# Solving for x in radicals

• October 19th 2011, 04:41 PM
vaironxxrd
I don't understand too well how this works

$\sqrt[3]{x+1=7}$

If I did that wrong please correct me . I saw it on my test today and did not understand it.
• October 19th 2011, 04:48 PM
skeeter
Re: Solving for x in radicals
Quote:

Originally Posted by vaironxxrd
I don't understand too well how this works

$\sqrt[3]{x+1=7}$

If I did that wrong please correct me . I saw it on my test today and did not understand it.

are you sure the entire equation was under the radical? or did you mean this ...

$\sqrt[3]{x+1}=7$ ???
• October 19th 2011, 04:57 PM
vaironxxrd
Re: Solving for x in radicals
Quote:

Originally Posted by skeeter
are you sure the entire equation was under the radical? or did you mean this ...

$\sqrt[3]{x+1}=7$ ???

Your method is the right one , wouldn't x = 6?
• October 19th 2011, 05:03 PM
skeeter
Re: Solving for x in radicals
Quote:

Originally Posted by vaironxxrd
Your method is the right one , wouldn't x = 6?

no.

cube both sides to eliminate the radical, then solve.
• October 19th 2011, 05:17 PM
vaironxxrd
Re: Solving for x in radicals
Quote:

Originally Posted by skeeter
no.

cube both sides to eliminate the radical, then solve.

$\sqrt[3]{x+1} = 7$

$(\sqrt[3]{x+1})^3 = (7)^3$

X+1 = 343
X= 342
• October 19th 2011, 05:27 PM
skeeter
Re: Solving for x in radicals
that's all there is to it.
• October 19th 2011, 06:32 PM
vaironxxrd
Re: Solving for x in radicals
Quote:

Originally Posted by skeeter
that's all there is to it.

That was fairly simple.
• October 19th 2011, 06:35 PM
psolaki
Re: Solving for x in radicals
The question now is why should not have in mathematics formulas like:

$\sqrt[3]{x+1=7}$ and then taking to the cube the whole equation get an x=6??

Or even better solve 1st for x under the sqrt and then take 1/3 of 6 ??
• October 19th 2011, 06:42 PM
Wilmer
Re: Solving for x in radicals
Quote:

Originally Posted by psolaki
The question now is why should not have in mathematics formulas like:
$\sqrt[3]{x+1=7}$ and then taking to the cube the whole equation get an x=6??
Or even better solve 1st for x under the sqrt and then take 1/3 of 6 ??

You got to be joking....
• October 19th 2011, 07:01 PM
vaironxxrd
Re: Solving for x in radicals
Quote:

Originally Posted by Wilmer
You got to be joking....

Even after I myself failed solving this.
• October 20th 2011, 05:31 AM
HallsofIvy
Re: Solving for x in radicals
Quote:

Originally Posted by psolaki
The question now is why should not have in mathematics formulas like:

$\sqrt[3]{x+1=7}$ and then taking to the cube the whole equation get an x=6??

Or even better solve 1st for x under the sqrt and then take 1/3 of 6 ??

Because functions, such as the square root, are defined on numbers, not equations. We sometimes talk about "taking the cube root" (or whatever other function you like) but what we mean is "take the cube root of each side".

You can for example, start with the equation, x+ 1= 7 and then take the cube root of both sides but that gives $\sqrt[3]{x+1}= \sqrt[3]{7}$, not $\sqrt[3]{x+ 1= 7}$.
• October 20th 2011, 05:47 AM
Wilmer
Re: Solving for x in radicals
Quote:

Originally Posted by HallsofIvy
...not $\sqrt[3]{x+ 1= 7}$.

Ah geeee Halls...yer spoilin' everybody's fun !
• October 20th 2011, 06:05 AM
Youkla
Re: Solving for x in radicals
Quote:

Originally Posted by HallsofIvy
Because functions, such as the square root, are defined on numbers, not equations. We sometimes talk about "taking the cube root" (or whatever other function you like) but what we mean is "take the cube root of each side".

You can for example, start with the equation, x+ 1= 7 and then take the cube root of both sides but that gives $\sqrt[3]{x+1}= \sqrt[3]{7}$, not $\sqrt[3]{x+ 1= 7}$.

HallsofIvy, please tell me you did not write all this with a straight face??(Rofl)