Solving equations with radicals and exponents

Question

In Algebra 2, I was given the following problems, with the only directions being "Solve:"

 

5 - √x² - 4x + 32√ = x + 5  (The section between the square root signs is under one square root sign.)

 

My textbook provides no examples with a number in front of the radical as the five is in mine. What should I do to solve the equation?

Answer 1

5 - √(x² - 4x + 32) = x + 5

 

Isolate the radical:  √(x² - 4x + 32) = x

 

Square both sides:  x² - 4x + 32 = x²

 

                                       -4x + 32 = 0

 

                                                 -4x = -32

 

                                                    x = 8

 

 

 

 

 

Answer 2

James, I'm gonna do what you did with the square root signs.

 

First, you need to factor what's under the radical. This will give you:

 

5 - √(x - 8)(x + 4)√ = x + 5

 

Next, subtract five from both sides:

 

- √(x - 8)(x + 4)√ = x

 

Now square both sides. You can square the entire radical to leave just what's inside, and then square x on the right as well. Squaring a negative always results in a positive, too. Doing that gives you.

 

(x - 8)(x + 4) = x²      Multiply out the left side using the FOIL method, or whichever way you like.

 

x² - 4x - 32 = x²         The x-squareds on both sides will cancel, giving you:

 

-4x - 32 = 0              Add 32 to both sides:

 

-4x = 32                 Divide by -4.

 

x = 8                     Final answer.