### All College Algebra Resources

## Example Questions

### Example Question #1 : Equations Reducible To Quadratic Form

Solve

**Possible Answers:**

There are no solutions

**Correct answer:**

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.

### Example Question #2 : Equations Reducible To Quadratic Form

Find all real roots of the polynomial function

**Possible Answers:**

(There are no nonzero solutions)

**Correct answer:**

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,

### Example Question #1 : Equations Reducible To Quadratic Form

Give the complete solution set for the equation:

**Possible Answers:**

**Correct answer:**

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 .

### Example Question #471 : College Algebra

Give the complete set of real solutions for the equation:

**Possible Answers:**

The equation has no solutions.

**Correct answer:**

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 .