Question

Need help with this homework problem asap please :)

Answer 1

Answers

Part 1: Solve for the variable

x = 5

Part 2: Is the answer extraneous or not?

5 is as an extraneous solution because when it is plugged back into the equation it results in a false statement

Part 1

Step-By-Step

Part 1:

Isolate x

Subtract 8 to both sides

`√(3x+1)` + 8 = 4

`√(3x+1)` = -4

Square both sides of the equation (this will remove the radical sign)

3x + 1 = 16

Subtract 1 to both sides

3x = 15

Divide 3 on both sides

x = 5

Part 2

Define Key Words

What is an extraneous solution?

An extraneous solution is an answer you find when solving for a variable in an equation (x), but when you plug the x back into the equation, the equation results in a false conclusion. This sometimes occurs when you have an equation with a square root.

Step-By-Step

Plug x (5) into the original equation

`√(3(5)+1)` + 8 = 4

Solve using PEMDAS

`√(15+1)` + 8 = 4

`√(16)` + 8 = 4

4 + 8 = 4

12
`\\eq` 4

This is false, thus 5 is an extraneous solution

Keywords

extraneous solution

Answer 2

x=5 :)

Explanation:

1) You have to move all the terms which do not include x to the opposite side of the equation so you can solve x for a simple variable.

To do this subtract 8 from both sides!

`√(3x+1)+8=4` will become
`√(3x+1)=-4`

2) Get rid of the square root

To do this square both sides of the equation

`√(3x+1) = 3x+1`

`(-4)^2=16`

You will now be left with 3x+1 = 16

3) Get rid of 1 from both sides (subtract)

`3x+1-1=3x`

`16-1=15`

You will now be left with 3=15

4) Divide both sides by 3 to get the result for x

`3x` ÷
`3`
`=x`

`15` ÷
`3=5`

You are left with x=5!!!

Hope this helps, have a great day!!!!