SOLUTION: 30.) Solve each equation and check for extraneous solutions. sqrt(a-1 -5 = 1)

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Question 165053This question is from textbook Elementary and Intermediate
: 30.) Solve each equation and check for extraneous solutions.
sqrt(a-1 -5 = 1)
 This question is from textbook Elementary and Intermediate

Answer by midwood_trail(310) About Me  (Show Source):
You can put this solution on YOUR website!
The same idea applies here as the other two questions.

Always isolate the radical first. Think of the radical as a loner.

sqrt(a - 1) - 5 =1

Add 5 to both sides.

sqrt{a - 1} = 1 + 5

sqrt{a - 1} = 6

We now square both sides.

[sqrt{a - 1}]^2 = (6)^2

a - 1 = 36

Add 1 to both sides.

a = 36 + 1

a = 37

Go back to your original question and replace a with 37 and then simplify.

sqrt{a - 1} - 5 = 1....Original question

Let a = 37

sqrt{37 - 1} - 5 = 1

sqrt{36} - 5 = 1

6 - 5 = 1

1 = 1....It checks!

The answer is a = 37

Understand?

This type of equation is called a radical equation. Later on, you will learn how to graph radical equations.