All GMAT Math Resources
Example Questions
Example Question #1471 : Problem Solving Questions
How many real solutions and how many imaginary solutions are there to the following equation?
One real solution, no imaginary solutions
One real solution, two imaginary solutions.
No real solutions, three imaginary solutions
No real solutions, two imaginary solutions
Three real solutions, no imaginary solutions
One real solution, two imaginary solutions.
, so this is the difference of cubes.
The difference of cubes can be factored as follows:
Using this pattern and replacing , this becomes:
or
The equation becomes
We set each factor equal to 0and examine the nature of the solutions.
, so the equation has at least one real solution.
is a quadratic equation, so we can explore these two solutions by looking at its discriminant .
We set , so the discriminant is
This negative disciminant indicates that there are two imaginary solutions in addition to the real one we found earlier.
Example Question #1472 : Problem Solving Questions
For what value of does the equation have exactly one solution?
For the equation to have exactly one solution, it must hold that the discriminant , so we solve for this equation setting :
Example Question #1481 : Gmat Quantitative Reasoning
For what value(s) of does this quadratic equation have exactly one solution?
For the equation to have exactly one solution, it must hold that discriminant , so we solve for in this equation, setting :
or
Example Question #1482 : Gmat Quantitative Reasoning
For what value of does this equation have exactly one solution?
For the equation to have exactly one solution, it must hold that discriminant .
Solve for , setting :
Example Question #1483 : Gmat Quantitative Reasoning
For what value(s) of does this equation have exactly one solution?
The discriminant of a quadratic expression is .
For the equation to have exactly one solution, it must hold that discriminant .
Solve for , setting :
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.