Equations Reducible to Quadratic Form - College Algebra

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 .

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.

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 .

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 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 .

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.

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 .

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 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 .

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.

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 .

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 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 .