### All GRE Math Resources

## Example Questions

### Example Question #1 : How To Find The Solution To A Quadratic Equation

Solve for x: x^{2} + 4x = 5

**Possible Answers:**

-1

None of the other answers

-5

-1 or 5

-5 or 1

**Correct answer:**

-5 or 1

Solve by factoring. First get everything into the form Ax^{2} + Bx + C = 0:^{}

x^{2} + 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

### Example Question #2 : How To Find The Solution To A Quadratic Equation

Solve for x: (x^{2} – x) / (x – 1) = 1

**Possible Answers:**

No solution

x = 2

x = 1

x = -1

x = -2

**Correct answer:**

No solution

Begin by multiplying both sides by (x – 1):

x^{2} – x = x – 1

Solve as a quadratic equation: x^{2} – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore, x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator. Therefore, there is no solution.

### Example Question #1 : How To Find The Solution To A Quadratic Equation

x^{2} – 3x – 18 = 0

Quantity A: *x*

Quantity B: 6

**Possible Answers:**

B. Quantity B is greater.

C. The two quantities are equal.

A. Quantity A is greater

D. The relationship cannot be determined from the information given.

**Correct answer:**

D. The relationship cannot be determined from the information given.

*x*^{2} – 3*x* – 18 = 0 can be factored to (*x –* 6)(x + 3) = 0

Therefore, x can be either 6 or –3.

### Example Question #25 : How To Find The Solution To A Quadratic Equation

A farmer has 44 feet of fence, and wants to fence in his sheep. He wants to build a rectangular pen with an area of 120 square feet. Which of the following is a possible dimension for a side of the fence?

**Possible Answers:**

**Correct answer:**

Set up two equations from the given information:

and

Substitute into the second equation:

Multiply through by .

Then divide by the coefficient of 2 to simplify your work:

Then since you have a quadratic setup, move the term to the other side (via subtraction from both sides) to set everything equal to 0:

As you look for numbers that multiply to positive 120 and add to -22 so you can factor the quadratic, you might recognize that -12 and -10 fit the bill. This makes your factorization:

This makes the possible solutions 10 and 12. Since 12 does not appear in the choices, is the only possible correct answer.

### Example Question #1 : Quadratic Equation

What is the sum of all the values of that satisfy:

**Possible Answers:**

**Correct answer:**

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

OR

The sum of these values is:

### Example Question #3 : Quadratic Equation

**Quantity A: **

**Quantity B:**

**Possible Answers:**

A comparison cannot be detemined from the given information.

Quantity A is larger.

The two quantities are equal.

Quantity B is larger.

**Correct answer:**

Quantity B is larger.

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

Now, while you could use the quadratic formula to solve this problem, the easiest way to work this question is to factor the left side of the equation. This gives you:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

OR

Since both of your answers are less than , quantity B is larger than quantity A.

### Example Question #4 : Quadratic Equation

If f(x) = -x^{2} + 6x - 5, then which could be the value of a if f(a) = f(1.5)?

**Possible Answers:**

**Correct answer:**4.5

We need to input 1.5 into our function, then we need to input "a" into our function and set these results equal.

f(a) = f(1.5)

f(a) = -(1.5)^{2} +6(1.5) -5

f(a) = -2.25 + 9 - 5

f(a) = 1.75

-a^{2} + 6a -5 = 1.75

Multiply both sides by 4, so that we can work with only whole numbers coefficients.

-4a^{2} + 24a - 20 = 7

Subtract 7 from both sides.

-4a^{2} + 24a - 27 = 0

Multiply both sides by negative one, just to make more positive coefficients, which are usually easier to work with.

4a^{2} - 24a + 27 = 0

In order to factor this, we need to mutiply the outer coefficients, which gives us 4(27) = 108. We need to think of two numbers that multiply to give us 108, but add to give us -24. These two numbers are -6 and -18. Now we rewrite the equation as:

4a^{2} - 6a -18a + 27 = 0

We can now group the first two terms and the last two terms, and then we can factor.

(4a^{2} - 6a )+(-18a + 27) = 0

2a(2a-3) + -9(2a - 3) = 0

(2a-9)(2a-3) = 0

This means that 2a - 9 =0, or 2a - 3 = 0.

2a - 9 = 0

2a = 9

a = 9/2 = 4.5

2a - 3 = 0

a = 3/2 = 1.5

So a can be either 1.5 or 4.5.

The only answer choice available that could be a is 4.5.

### Example Question #141 : Equations / Inequalities

Solve for x: 2(x + 1)^{2} – 5 = 27

**Possible Answers:**

–2 or 5

3 or 4

–2 or 4

–3 or 2

3 or –5

**Correct answer:**

3 or –5

Quadratic equations generally have two answers. We add 5 to both sides and then divide by 2 to get the quadratic expression on one side of the equation: (x + 1)^{2} = 16. By taking the square root of both sides we get x + 1 = –4 or x + 1 = 4. Then we subtract 1 from both sides to get x = –5 or x = 3.

### Example Question #1 : How To Find The Solution To A Quadratic Equation

Two consecutive positive multiples of three have a product of 54. What is the sum of the two numbers?

**Possible Answers:**

6

15

3

12

9

**Correct answer:**

15

Define the variables to be x = first multiple of three and x + 3 = the next consecutive multiple of 3.

Knowing the product of these two numbers is 54 we get the equation x(x + 3) = 54. To solve this quadratic equation we need to multiply it out and set it to zero then factor it. So x^{2} + 3x – 54 = 0 becomes (x + 9)(x – 6) = 0. Solving for x we get x = –9 or x = 6 and only the positive number is correct. So the two numbers are 6 and 9 and their sum is 15.

### Example Question #311 : Algebra

Solve 3x^{2} + 10x = –3

**Possible Answers:**

x = –2/3 or –2

x = –1/3 or –3

x = –4/3 or –1

x = –1/9 or –9

x = –1/6 or –6

**Correct answer:**

x = –1/3 or –3

Generally, quadratic equations have two answers.

First, the equations must be put in standard form: 3x^{2} + 10x + 3 = 0

Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.

Third, check the answer by plugging the answers back into the original equation.