–1

Take the square root of both sides to begin solving this polynomial. So, 2x + 2 = 0. Solve for x and enter in –1 as the final answer.

This is one of the common quadratic equations: (3 – 2)(3 + 2) = 3² – 2².The answer is choice (C). If you don’t recognize the common quadratic, you can just do the arithmetic and discover that 9 – 4 = (1)(5).

Try FOILing. For Quantity A, you get 3a² + 6a + 6a + 12, or 3a2 + 12a + 12. For Quantity B, you get 3a² + 18a + 2a + 12, or 3a² + 20a + 12.Remember to compare, not calculate. Notice that the only difference between the quantities is that between 20a and 12a.Because a is positive, 20a must be greater than 12a, thus, Quantity B will always be greater.

Evaluate the relationship between the quantities by plugging in values for p:Try p = 2. Quantity A is 7 × 5 = 35, and Quantity B is 9 × 4 = 36; Quantity B is greater, so eliminate choices (A) and (C). Any value gives the same outcome, so select answer choice (B). Algebraically, you could either FOIL Quantity A or recognize the common quadratics—either way, Quantity A simplifies to 9p² – 1, which is always exactly 1 less than Quantity B.

135

First solve for x: Add 2 to both sides to yield x^–2 = 0.04, so \(\frac{1}{x^{2}}=\frac{1}{25}\), and x =±5. Next, add 5 to both sides of the other equation to yield y^1/3 = 3, so \(\sqrt[3]{y}\) = 3, and y = 27. Since y is positive, you’ll need to use the positive value of x as well, so 5 × 27 = 135.

A,B, and C

The lower boundary for xy is –12 × 6, or –72, and the upper boundary is –2 × 3 = –6. Any values between –72 and –6 work.Be careful about choice (D):The value has to be greater than –6, so –6 itself doesn’t count.

35

Wouldn’t this be a lot easier if you knew how many students there are? The question doesn’t say, so plug in your own number.The question deals in percentages, so let’s say there are 100 students in the class.That means that there are 80 boys and 20 girls, and 50 students who are 7 and 50 students who are 8. 25 percent of 20 is 5, which leaves 15 girls who are 7. Subtract 15 from 50 and you are left with 35 boys who are 7.You can also employ the group grid.

C

and E Try plugging in values for m and n to the inequality in the question. If m = 5 and n = 1, both numbers are integers, so choices (A) and (B) work as do all of the other choices.But what if m and n are negative decimals, say –2.3 and –6.3? Then choices (A) and (B) don’t work, since neither number is an integer; eliminate them.Choice (C) still works.Choice (D) is out, since the sum of the two numbers is negative, but the difference is positive, leaving choice (E) in.Choices (F) and (G) also still work. However, by switching m to a positive value, such as 5, you can knock out choices (F) and (G). Only choices (C) and (E) will work no matter what numbers you plug in.

12

You are being asked to subtract the a and b terms.Be careful that you don’t just combine the largest value of a with the largest value of b to get a – b = 9 – 2 = 7, the wrong answer. When you combine inequalities, you have to make four calculations to check the four possibilities. Subtract the smallest values of a and b: –4 – (–3) = –1. Subtract the largest values of a and b: 9 – 2 = 7. Subtract the smallest value of a and the largest value of b: –4 – 2 = 6. Subtract the largest value of a and the smallest value of b: 9 – (–3) = 12. Of the four possible values above, the greatest possible value is 12.

By rearranging the first two equations, you obtain c = 7 – ab and b = 5 − ac.By substituting these relationships into the third equation, a + (5 − ac) + (7 − ab) = 6 is obtained. Manipulating the equation yields a − ac − ab + 12 = 6, or a − a (c + b) + 6 = 0. Since a + b + c = 6, b + c = 6 − a. Substituting this into the previous equation yields a − a (6 − a) + 6 = a − 6a + a² + 6 = 0, or a² − 5a + 6 = 0. Factoring and finding the roots gives you a = 2 or a = 3. If a = 2—by substituting the value backinto the equations—then b = 2 and c = 1. If a = 3, then b = 3 and c = 1.You are looking for abc, and either way, the value is 3 × 2 × 1 = 6.The answer is choice (B).