2.6 Other types of equations (Page 4/10)

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

$x = −1,$ $x = \frac{3}{2}$

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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 $x^{\frac{2}{3}} + 4 x^{\frac{1}{3}} + 2 = 0.$ 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.

A General Note

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.

How To

Given an equation quadratic in form, solve it.

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

\begin{matrix} 3 u^{2} - 2 u - 1 & = & 0 \\ (3 u + 1) (u - 1) & = & 0 \end{matrix}

Solve each factor and replace the original term for u.

\begin{matrix} 3 u + 1 & = & 0 \\ 3 u & = & −1 \\ u & = & - \frac{1}{3} \\ x^{2} & = & - \frac{1}{3} \\ x & = & \pm i \sqrt{\frac{1}{3}} \end{matrix}

\begin{matrix} u - 1 & = & 0 \\ u & = & 1 \\ x^{2} & = & 1 \\ x & = & ±1 \end{matrix}

The solutions are $x = \pm i \sqrt{\frac{1}{3}}$ and $x = \pm 1.$

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Try It

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.

\begin{matrix} u^{2} + 11 u - 12 & = & 0 \\ (u + 12) (u - 1) & = & 0 \end{matrix}

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

\begin{matrix} u + 12 & = & 0 \\ u & = & −12 \\ x + 2 & = & −12 \\ x & = & −14 \end{matrix}

The second factor results in

\begin{matrix} u - 1 & = & 0 \\ u & = & 1 \\ x + 2 & = & 1 \\ x & = & −1 \end{matrix}

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

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Try It

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

find the equation of the tangent to the curve y=2x³-x²+3x+1 at the points x=1 and x=3

Esther Reply

derivative of logarithms function

Iqra Reply

how to solve this question

sidra

ex 2.1 question no 11

khansa

anyone can help me

khansa

question please

Rasul

ex 2.1 question no. 11

khansa

i cant type here

khansa

Find the derivative of g(x)=−3.

Abdullah Reply

any genius online ? I need help!!

Guzorochi Reply

how can i help you?

Pina

need to learn polynomial

Zakariya

i will teach...

nandu

I'm waiting

Zakariya

plz help me in question

Abish

How can I help you?

Tlou

evaluate the following computation (x³-8/x-2)

Murtala Reply

teach me how to solve the first law of calculus.

Uncle Reply

teach me also how to solve the first law of calculus

Bilson

what is differentiation

Ibrahim Reply

only god knows😂

abdulkadir

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)

Naufal Reply

fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5

diron

pweding paturo nsa calculus?

jimmy

how to use fundamental theorem to solve exponential

JULIA Reply

find the bounded area of the parabola y^2=4x and y=16x

Omar Reply

what is absolute value means?

Geo Reply

Chicken nuggets

Hugh

🐔

🐔🦃 nuggets

(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.

Ismael

find integration of loge x

Game Reply

find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3

Godwin Reply

how does this work

Brad Reply

Can calculus give the answers as same as other methods give in basic classes while solving the numericals?

Cosmos Reply

log tan (x/4+x/2)

Rohan

please answer

Rohan

y=(x^2 + 3x).(eipix)

Claudia

is this a answer

Ismael

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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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