3x = Square root of 3 + x

**el123** · July 31st 2009, 04:48 PM

Can someone please let me know if i done this correctly?
And if there is a faster way I'd love to know.
$3x = \sqrt{3+x}$

$3x^2 = (3+x)^2$

$3x^2 = 3^2 + x^2$

$3x^2 - x^2 = 3^2$

$2x^2 = 3^2$

$\frac{2x^2}{2} = \frac{9}{2}$

$x^2 = 4.5$

$x = 2.121$

**VonNemo19** · July 31st 2009, 05:03 PM

Originally Posted by el123

Can someone please let me know if i done this correctly?...

No, you have not.

Your second step is wrong. It should be

$(3x)^2=(\sqrt{3+x})^2$

$9x^2=3+x$

S0me of your remaining steps are questionable as well.

**yeongil** · July 31st 2009, 05:04 PM

(Psst, this should have been posted in the Pre-Algebra/Algebra subforum.)

Originally Posted by el123

Can someone please let me know if i done this correctly?
And if there is a faster way I'd love to know.
$3x = \sqrt{3+x}$

${\color{red}3x^2 = (3+x)^2}$

Oops! That's not how to do it. You have to square both sides this way:

$(3x)^2 = (\sqrt{3+x})^2$

On the left side, both the 3 and x get squared. On the right side, the square-root and the power of 2 "cancel-out":

$9x^2 = 3 + x$

$9x^2 - x - 3 = 0$

Then use the quadratic formula to solve for x.

01

EDIT: Beaten to it...

**Wilmer** · July 31st 2009, 07:24 PM

Originally Posted by el123

$3x = \sqrt{3+x}$
$3x^2 = (3+x)^2$

Hmmm....do you have a math teacher? Or are you self-learning?

The question is asked for this reason:
it is easier to help someone if I know whether the person asking for help
is a classroom student or someone learning on his/her own.

**AlephZero** · July 31st 2009, 08:31 PM

Originally Posted by el123

$x^2 = 4.5$

$x = 2.121$

Even though others have pointed out an earlier mistake you made, I just want to let you know that there's a problem with this step, also. When taking the square root of both sides of an equation, you need to use the plus-or-minus sign, unless the question specifically tells you the answer is positive. Hence

$x^2=4.5 \implies x\approx\pm 2.121$ .

Of course, as I and others have said, this is still not the correct value of $x$ for your problem; I just wanted to point out another place you went wrong in the hope of being helpful.

**yeongil** · July 31st 2009, 08:39 PM

Originally Posted by el123

$3x^2 = (3+x)^2\;\;\;\;$ (2)

$3x^2 = 3^2 + x^2\;\;\;\;$ (3)

Also, ignoring the fact that you arrived at step 2 incorrectly, when going from step 2 to step 3, you need to FOIL.
$(3 + x)^2 \neq 3^2 + x^2$

$(3 + x)^2 = 3^2 + 3x + 3x + x^2 = 9 + 6x + x^2$

For the love of (insert deity here), please remember this next time. I can't tell you how many times I've seen people make the same mistake. It's a pet peeve of mine. Arrrrgh!

01

**el123** · August 1st 2009, 01:56 AM

Thanks for the help. Was a mistake , i should've seen what i had done wrong. Haven't done math in such a long time. Sometimes i just can't see where i've gone wrong. Hence why i asked for help.

**songoku** · August 1st 2009, 02:21 AM

Hi el123

That's ok. This is indeed where you can seek for help ^^

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