Check my solution to a radical equation

[scientist]

Could a tutor check my solution to my radical question? It is enclosed in this attachment as a word document.

 

[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.

 

[HallsofIvy]

Why don't check your own answer? Just put your answer back into the original equation and do the arithmetic.

 

[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.

 

[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?

 

[timon]

3x + x√5 = 2

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

i get the same

 

[HallsofIvy]

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?

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]<br /> <br /> By the way, none of these are really "radical" equations since they don't involve roots of the <b>unknown</b> number. $\sqrt{5}$ is just another number!<br /> <br /> Oh, and there are no "tutors" here. Just us folks.[/tex]