2.6 Other types of equations

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In this section you will:

Solve equations involving rational exponents.
Solve equations using factoring.
Solve radical equations.
Solve absolute value equations.
Solve other types of equations.

We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.

Solving equations involving rational exponents

Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, $16^{\frac{1}{2}}$ is another way of writing $\sqrt{16};$ $8^{\frac{1}{3}}$ is another way of writing $ \sqrt[3]{8} .$ The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.

We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, $\frac{2}{3} (\frac{3}{2}) = 1,$ $3 (\frac{1}{3}) = 1,$ and so on.

A General Note

Rational exponents

A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:

a^{\frac{m}{n}} = {(a^{\frac{1}{n}})}^{m} = {(a^{m})}^{\frac{1}{n}} = \sqrt[n]{a^{m}} = {(\sqrt[n]{a})}^{m}

Evaluating a number raised to a rational exponent

Evaluate $8^{\frac{2}{3}} .$

Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite $8^{\frac{2}{3}}$ as ${(8^{\frac{1}{3}})}^{2} .$

\begin{matrix} {(8^{\frac{1}{3}})}^{2} & = & {(2)}^{2} \\ = & 4 \end{matrix}

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Evaluate $64^{- \frac{1}{3}} .$

$\frac{1}{4}$

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Solve the equation including a variable raised to a rational exponent

Solve the equation in which a variable is raised to a rational exponent: $x^{\frac{5}{4}} = 32.$

The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of $\frac{5}{4},$ which is $\frac{4}{5} .$

\begin{matrix} x^{\frac{5}{4}} & = & 32 \\ {(x^{\frac{5}{4}})}^{\frac{4}{5}} & = & {(32)}^{\frac{4}{5}} \\ x & = & {(2)}^{4} & The fifth root of 32 is 2. \\ = & 16 \end{matrix}

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Solve the equation $x^{\frac{3}{2}} = 125.$

$25$

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Solving an equation involving rational exponents and factoring

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

This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.

\begin{matrix} 3 x^{\frac{3}{4}} - (x^{\frac{1}{2}}) & = & x^{\frac{1}{2}} - (x^{\frac{1}{2}}) \\ 3 x^{\frac{3}{4}} - x^{\frac{1}{2}} & = & 0 \end{matrix}

Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite $x^{\frac{1}{2}}$ as $x^{\frac{2}{4}} .$ Then, factor out $x^{\frac{2}{4}}$ from both terms on the left.

\begin{matrix} 3 x^{\frac{3}{4}} - x^{\frac{2}{4}} & = & 0 \\ x^{\frac{2}{4}} (3 x^{\frac{1}{4}} - 1) & = & 0 \end{matrix}

Where did $x^{\frac{1}{4}}$ come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply $x^{\frac{2}{4}}$ back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to $\frac{2}{4}$ equals $\frac{3}{4} .$ Thus, the exponent on x in the parentheses is $\frac{1}{4} .$

Let us continue. Now we have two factors and can use the zero factor theorem.

\begin{matrix} x^{\frac{2}{4}} (3 x^{\frac{1}{4}} - 1) & = & 0 \\ x^{\frac{2}{4}} & = & 0 \\ x & = & 0 \\ 3 x^{\frac{1}{4}} - 1 & = & 0 \\ 3 x^{\frac{1}{4}} & = & 1 \\ x^{\frac{1}{4}} & = & \frac{1}{3} & Divide both sides by 3 . \\ {(x^{\frac{1}{4}})}^{4} & = & {(\frac{1}{3})}^{4} & Raise both sides to the reciprocal of \frac{1}{4} . \\ x & = & \frac{1}{81} \end{matrix}

The two solutions are $0$ and $\frac{1}{81} .$

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Questions & Answers

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

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derivative of logarithms function

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

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

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(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.

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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