C and D

Try plugging in each of the answers rather than solving the quadratic:

x = -√2 \((-\sqrt{2})^{2}-(-\sqrt{2})\sqrt{2}+3(-\sqrt{2})\sqrt{3}=2+2-3\sqrt{6}\neq \sqrt{54}\)

x = -3√2 \((-3\sqrt{2})^{2}-(-3\sqrt{2})\sqrt{2}+3(-3\sqrt{2})\sqrt{3}=9\times 2+3\times 2-9\sqrt{6}=24-9\sqrt{6}\neq \sqrt{54}\)

x = -3√3 \((-3\sqrt{3})^{2}-(-3\sqrt{3})\sqrt{2}+3(-3\sqrt{3})\sqrt{3}=9\times 3+3\sqrt{6}-9\times 3\neq \sqrt{54}\)

x = √2 \((\sqrt{2})^{2}-(\sqrt{2})\sqrt{2}+3(\sqrt{2})\sqrt{3}=2-2+3\sqrt{6}=\sqrt{54}\)

This problem is simplified when you recognize that this is actually a common quadratic equation of the formula x² – y² = x² – 2xy + y².Therefore \(\left ( \sqrt{245}-\sqrt{75} \right )^{2}=245-2\sqrt{(245)(\sqrt{75})+75}\) simplifies to 320 – 2 Rather than multiplying out, check to see if these large numbers simplify to the multiples of perfect squares. As is usually the case on the GRE, they do. \(\sqrt{345}=\sqrt{5\times 49}=7\sqrt{5}\) and \(\sqrt{75}=\sqrt{3\times 25}=5\sqrt{3}\), so \(2(\sqrt{245})(\sqrt{75})=2(7\sqrt{5})(5\sqrt{3})=70(\sqrt{15})\).Therefore,

Rewrite 72⁴ in terms of prime factors. 72⁴ = (9 × 8)⁴ = (3² × 2³)⁴ .You can now distribute the 4 to get 3⁸ × 2¹². Now break down the right side of the equation. 16 = 2⁴, so you can rewrite the right side as 2⁴ × 6ⁿ or 2⁴ × (2 × 3)ⁿ.The whole equation is therefore 3⁸ × 2¹² = 2⁴ × 2ⁿ × 3ⁿ.That means that 3⁸ = 3ⁿ and 2² = 24 × 3ⁿ, so n must equal 8.

¹⁄₂

When working with exponents, everything must have the same base.Express 9 as 3³. Now the given expression is (3²) = 3^{2y + 5}. When raising a power to another power, you multiply the exponents.This gives you 3⁶ = 3^{2y + 5}.The bases are the same, so now you can set the exponents equal to each other and solve for y: 6 = 2y + 5.The correct answer is ¹⁄₂.

If you’re extremely comfortable working with exponents, start by converting everything to the same base so you can use the basic exponent rules:\(\frac{8^{r}}{4^{s}}=\frac{(2^{3})^{r}}{(2^{2})^{s}}=\frac{2^{3r}}{2^{2s}}=2^{3r-2}\).Thus 2^{3r – 2s} = 2^t, and 3r – 2s = t; solve for r, and r = \(\frac{2s+r}{3}\). Alternately, you could dispense with all the algebra and Plug In numbers to make the equation true: If r = 2 and s = 3, for instance,\(\frac{64}{64}\) = 2^t, so t = 0. Plug your values for s and t into the answers, and only choice (C) hits your target answer of 2.

³⁄₄ or any equivalent value, such as \(\frac{6}{8},\frac{9}{12},\frac{12}{16}\) etc.

Remember that a negative exponent means “positive reciprocal,” and thus 2^-5 = \(\frac{1}{2^{5}}\). A fractional exponent asks you to find the root,so4^1⁄₂ = √2 = 2.Therefore, the initial equation can be rewritten as \(\frac{x}{12}=\frac{1}{2^{5}}\times 2\) or \(\frac{x}{12}=\frac{1}{2^{5}}\times 2\). Simplifying again,\(\frac{x}{12}=\frac{1}{16}\).

Multiply both sides by 12 to find that x = \(\frac{12}{16}=\frac{3}{4}\), the final answer.

24

Don’t calculate that big number under the radical, since your on-screen calculator won’t do cube roots anyway. Instead, remember that \(\sqrt[3]{8\times 27\times 64}=\sqrt[3]{8}\times \sqrt[3]{27}\times \sqrt[3]{64}=2\times 3\times 4\), so the answer is 24.

From the restriction in this problem, x could equal a positive fraction, 0, or any negative number. If you raise 1 to any power, it remains equal to 1.This also applies if x equals 0, because any number raised to the 0 power equals.1.Therefore, only possible correct answer is choice (D).

The only thing to worry about here is not getting caught in the trap answer choice (D).Remember that you cannot take the roots before you add up what’s under the root sign. If you add first, you’ll find that you’re looking for \(\sqrt{150}\). 150 has factors of 25 and 6. You can take the square root of 25 (which is 5), but not of 6, so leave it under the root sign. Select choice (B).

If (x + y)(x − y) = 0 and xy ≠ 0, then either x + y = 0 or x – y = 0; hence, x = y or x = –y. Fortunately, the variables will be positive and will work out the same either way. Plug in values for x and y to simplify the comparison:Try making both x and y equal to 2. Now Quantity A is \(\sqrt[6]{\frac{19}{2(2)^{2}}}\), or \(\sqrt[6]{\frac{19}{2^{3}}}\); Quantity B is \(\sqrt{\frac{342}{4}}\). At this point, manipulate Quantity B to make it look like Quantity A. Since Quantity A contains 19, test 19 as a factor of 342 in Quantity B:\(\sqrt{\frac{342}{4}}\); multiplying by ²⁄₂ under the radical yields \(\sqrt{\frac{19\times 9\times 2\times 2}{2^{2}\times 2}}\), or \(\sqrt{\frac{19\times 9\times 2\times 2}{2^{2}\times 2}}\). Moving the perfect square 36 outside the radical yields \(\sqrt{\frac{19\times 9\times 2\times 2}{2^{2}\times 2}}\).