Solving Equations and Inequallities - College Algebra

We have to set up two equations, which are.

Now lets solve for x in each equation.

So the solutions are and

We need to set up two equations since we are dealing with absolute value

We solve for , in each equation to get the solutions.

We need to set up two equations since we are dealing with absolute value

We solve for , in each equation to get the solutions.

To solve, you must split the absolute value into the two following equations.

and

Now, solve for x and right it in interval form.

We first need to get rid of the absolute value symbol to solve the equation. TO break this absolute value, we assign two values to the right hand side, as shown below.

We now proceed to solve each equation independently.

Starting with the first equation, we get

Now for the second equation,

To simplify the radical, break the radical in the numerator down into its factors. When doing so, the radical in the bottom will call with one from the numerator.

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

Recall that the absolute value can come from either a negative or positive value therefore two possible equations are set up.

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

Since absolute values can come from either negative or positive values, two equations need to be set up and solved for.

To solve we need to find a way to reverse the operation of taking the absolute value. What we need to do is think about what the absolute value operation does to an expression. Since it makes everything positive, . So actually solving the original equation comes down to solving the following two equations:

So we get the two solutions as:

First, isolate the radical on one side of the equation. Start by adding to both sides of the equation.

Now, square both sides of the equation.

Expand the right side of the equation.

Collect all the terms to one side of the equation and simplify to create a quadratic equation equal to zero.

Quadratic equations can be written in the following generic form:

We need to find two numbers whose product equals a multiplied by c and whose sum equals b; therefore, the product of the factors must be -12 and their sum must equal -1. Notice the following:

Write the two quantities. When c is negative, one quantity needs to have a plus sign and one needs to have a minus sign. When b is negative, the larger number should be associated with the minus sign. Write the following quadratic factorization:

The solutions for this quadratic are:

One common mistake for students is to assume that these solutions are—in fact—solutions to the original equation. Whenever you work with absolute value equations, or radical equations, you must check the solution carefully to make sure the solution actually works. As you will see, only one of the two solutions above actually works in this particular case.

We started with the following equation:

Now, substitute the solution into the equation.

We can see that -3 is not a solution. Now, substitute the solution into the equation.

The correct answer is 4.

Solve for ,

First isolate one of the radicals; the easiest would be the one with more than one term.

Square both sides of the equation,

Expand the right side,

Now collect terms and isolate the remaining radical expression; note the the 's on the left are right sides cancel.

Square both sides,

CHECK THE SOLUTION

We have done all of the algebra correctly, but we can still end up with an erroneous solution due to the squaring operation (a very similar problem arises when dealing with absolute value equations). Once you arrive at a solution, make sure you check that the solution works. If it does not work, and you know your algebra was right, then there are no real solutions.

Therefore there are no real solutions.

using the quadratic formula we get

so the possible solutions are and . However, is not an actual solution because it is negative and the equation can only be satisfied by a positive value.

Square both sides to eliminate the radical.

Divide by three on both sides to isolate the x.

The answer is:

Square both sides.

Add 2 on both sides.

Divide by 3 on both sides.

The answer is:

To make this problem easier, lets start off by doing a u-substitution.

Let .

Now we can factor the left hand side.

We have two solutions for , now we can plug those into , to get all the solutions.

Find the roots of the polynomial,

Set equal to

Factor out ,

Notice that the the factor is a quadratic even though it might not seem so at first glance. One way to think of this is as follows:

Let

Then we have , substitute into to get,

Notice that the change in variable from to has resulted in a quadratic equation that can be easily factored due to the fact that it is a square of a simple binomial:

The solution for is,

Because we go back to the variable ,

Therefore, the roots of the factor are,

The other root of is since the function clearly equals when .

The solution set is therefore,

Below is a plot of . You can see where the function intersects the -axis at points corresponding to our solutions.

Further Discussion

The change of variable was a tool we used to write the quadratic factor in a more familiar form, but we could have just factored the original function in terms of as follows,

Setting this to zero gives the same solution set,

can be rewritten in quadratic form by setting , and, consequently, ; the resulting equation is as follows:

By the reverse-FOIL method we can factor the trinomial at left. We are looking for two integers with sum 8 and product 12; they are 2 and 6, so the equation becomes

Setting both binomials equal to 0, it follows that

or .

Substituting for , we get

,

in which case

,

or

in which case

.

The solution set is .

can be rewritten in quadratic form by setting , and, consequently, ; the resulting equation is as follows:

By the reverse-FOIL method we can factor the trinomial at left. We are looking for two integers with sum 8 and product 12; they are and , so the equation becomes

Setting both binomials equal to 0, it follows that

or .

Substituting for , we get

in which case

,

and

,

in which case

The set of real solutions is therefore .

In order to solve this, we need to figure out what values of x makes the numerator and denominator equal zero.

These values are , and .

Then test a few numbers greater or lesser than the above values to figure out what range of values make the inequality true.

The boundary points of a rational inequality are the zeroes of the numerator and the denominator.

First, set the numerator equal to 0 and solve for :

Now set the denominator equal to 0 and solve for :

Factor the trinomial using the reverse-FOIL method. Look for two integers whose sum is and whose product is 7; these are , and , so the equation can be rewritten as

Set each factor to zero and solve for :

Therefore, the boundary points are , which divide the real numbers into four intervals. Choose any value from each interval as a test point, setting to that value and determining whether the inequality is true.

The four intervals are listed below, along with their arbitrary test points.

: Set

True; include .

: Set

False; exclude .

: Set :

True; include

: Set :

False; exclude .

Since the inequality symbol is the "is greater than or equal to" symbol, include the zero, 0, of the numerator - but not the other two boundaries, the zeroes of the denominator. This makes the solution set

.