All GMAT Math Resources
Example Questions
Example Question #11 : Solving Quadratic Equations
Solve the following for x:
x= -6 or x=1
x= -2 or x=3
x= 2 or x=-3
x= 2 or x=3
x= 1 or x=-6
x= -2 or x=3
There are 2 ways to solve this - factoring or using the quadratic formula.
First, for factoring we wish to find a way to factor this into an expression that looks like (cx+m)(dx+n)= 0. This way, we know the possible values for x will be -m/c and -n/d. (You can check by plugging in and getting the result algebraically.)
To do this, we need to look at the terms of the quadratic. Quadratics are expressed as . In this case, a = 1, b = -1, and c = -6.
When solving for factors, the easiest starting point is to find the factors of a and c. Since a= 1, the only factor is 1, Therefore we know our c and d value will be 1. So our factored expression will look like (1x+m)(1x+n).
If we FOIL (x+m)(x+n) we see that we get
So to find m and n, we look at our factors for c. Since c=-6, our possible factors are 1,2,3, and 6. Since we also know that m+n must equal b, we can try and find a fitting m and n so that the sum is -1. However, we know that c is negative; therefore we know that one of our factors will be positive and one will be negative. Therefore, we are looking to find a DIFFERENCE of 1 between our factors, which we can find between 2 and 3.
So we know that -2*3 = -6, and 2*-3 = -6. We also know that m+n =b, so -2+3 =1 (NOT WHAT WE WANT!) and 2+-3 = -1 (SCORE!). Since we have all the pieces to the puzzle, we see that this factors into (x+2)(x-3). Thus, solving our equation of (x+2)(x-3) = 0, we solve x+2=0 AND x-3=0. Thus our answer: x is -2 or 3.
ALTERNATE SOLUTIONS
If you are having trouble with the factoring, or if you find a quadratic that is not factorable, you can always use the quadratic equation. This says that when , x =
Plugging in we get x=
So x = -2 or x=3.
Example Question #12 : Solving Quadratic Equations
Solve the following expression for :
Let us start by simplifying the equation:
Now that we have the equation simplified into two products, we solve for each case:
or
or
So the answer is the set
Example Question #402 : Algebra
How many real number solutions are there to the equation when solving for ?
Since the problem didn't ask for what actually is, we don't have to solve for it. Instead we check the discriminant of the quadratic equation;
First we put our equation in standard form, and we then see that . Substituting these into the discriminant gives us . Since the number inside the square root is greater than , our original equation has real roots.
Example Question #3 : How To Use The Quadratic Function
The length of a rectangular piece of land is two feet more than three times its width. If the area of the land is , what is the width of that piece of land?
The area of a rectangle is the product of its length by its width, which we know to be equal to in our problem. We also know that the length is equal to , where represents the width of the land. Therefore, we can write the following equation:
Distributing the outside the parentheses, we get:
Subtracting from each side of the equation, we get:
We get a quadratic equation, and since there is no factor of and that adds up to , we use the quadratic formula to solve this equation.
We can first calculate the discriminant (i.e. the part under the square root)
We replace that value in the quadratic formula, solving both the positive version of the formula (on the left) and the negative version of the formula (on the right):
Breaking down the square root:
We can pull two of the twos out of the square root and place a outside of it:
We can then multiply the and the :
At this point, we can reduce the equations, since each of the component parts of their right sides has a factor of :
Since width is a positive value, the answer is:
The width of the piece of land is approximately .
Example Question #13 : Solving Quadratic Equations
How many real solutions and how many imaginary solutions are there to the following equation?
Three real solutions, no imaginary solutions
No real solutions, two imaginary solutions
One real solution, two imaginary solutions
No real solutions, three imaginary solutions
Two real solutions, no imaginary solutions
Three real solutions, no imaginary solutions
Write the equation in standard form:
Factor out the greatest common factor of :
Factor the trinomial by writing , replacing the question marks with two integers with product and sum . These integers are , so the above becomes
Since the expression factors out to three distinct linear expressions, the expression has three real zeros,and the correct choice is three real solutions.
Example Question #1488 : Gmat Quantitative Reasoning
Solve for :
We start by splitting the numerator into two expressions, and factoring them:
From there, we can pull (z-15) out of the numerator:
so we are left with z-15.
Example Question #1 : Solving By Factoring
Factor .
To factor this, we need two numbers that multiply to and sum to . The numbers and work.
Example Question #2 : Solving By Factoring
Factor .
is a difference of squares. The difference of squares formula is
So .
Then, .
Example Question #1 : Understanding Factoring
Solve .
Let's factor the expression: .
We need to look at the behavior of the function to the left and right of 1 and 5. To the left of ,
You can check this by plugging in any value smaller than 1. For example, if ,
,
which is greater than 0.
When takes values in between 1 and 5, . Again we can check this by plugging in a number between 1 and 5.
, which is less than 0, so no numbers between 1 and 5 satisfy the inequality.
When takes values greater than 5, .
To check, let's try . Then:
so numbers greater than 5 also satisfy the inequality.
Therefore .
Example Question #401 : Algebra
Solve .
First let's factor:
x < -8: Let's try -10. (-10 + 8)(-10 - 1) = 22, so values less than -8 don't satisfy the inequality.
-8 < x < 1: Let's try 0. (0 + 8)(0 - 1) = -8, so values in between -8 and 1 satisfy the inequality.
x > 1: Let's try 2. (2 + 8)(2 - 1) = 10, so values greater than 1 don't satisfy the inequality.
Therefore the answer is -8 < x < 1.