All GRE Math Resources
Example Questions
Example Question #1 : How To Find The Solution To A Quadratic Equation
Solve for x: x2 + 4x = 5
-1
-1 or 5
None of the other answers
-5 or 1
-5
-5 or 1
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
Example Question #2 : How To Find The Solution To A Quadratic Equation
Solve for x: (x2 – x) / (x – 1) = 1
x = -1
x = 2
No solution
x = -2
x = 1
No solution
Begin by multiplying both sides by (x – 1):
x2 – x = x – 1
Solve as a quadratic equation: x2 – 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 #3 : How To Find The Solution To A Quadratic Equation
x2 – 3x – 18 = 0
Quantity A: x
Quantity B: 6
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
A. Quantity A is greater
D. The relationship cannot be determined from the information given.
x2 – 3x – 18 = 0 can be factored to (x – 6)(x + 3) = 0
Therefore, x can be either 6 or –3.
Example Question #33 : Quadratic Equations
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?
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 #4 : How To Find The Solution To A Quadratic Equation
What is the sum of all the values of that satisfy:
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 #162 : Gre Quantitative Reasoning
Quantity A:
Quantity B:
A comparison cannot be detemined from the given information.
Quantity A is larger.
The two quantities are equal.
Quantity B is larger.
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 #11 : Quadratic Equations
If f(x) = -x2 + 6x - 5, then which could be the value of a if f(a) = f(1.5)?
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
-a2 + 6a -5 = 1.75
Multiply both sides by 4, so that we can work with only whole numbers coefficients.
-4a2 + 24a - 20 = 7
Subtract 7 from both sides.
-4a2 + 24a - 27 = 0
Multiply both sides by negative one, just to make more positive coefficients, which are usually easier to work with.
4a2 - 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:
4a2 - 6a -18a + 27 = 0
We can now group the first two terms and the last two terms, and then we can factor.
(4a2 - 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 #2 : Quadratic Equations
Solve for x: 2(x + 1)2 – 5 = 27
3 or 4
–2 or 5
3 or –5
–3 or 2
–2 or 4
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 #5 : 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?
12
9
15
6
3
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 x2 + 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 #2 : How To Find The Solution To A Quadratic Equation
Solve 3x2 + 10x = –3
x = –1/9 or –9
x = –1/3 or –3
x = –4/3 or –1
x = –2/3 or –2
x = –1/6 or –6
x = –1/3 or –3
Generally, quadratic equations have two answers.
First, the equations must be put in standard form: 3x2 + 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.