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Solving a rational equation leading to a quadratic

Solve the following rational equation: 4 x x 1 + 4 x + 1 = 8 x 2 1 .

We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, x 2 −1 = ( x + 1 ) ( x 1 ) . Then, the LCD is ( x + 1 ) ( x 1 ) . Next, we multiply the whole equation by the LCD.

( x + 1 ) ( x 1 ) [ −4 x x 1 + 4 x + 1 ] = [ −8 ( x + 1 ) ( x 1 ) ] ( x + 1 ) ( x 1 ) −4 x ( x + 1 ) + 4 ( x 1 ) = −8 −4 x 2 4 x + 4 x 4 = −8 −4 x 2 + 4 = 0 −4 ( x 2 1 ) = 0 −4 ( x + 1 ) ( x 1 ) = 0 x = −1 x = 1

In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.

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Solve 3 x + 2 x 2 + 1 x = 2 x 2 2 x .

x = −1 , x = 0 is not a solution.

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Access these online resources for additional instruction and practice with different types of equations.

Key concepts

  • Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See [link] , [link] , and [link] .
  • Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See [link] and [link] .
  • We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See [link] and [link] .
  • To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See [link] .
  • Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See [link] and [link] .
  • Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See [link] .

Section exercises

Verbal

In a radical equation, what does it mean if a number is an extraneous solution?

This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

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Explain why possible solutions must be checked in radical equations.

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Your friend tries to calculate the value 9 3 2 and keeps getting an ERROR message. What mistake is he or she probably making?

He or she is probably trying to enter negative 9, but taking the square root of −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27.

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Explain why | 2 x + 5 | = −7 has no solutions.

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Explain how to change a rational exponent into the correct radical expression.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

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Algebraic

For the following exercises, solve the rational exponent equation. Use factoring where necessary.

( x 1 ) 3 4 = 8

x = 17

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x 2 3 5 x 1 3 + 6 = 0

x = 8 ,     x = 27

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x 7 3 3 x 4 3 4 x 1 3 = 0

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For the following exercises, solve the following polynomial equations by grouping and factoring.

x 3 + 2 x 2 x 2 = 0

x = −2 , 1 , −1

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3 x 3 6 x 2 27 x + 54 = 0

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4 y 3 9 y = 0

y = 0 ,     3 2 ,     3 2

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x 3 + 3 x 2 25 x 75 = 0

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m 3 + m 2 m 1 = 0

m = 1 , −1

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5 x 3 + 45 x = 2 x 2 + 18

x = 2 5

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For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

3 x + 7 + x + 2 = 1

x = −2

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For the following exercises, solve the equation involving absolute value.

| 3 x 4 | = 8

x = 4 , −4 3

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| 1 4 x | 1 = 5

x = 5 4 , 7 4

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| 2 x 1 | 7 = −2

x = 3 , −2

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| 2 x + 1 | 2 = −3

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For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

x 4 10 x 2 + 9 = 0

x = 1 , −1 , 3 , −3

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4 ( t 1 ) 2 9 ( t 1 ) = −2

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( x 2 1 ) 2 + ( x 2 1 ) 12 = 0

x = 2 , −2

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( x + 1 ) 2 8 ( x + 1 ) 9 = 0

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( x 3 ) 2 4 = 0

x = 1 , 5

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Extensions

For the following exercises, solve for the unknown variable.

x −2 x −1 12 = 0

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| x | 2 = x

All real numbers

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| x 2 + 2 x 36 | = 12

x = 4 , 6 , −6 , −8

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Real-world applications

For the following exercises, use the model for the period of a pendulum, T , such that T = 2 π L g , where the length of the pendulum is L and the acceleration due to gravity is g .

If the acceleration due to gravity is 9.8 m/s 2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).

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If the gravity is 32 ft/s 2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.

10 in.

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For the following exercises, use a model for body surface area, BSA, such that B S A = w h 3600 , where w = weight in kg and h = height in cm.

Find the height of a 72-kg female to the nearest cm whose B S A = 1.8.

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Find the weight of a 177-cm male to the nearest kg whose B S A = 2.1.

90 kg

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Questions & Answers

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Practice Key Terms 5

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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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