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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 1 2 is another way of writing 16 ; 8 1 3 is another way of writing 8 3 . 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, 2 3 ( 3 2 ) = 1 , 3 ( 1 3 ) = 1 , and so on.

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 m n = ( a 1 n ) m = ( a m ) 1 n = a m n = ( a n ) m

Evaluating a number raised to a rational exponent

Evaluate 8 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 2 3 as ( 8 1 3 ) 2 .

( 8 1 3 ) 2 = ( 2 ) 2 = 4
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Evaluate 64 1 3 .

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 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 5 4 , which is 4 5 .

x 5 4 = 32 ( x 5 4 ) 4 5 = ( 32 ) 4 5 x = ( 2 ) 4 The fifth root of 32 is 2. = 16
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Solve the equation x 3 2 = 125.

25

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Solving an equation involving rational exponents and factoring

Solve 3 x 3 4 = x 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.

3 x 3 4 ( x 1 2 ) = x 1 2 ( x 1 2 ) 3 x 3 4 x 1 2 = 0

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 1 2 as x 2 4 . Then, factor out x 2 4 from both terms on the left.

3 x 3 4 x 2 4 = 0 x 2 4 ( 3 x 1 4 1 ) = 0

Where did x 1 4 come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply x 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 2 4 equals 3 4 . Thus, the exponent on x in the parentheses is 1 4 .

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

x 2 4 ( 3 x 1 4 1 ) = 0 x 2 4 = 0 x = 0 3 x 1 4 1 = 0 3 x 1 4 = 1 x 1 4 = 1 3 Divide both sides by 3 . ( x 1 4 ) 4 = ( 1 3 ) 4 Raise both sides to the reciprocal of  1 4 . x = 1 81

The two solutions are 0 and 1 81 .

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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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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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