Solving Quadratic Equations - GMAT Quantitative

The discriminant of the expression is . For the equation to have two real solutions, this discriminant must be positive, so:

One of two things happens:

Case 1:

and

and

But this is the same as simply saying

Case 2:

and

and

But this is the same as simply saying

Therefore, the equation has two real solutions if and only if either or

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.

Distribute:

We find the -coordinate of the vertex of the parabola for . First, we find its -coordinate using the formula

, setting .

is the -coordindate of the vertex, and, subsequently, the minimum value of :

This can be found by factoring the equation. Doing this we get

We can solve this equation happen when or So the roots are .

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.

Write the equation in standard form:

Evaluate the discriminant , using .

The discriminant is positive, so the equation has two distinct real 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.

For the equation to have exactly one solution, it must hold that discriminant , so we solve for in this equation, setting :

or

For the equation to have exactly one solution, it must hold that the discriminant , so we solve for this equation setting :

The discriminant of a quadratic expression is .

For the equation to have exactly one solution, it must hold that discriminant .

Solve for , setting :

For the equation to have exactly one solution, it must hold that discriminant .

Solve for , setting :

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

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.

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 .

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.

Subsitute in the defintion, and set it equal to 21:

This sets up a quadratic equation, Move all terms to the left, factor the expression, set each factor to 0, and solve separately.

or

The solution set is

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then colleciting all of the terms on the left side:

Now factor the quadratic expression to two binomial factors , replacing the question marks with two integers whose product is 36 and whose sum is . These numbers are , so:

or

The solution set is

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

Now factor the quadratic expression into two binomial factors , replacing the question marks with two integers whose product is 16 and whose sum is . These numbers are , so:

or

The solution set is .