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Solve the absolute value equation: | 1 4 x | + 8 = 13.

x = −1 , x = 3 2

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Solving other types of equations

There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic.

Solving equations in quadratic form

Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include x 4 5 x 2 + 4 = 0 , x 6 + 7 x 3 8 = 0 , and x 2 3 + 4 x 1 3 + 2 = 0. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.

Quadratic form

If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form , which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.

Given an equation quadratic in form, solve it.

  1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.
  2. If it is, substitute a variable, such as u , for the variable portion of the middle term.
  3. Rewrite the equation so that it takes on the standard form of a quadratic.
  4. Solve using one of the usual methods for solving a quadratic.
  5. Replace the substitution variable with the original term.
  6. Solve the remaining equation.

Solving a fourth-degree equation in quadratic form

Solve this fourth-degree equation: 3 x 4 2 x 2 1 = 0.

This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let u = x 2 . Rewrite the equation in u .

3 u 2 2 u 1 = 0

Now solve the quadratic.

3 u 2 2 u 1 = 0 ( 3 u + 1 ) ( u 1 ) = 0

Solve each factor and replace the original term for u.

3 u + 1 = 0 3 u = −1 u = 1 3 x 2 = 1 3 x = ± i 1 3
u 1 = 0 u = 1 x 2 = 1 x = ±1

The solutions are x = ± i 1 3 and x = ± 1.

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Solve using substitution: x 4 8 x 2 9 = 0.

x = −3 , 3 , i , i

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Solving an equation in quadratic form containing a binomial

Solve the equation in quadratic form: ( x + 2 ) 2 + 11 ( x + 2 ) 12 = 0.

This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting u = x + 2. Then rewrite the equation in u.

u 2 + 11 u 12 = 0 ( u + 12 ) ( u 1 ) = 0

Solve using the zero-factor property and then replace u with the original expression.

u + 12 = 0 u = −12 x + 2 = −12 x = −14

The second factor results in

u 1 = 0 u = 1 x + 2 = 1 x = −1

We have two solutions: x = −14 , x = −1.

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Solve: ( x 5 ) 2 4 ( x 5 ) 21 = 0.

x = 2 , x = 12

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Solving rational equations resulting in a quadratic

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

Questions & Answers

An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
Practice Key Terms 5

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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