Solving for x in radicals

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October 19th 2011, 04:41 PM
vaironxxrd

Solving for x in radicals

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....

I had to laugh... (Rofl).
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)