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Math Help - Solving for x in radicals

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    Senior Member vaironxxrd's Avatar
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    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.
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    Re: Solving for x in radicals

    Quote Originally Posted by vaironxxrd View Post
    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 ???
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    Senior Member vaironxxrd's Avatar
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    Re: Solving for x in radicals

    Quote Originally Posted by skeeter View Post
    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?
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    Re: Solving for x in radicals

    Quote Originally Posted by vaironxxrd View Post
    Your method is the right one , wouldn't x = 6?
    no.

    cube both sides to eliminate the radical, then solve.
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    Re: Solving for x in radicals

    Quote Originally Posted by skeeter View Post
    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
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    Re: Solving for x in radicals

    that's all there is to it.
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    Senior Member vaironxxrd's Avatar
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    Re: Solving for x in radicals

    Quote Originally Posted by skeeter View Post
    that's all there is to it.
    That was fairly simple.
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    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 ??
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    Re: Solving for x in radicals

    Quote Originally Posted by psolaki View Post
    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....
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    Senior Member vaironxxrd's Avatar
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    Re: Solving for x in radicals

    Quote Originally Posted by Wilmer View Post
    You got to be joking....
    I had to laugh... .
    Even after I myself failed solving this.
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    Re: Solving for x in radicals

    Quote Originally Posted by psolaki View Post
    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}.
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    Re: Solving for x in radicals

    Quote Originally Posted by HallsofIvy View Post
    ...not \sqrt[3]{x+ 1= 7}.
    Ah geeee Halls...yer spoilin' everybody's fun !
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    Re: Solving for x in radicals

    Quote Originally Posted by HallsofIvy View Post
    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??
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