College Algebra : Equations Reducible to Quadratic Form

Study concepts, example questions & explanations for College Algebra

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Example Questions

Example Question #1 : Equations Reducible To Quadratic Form

Solve 

Possible Answers:

There are no solutions

Correct answer:

Explanation:

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 #1 : Equations Reducible To Quadratic Form

Find all real roots of the polynomial function 

 

Possible Answers:

  (There are no nonzero solutions) 

Correct answer:

Explanation:

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. 

Problem 1 precalc plot of polynomial showing the roots

 

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 #2 : Equations Reducible To Quadratic Form

Give the complete solution set for the equation:

Possible Answers:

Correct answer:

Explanation:

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 #3 : Equations Reducible To Quadratic Form

Give the complete set of real solutions for the equation:

Possible Answers:

The equation has no solutions.

Correct answer:

Explanation:

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 .

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