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Thread: Algebra Question. Involving Exponents

  1. #1
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    Unhappy Algebra Question. Involving Exponents

    I Got this Question from my maths teacher.. Plz help.!!
    X^1/3 + x^1/2 =12

    I got the answer 64.. but by hit n trial, a meaningful one at that..but still show me how to solve it..PLZ!

    TY!
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  2. #2
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    Hello, umangarora!

    $\displaystyle \text{Solve for }x\!:\;\;x^{\frac{1}{2}} + x^{\frac{1}{3}}\:=\:12$

    Let $\displaystyle x^{\frac{1}{6}} \:=\:u \quad\Rightarrow\quad x^{\frac{1}{2}} \:=\:u^3 \quad\Rightarrow\quad x^{\frac{1}{3}} \:=\:u^2$

    Then we have: .$\displaystyle u^3 + u^2 - 12 \:=\:0$


    If this cubic has any rational roots,
    . . they must be among: .$\displaystyle \pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12$

    We find that $\displaystyle u = 2$ is a root.

    Then: .$\displaystyle u^3 + u^2 - 12 \:=\:(u-2)(u^2 + 3u + 6)\:=\:0$
    . . and the quadratic factor has no real roots.

    So we have: .$\displaystyle u \,=\,2 \quad\Rightarrow\quad x^{\frac{1}{6}} \,=\,2 \quad\Rightarrow\quad x \:=\:64$
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  3. #3
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    Quote Originally Posted by umangarora View Post
    I Got this Question from my maths teacher.. Plz help.!!
    X^1/3 + x^1/2 =12

    I got the answer 64.. but by hit n trial, a meaningful one at that..but still show me how to solve it..PLZ!

    TY!
    A slightly simpler way (than working with rational powers) to solve is...

    $\displaystyle \displaystyle\ x^{\frac{1}{3}}+x^{\frac{1}{2}}=12$

    $\displaystyle \displaystyle\ y=x^{\frac{1}{3}}$

    $\displaystyle \displaystyle\ y+y^{\frac{3}{2}}=12$

    $\displaystyle \displaystyle\ y(1+\sqrt{y})=12$

    $\displaystyle w=\sqrt{y}$

    $\displaystyle w^2(1+w)=12$

    $\displaystyle f(w)=w^3+w^2-12=0$

    $\displaystyle f(2)=0$

    $\displaystyle w=2\Rightarrow\ y=w^2=4\Rightarrow\ x=y^3=64$
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  4. #4
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    exactly Guys..
    what u have done is factored.. but u did use hit and trial somewhere.. is there NO proof which boils down to x=64..?
    Do u have to put 2 as u or x^1/3
    to get the answer :S ?
    P.s. i did the same thing.
    what i did is:
    x^1/2 + x^1/3 =12
    since rhs is rational lhs has to be rational.
    there for x has to be m^3 and n^ 2
    min. value is when x=m^6
    trying 1, we get 2.
    trying 2 we get the answer.
    trying 3, we exceed.

    But since the Q includes cube root, we must try negative value,
    trying -1, we get 0
    trying -2, we get -4
    trying -3, we exceed.
    so 2 is the only real ans.
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  5. #5
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    Quote Originally Posted by umangarora View Post
    exactly Guys..
    what u have done is factored.. but u did use hit and trial somewhere.. is there NO proof which boils down to x=64..?
    Do u have to put 2 as u or x^1/3
    to get the answer :S ?
    P.s. i did the same thing.
    what i did is:
    x^1/2 + x^1/3 =12
    since rhs is rational lhs has to be rational.
    there for x has to be m^3 and n^ 2
    min. value is when x=m^6
    trying 1, we get 2.
    trying 2 we get the answer.
    trying 3, we exceed.

    But since the Q includes cube root, we must try negative value,
    trying -1, we get 0
    trying -2, we get -4
    trying -3, we exceed.
    so 2 is the only real ans.
    Apart from the techniques to solve cubics,

    using a little logic, we can say for $\displaystyle w^3+w^2=12$

    $\displaystyle |w|>1$

    therefore if $\displaystyle w$ is negative, then $\displaystyle w(w^2+w)$ is negative.
    Negative solutions do not apply.

    Then, if you look at $\displaystyle w(w^2+w)$

    you see that if $\displaystyle w<2,$ the product is $\displaystyle <12$

    and if $\displaystyle w>2,$ the product is $\displaystyle >12$
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