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Solving an equation with one radical

Solve 15 2 x = x .

The radical is already isolated on the left side of the equal side, so proceed to square both sides.

15 2 x = x ( 15 2 x ) 2 = ( x ) 2 15 2 x = x 2

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

0 = x 2 + 2 x 15 = ( x + 5 ) ( x 3 ) x = −5 x = 3

The proposed solutions are −5 and 3. Let us check each solution back in the original equation. First, check x = −5.

15 2 x = x 15 2 ( 5 ) = −5 25 = −5 5 −5

This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.

Check x = 3.

15 2 x = x 15 2 ( 3 ) = 3 9 = 3 3 = 3

The solution is 3.

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Solve the radical equation: x + 3 = 3 x 1

x = 1 ; extraneous solution x = 2 9

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Solving a radical equation containing two radicals

Solve 2 x + 3 + x 2 = 4.

As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.

2 x + 3 + x 2 = 4 2 x + 3 = 4 x 2 Subtract  x 2  from both sides . ( 2 x + 3 ) 2 = ( 4 x 2 ) 2 Square both sides .

Use the perfect square formula to expand the right side: ( a b ) 2 = a 2 −2 a b + b 2 .

2 x + 3 = ( 4 ) 2 2 ( 4 ) x 2 + ( x 2 ) 2 2 x + 3 = 16 8 x 2 + ( x 2 ) 2 x + 3 = 14 + x 8 x 2 Combine like terms . x 11 = −8 x 2 Isolate the second radical . ( x 11 ) 2 = ( −8 x 2 ) 2 Square both sides . x 2 22 x + 121 = 64 ( x 2 )

Now that both radicals have been eliminated, set the quadratic equal to zero and solve.

x 2 22 x + 121 = 64 x 128 x 2 86 x + 249 = 0 ( x 3 ) ( x 83 ) = 0 Factor and solve . x = 3 x = 83

The proposed solutions are 3 and 83. Check each solution in the original equation.

2 x + 3 + x 2 = 4 2 x + 3 = 4 x 2 2 ( 3 ) + 3 = 4 ( 3 ) 2 9 = 4 1 3 = 3

One solution is 3.

Check x = 83.

2 x + 3 + x 2 = 4 2 x + 3 = 4 x 2 2 ( 83 ) + 3 = 4 ( 83 2 ) 169 = 4 81 13 5

The only solution is 3. We see that x = 83 is an extraneous solution.

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Solve the equation with two radicals: 3 x + 7 + x + 2 = 1.

x = −2 ; extraneous solution x = −1

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Solving an absolute value equation

Next, we will learn how to solve an absolute value equation    . To solve an equation such as | 2 x 6 | = 8 , we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is 8 or −8. This leads to two different equations we can solve independently.

2 x 6 = 8 or 2 x 6 = −8 2 x = 14 2 x = −2 x = 7 x = −1

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

Absolute value equations

The absolute value of x is written as | x | . It has the following properties:

If  x 0 ,  then  | x | = x . If  x < 0 ,  then  | x | = x .

For real numbers A and B , an equation of the form | A | = B , with B 0 , will have solutions when A = B or A = B . If B < 0 , the equation | A | = B has no solution.

An absolute value equation    in the form | a x + b | = c has the following properties:

If   c < 0 , | a x + b | = c  has no solution . If   c = 0 , | a x + b | = c  has one solution . If   c > 0 , | a x + b | = c  has two solutions .

Given an absolute value equation, solve it.

  1. Isolate the absolute value expression on one side of the equal sign.
  2. If c > 0 , write and solve two equations: a x + b = c and a x + b = c .

Solving absolute value equations

Solve the following absolute value equations:

  • (a) | 6 x + 4 | = 8
  • (b) | 3 x + 4 | = −9
  • (c) | 3 x 5 | 4 = 6
  • (d) | −5 x + 10 | = 0
  • (a) | 6 x + 4 | = 8

    Write two equations and solve each:

    6 x + 4 = 8 6 x + 4 = −8 6 x = 4 6 x = −12 x = 2 3 x = −2

    The two solutions are 2 3 and −2.

  • (b) | 3 x + 4 | = −9

    There is no solution as an absolute value cannot be negative.

  • (c) | 3 x 5 | 4 = 6

    Isolate the absolute value expression and then write two equations.

    | 3 x 5 | 4 = 6 | 3 x 5 | = 10 3 x 5 = 10 3 x 5 = −10 3 x = 15 3 x = −5 x = 5 x = 5 3

    There are two solutions: 5 , and 5 3 .

  • (d) | −5 x + 10 | = 0

    The equation is set equal to zero, so we have to write only one equation.

    −5 x + 10 = 0 −5 x = −10 x = 2

    There is one solution: 2.

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Questions & Answers

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(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
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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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