Equations / Inequalities - GRE Quantitative Reasoning

The quantity can be regrouped to be –3 = a – b. Thus, (a – b)2 = (–3)2 = 9.

To solve for x:

bx + c = e – ax

bx + ax = e – c

x(b+a) = e-c

x = (e-c) / (b+a)

Start by setting up an equation using Quantity A and Quantity B. In other words, you can solve an inequality where Quantity A > Quantity B. You would have one of four outcomes:

1. Quantity A = Quantity B: the two quantities are equal.
2. The inequality is always satisfied: Quantity A is always larger.
3. The inequality is never satisfied (but the two are unequal): Quantity B is always larger.
4. The inequality is not always correct or incorrect: the relationship cannot be determined.

So solve:

–5x + 4 > 8 – 2x (Quantity A > Quantity B)

+2x +2x

–3x + 4 > 8

–4 –4

–3x > 4 or x < –4/3

*remember to switch the direction of the inequality when you divide by a negative number

As the inequality \[x < –4/3\] is always false for \[x>0\], Quantity B is always larger.

First we need the area or the rectangle. 24 * 8 = 192. So now we know that 10 gallons will cover 75 ft2 and x gallons will cover 192 ft2. We set up a simple ratio and cross multiply to find that 75_x_ = 1920.

x = 25.6

First, solve for x:

11 + 3_x_ = 29

29 – 11 = 3_x_

18 = 3_x_

x = 6

Then, solve for 2_x_:

2_x_ = 2 * 6 = 12

For this type of problem, we use the formula:

When Karen catches up with Jen, their distances are equivalent. Thus,

We then make a variable for Jen's time, . Thus we know that Karen's time is (since we are working in hours).

Thus,

There's a logical shortcut you can use on "catching up" distance/rate problems. The difference between the faster (Karen at 60mph) and slower (Jen at 45mph) drivers is 15mph. Which means that every one hour, the faster driver, Karen, gains 15 miles on Jen. We know that Jen gets a 1/2 hour head start, which at 45mph means that she's 22.5 miles ahead when Karen gets started. So we can calculate the number of hours (H) of the 15mph of Karen's "catchup speed" (the difference between their speeds) it will take to make up the 22.5 mile gap:

15H = 22.5

So H = 1.5.

Given a quadratic equation equal to zero you can factor the equation and set each factor equal to zero. To factor you have to find two numbers that multiply to make 9 and add to make 6. The number is 3. So the factored form of the problem is (x+3)(x+3)=0. This statement is true only when x+3=0. Solving for x gives x=-3. Since this problem is multiple choice, you could also plug the given answers into the equation to see which one works.

x2 – x = 72. Solve for x using the quadratic formula and x = 9 and –8. Only 9 satisfies the restrictions.

Solve by factoring. First get everything into the form Ax2 + Bx + C = 0:

x2 + 4x - 5 = 0

Then factor: (x + 5) (x - 1) = 0

Solve each multiple separately for 0:

X + 5 = 0; x = -5

x - 1 = 0; x = 1

Therefore, x is either -5 or 1

There are two values where _x_2 = x, namely x = 0 and x = 1. All values between 0 and 1 get smaller after squaring. All values greater than 1 get larger after squaring.

The equation in the problem is quadratic, so we can use the quadratic formula to solve it. If an equation is in the form _ax_2 + bx + c = 0, where a, b, and c are constants, then the quadratic formula, given below, gives us the solutions of x.

In this particular problem, a = 2, b = –6, and c = 3.

The value under the square-root, b_2 – 4_ac, is called the discriminant, and it gives us important information about the nature of the solutions of a quadratic equation.

If the discriminant is less than zero, then the roots are not real, because we would be forced to take the square root of a negative number, which yields an imaginary result. The discriminant of the equation we are given is (–6)2 – 4(2)(3) = 36 – 24 = 12 > 0. Because the discriminant is not negative, the solutions to the equation will be real. Thus, option I is correct.

The discriminant can also tell us whether the solutions of an equation are rational or not. If we take the square root of the discriminant and get a rational number, then the solutions of the equation must be rational. In this problem, we would need to take the square root of 12. However, 12 is not a perfect square, so taking its square root would produce an irrational number. Therefore, the solutions to the equation in the problem cannot be rational. This means that choice II is incorrect.

Lastly, the discriminant tells us if the roots to an equation are distinct (different from one another). If the discriminant is equal to zero, then the solutions of x become (–b + 0)/2_a_ and (–b – 0)/2_a_, because the square root of zero is 0. Notice that (–b + 0)/2_a_ is the same as (–b – 0)/2_a_. Thus, if the discriminant is zero, then the roots of the equation are the same, i.e. indistinct. In this particular problem, the discriminant = 12, which doesn't equal zero. This means that the two roots will be different, i.e. distinct. Therefore, choice III applies.

The answer is choices I and III only.

Begin by getting all factors to one side of the equal sign.

–3x + 3 = x(x – 1) → -3x + 3 = x2 – x → 0 = x2 +2x – 3.

Now, factor the right side: 0 = (x + 3)(x – 1).

Each of these factors can be set equal to 0 and solved for x. (x + 3) = 0; x = –3.

(x – 1) = 0 → x = 1.

However, the answer is not A, because if we return to the original problem, we must note that the denominator of the fraction is (x – 1); therefore, 1 is not a valid answer because this would cause a division by 0. Thus, –3 is the only acceptable answer.

We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y_2_ is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).

factor the equation:

(x – 3)(x – 6) = 0

set each equal to 0

x = 3, 6

Factor the equation by finding two numbers that add to -3 and multiply to -28.

Factors of 28: 1,2,4,7,14,28

-7 and 4 work.

(x-7)(x+4) = 0

Set each equal to zero:

x=7,-4

This is the difference of squares. You must know this formula for the GRE!

_a_2 – _b_2 = (a – b)(a + b)

Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).

To begin, let's factor the first two terms and the second two terms separately.

_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)

(z – 1) can be pulled out because it appears in both terms.

_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1) = (z – 1)(_z_2 – 9)

(_z_2 – 9) is a difference of squares, so we can use the formula _a_2 – _b_2 = (a – b)(a + b).

_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)

= (z – 1)(_z_2 – 9)

= (z – 1)(z – 3)(z + 3)

First pull out 3u from both terms.

3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v, so

3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]

= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)