1. Feb 3, 2008

### scientist

Could a tutor check my solution to my radical question? It is enclosed in this attachment as a word document.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

File size:
33 KB
Views:
73
2. Feb 3, 2008

### rocomath

Your attachment needs to be approved. Either way, it's still pretty dangerous to open word attachments.

Type it up and we'll take a look at it.

3. Feb 3, 2008

### HallsofIvy

Staff Emeritus

4. Feb 3, 2008

### kuahji

You can plug your answer back into the equation to see if it works out. Its absolutely correct & absolutely wrong all at the same time. Can you figure out why? Plug your values back in & see why I said what I said. Your first step contains an algebra error, also think factoring for the first step.

5. Feb 5, 2008

### scientist

Is this correct?

3x + x*sqrt(5) = 2

solution

(3x + x*sqrt(5)) = 2

I factored out the x
x (3 + sqrt(5)) = 2

Divide both sides by sqrt(5) + 3
x (3 + sqrt(5) / (sqrt(5) + 3) = (2) / (sqrt(5) + 3)

cancel

x = (2) / (sqrt(5) + 3) final answer

Is this correct?

Last edited: Feb 5, 2008
6. Feb 5, 2008

### timon

3x + x√5 = 2

²√5 = 5^½
3x + (5^½)x = 2
~5.24x=2
x=~0.38

i get the same

Last edited: Feb 5, 2008
7. Feb 6, 2008

### HallsofIvy

Staff Emeritus
Yes, that is correct. You could make it look nicer by "rationalizing the denominator": multiply both numerator and denominator by $\sqrt{5}- 3$ and you get
[tex]x= \frac{2\sqrt{5}- 6}{2}= \sqrt{5}- 3[/itex]

By the way, none of these are really "radical" equations since they don't involve roots of the unknown number. $\sqrt{5}$ is just another number!

Oh, and there are no "tutors" here. Just us folks.