2\sqrt{6}} = \sqrt{a} + \sqrt{b}$ where $a$ and $b$ are positive integers, what is the value of $ab$ ?
Answer: $6$ When you see a nested radical like $\sqrt{5 + 2\sqrt{6}}$ that needs to be written as $\sqrt{a} + \sqrt{b}$ , you're dealing with a denesting problem. The key insight is to assume the form works and use algebra to find the values. Start by squaring both sides: $5 + 2\sqrt{6} = (\sqrt{a} + \sqrt{b})^2$ Expanding the right side: $5 + 2\sqrt{6} = a + b + 2\sqrt{ab}$ For this equation to hold, the rational and irrational parts must match separately: - Rational parts: $a + b = 5$ - Irrational parts: $2\sqrt{ab} = 2\sqrt{6}$ , so $ab = 6$ Now you have a system: $a + b = 5$ and $ab = 6$ . These are the sum and product of two numbers, so you can find them using the quadratic whose roots are $a$ and $b$ : $x^2 - 5x + 6 = 0$ Factoring: $(x-2)(x-3) = 0$ , giving $a = 2$ and $b = 3$ (or vice versa). Therefore $ab = 6$ , which is choice D. Choice A (15) would come from incorrectly multiplying $a + b = 5$ by $3$ . Choice B (8) might result from computational errors when solving the quadratic. Choice C (12) could arise from doubling the correct answer or mishandling the coefficient $2$ in $2\sqrt{6}$ . Strategy tip: When denesting radicals, always square both sides and match rational/irrational parts separately. The resulting system of equations for sum and product leads directly to a quadratic equation.
Which of the following is equivalent to $\frac{2\sqrt{12} + \sqrt{27}}{\sqrt{3}}$ ?
Answer: $7$ When you encounter radical expressions in fractions, your goal is to simplify by factoring out perfect squares and combining like terms. This question tests your ability to simplify radicals and perform arithmetic with them. Start by simplifying each radical in the numerator. For $\sqrt{12}$ , factor out the perfect square: $12 = 4 \times 3$ , so $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ . For $\sqrt{27}$ , factor similarly: $27 = 9 \times 3$ , so $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ . Now substitute these simplified forms: $\frac{2\sqrt{12} + \sqrt{27}}{\sqrt{3}} = \frac{2(2\sqrt{3}) + 3\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3} + 3\sqrt{3}}{\sqrt{3}}$ Combine like terms in the numerator: $\frac{7\sqrt{3}}{\sqrt{3}}$ Since $\frac{\sqrt{3}}{\sqrt{3}} = 1$ , you get $7 \times 1 = 7$ . Choice A ( $9$ ) likely results from incorrectly adding coefficients without proper simplification. Choice B ( $5\sqrt{3}$ ) suggests you might have made an error when combining the coefficients 4 and 3, or forgotten to divide by $\sqrt{3}$ . Choice C ( $7\sqrt{3}$ ) is what you'd get if you correctly found the numerator but forgot the final division step. Study tip: When simplifying radical fractions, always factor out perfect squares first, then combine like terms, and finally perform the division. Remember that $\frac{\sqrt{a}}{\sqrt{a}} = 1$ for any positive value $a$ .
Which expression is equivalent to $\sqrt{50} - \sqrt{18} + \sqrt{8}$ ?
Answer: $4\sqrt{2}$ Simplify each radical separately: $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$ , $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$ , and $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$ . Therefore: $\sqrt{50} - \sqrt{18} + \sqrt{8} = 5\sqrt{2} - 3\sqrt{2} + 2\sqrt{2} = (5 - 3 + 2)\sqrt{2} = 4\sqrt{2}$ . Choice B results from the error $5 - 3 + 2 = 5$ instead of $4$ . Choice C comes from adding all coefficients: $5 + 3 + 2 = 10$ but miscalculating. Choice D results from incorrectly simplifying one of the radicals.

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