How to solve this basic C1 problem?

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    a) x^1/2 + 2x^-1/2 = 3

    b) x + x^1/2 = 12

    I know the answers because I did trial and error in my head. If I am do to lines of working out, what steps lead to the answer?
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    (Original post by Baaah)
    a) x^1/2 + 2x^-1/2 = 3

    b) x + x^1/2 = 12

    I know the answers because I did trial and error in my head. If I am do to lines of working out, what steps lead to the answer?
    Let A=x^{1/2}

    Your part a turns into A+2A^{-1}=3 then multiply through by A and solve the quadratic in A. Then once you have A=... turn it into x^{1/2}=... and get x.

    Same with part b. It would turn into A^2+A=12
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    (Original post by Baaah)
    a) x^1/2 + 2x^-1/2 = 3

    b) x + x^1/2 = 12

    I know the answers because I did trial and error in my head. If I am do to lines of working out, what steps lead to the answer?
    a) To solve x^1/2 + 2x^-1/2 = 3

    let Y = x^(1/2)

    remember 2x^(-1/2) = 2/x^(1/2)

    ; Y + 2/Y = 3

    *Y ; Y^2 + 2 = 3Y

    ; Y^2 - 3Y + 2 =0

    ; (Y - 1)(Y - 2) = 0

    ; Y = 1 , Y = 2

    Substitute Y = 1 and Y = 2 into Y = x^(1/2)

    ; 1 = x^(1/2) ; 2 = x^(1/2)

    ; x = 1 ; x = 4

    b) To solve x + x^1/2 = 12

    let U = x^1/2

    remember x = (x^1/2)^2

    therefore ; U^2 + U = 12

    ; U^2 + U - 12 = 0

    ; (U + 4)(U - 3)

    ; U = -4 , U = 3

    substitute U = -4 and U = 3 into U = x^1/2

    ; -4 = x^1/2 ; 3 = x^1/2

    ; x = no solution ; x = 9


    Hope this helps
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    (Original post by Pixel Warrior)
    a) To solve x^1/2 + 2x^-1/2 = 3

    let Y = x^(1/2)

    remember 2x^(-1/2) = 2/x^(1/2)

    ; Y + 2/Y = 3

    *Y ; Y^2 + 2 = 3Y

    ; Y^2 - 3Y + 2 =0

    ; (Y - 1)(Y - 2) = 0

    ; Y = 1 , Y = 2

    Substitute Y = 1 and Y = 2 into Y = x^(1/2)

    ; 1 = x^(1/2) ; 2 = x^(1/2)

    ; x = 1 ; x = 4

    b) To solve x + x^1/2 = 12

    let U = x^1/2

    remember x = (x^1/2)^2

    therefore ; U^2 + U = 12

    ; U^2 + U - 12 = 0

    ; (U + 4)(U - 3)

    ; U = -4 , U = 3

    substitute U = -4 and U = 3 into U = x^1/2

    ; -4 = x^1/2 ; 3 = x^1/2

    ; x = no solution ; x = 9


    Hope this helps
    Damn, can I get you to do my homework for me as well???
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    (Original post by Pixel Warrior)
    a) To solve x^1/2 + 2x^-1/2 = 3

    let Y = x^(1/2)

    remember 2x^(-1/2) = 2/x^(1/2)

    ; Y + 2/Y = 3

    *Y ; Y^2 + 2 = 3Y

    ; Y^2 - 3Y + 2 =0

    ; (Y - 1)(Y - 2) = 0

    ; Y = 1 , Y = 2

    Substitute Y = 1 and Y = 2 into Y = x^(1/2)

    ; 1 = x^(1/2) ; 2 = x^(1/2)

    ; x = 1 ; x = 4

    b) To solve x + x^1/2 = 12

    let U = x^1/2

    remember x = (x^1/2)^2

    therefore ; U^2 + U = 12

    ; U^2 + U - 12 = 0

    ; (U + 4)(U - 3)

    ; U = -4 , U = 3

    substitute U = -4 and U = 3 into U = x^1/2

    ; -4 = x^1/2 ; 3 = x^1/2

    ; x = no solution ; x = 9


    Hope this helps
    2x^(-1/2) = 2/x^(1/2)
    I thought 2x^(-1/2) = 1/2x^(1/2) ?
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    nope
    (Original post by Baaah)
    2x^(-1/2) = 2/x^(1/2)
    I thought 2x^(-1/2) = 1/2x^(1/2) ?
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    (Original post by Baaah)
    2x^(-1/2) = 2/x^(1/2)
    I thought 2x^(-1/2) = 1/2x^(1/2) ?
    No that's wrong

    by breaking it up you see that

    2x^(-1/2) = 2 * x^(-1/2)

    = 2 * 1/x^(1/2)

    = 2/x^(1/2)
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    (Original post by Pixel Warrior)
    No that's wrong

    by breaking it up you see that

    2x^(-1/2) = 2 * x^(-1/2)

    = 2 * 1/x^(1/2)

    = 2/x^(1/2)
    Oh I see

    I was confused because I thought a negative index meant I just have to make the whole base a reciprocal
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    (Original post by Baaah)
    2x^(-1/2) = 2/x^(1/2)
    I thought 2x^(-1/2) = 1/2x^(1/2) ?
    2x^{\frac{-1}{2}} = 2 \times x^{\frac{-1}{2}} = 2 \times \dfrac{1}{x^{\frac{1}{2}}} = \dfrac{2}{x^{\frac{1}{2}}}

    Ahh, beaten to it. Never mind...
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    (Original post by K-Man_PhysCheM)
    2x^{-1/2} = 2 \times x^{-1/2} = 2 \times \dfrac{1}{x^{1/2}} = \dfrac{2}{x^{1/2}}

    Ahh, beaten to it. Never mind...
    Well thanks anyway

    No wonder I kept getting it wrong...
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    (Original post by Pixel Warrior)
    a) To solve x^1/2 + 2x^-1/2 = 3

    let Y = x^(1/2)

    remember 2x^(-1/2) = 2/x^(1/2)

    ; Y + 2/Y = 3

    *Y ; Y^2 + 2 = 3Y

    ; Y^2 - 3Y + 2 =0

    ; (Y - 1)(Y - 2) = 0

    ; Y = 1 , Y = 2

    Substitute Y = 1 and Y = 2 into Y = x^(1/2)

    ; 1 = x^(1/2) ; 2 = x^(1/2)

    ; x = 1 ; x = 4

    b) To solve x + x^1/2 = 12

    let U = x^1/2

    remember x = (x^1/2)^2

    therefore ; U^2 + U = 12

    ; U^2 + U - 12 = 0

    ; (U + 4)(U - 3)

    ; U = -4 , U = 3

    substitute U = -4 and U = 3 into U = x^1/2

    ; -4 = x^1/2 ; 3 = x^1/2

    ; x = no solution ; x = 9


    Hope this helps
    Sorry to bother you again

    -4 = x^1/2
    Wouldn't that mean x has a solution of 16 because (-4)^2 = 16? Why is there no solution :?
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    (Original post by Baaah)
    Sorry to bother you again

    -4 = x^1/2
    Wouldn't that mean x has a solution of 16 because (-4)^2 = 16? Why is there no solution :?
    I guess that since sqrt(16) gives +/- 4 you are correct, but you should ask a teacher or something as its been a while since i've done this.

    And no probs, happy to help
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    (Original post by Baaah)
    Sorry to bother you again

    -4 = x^1/2
    Wouldn't that mean x has a solution of 16 because (-4)^2 = 16? Why is there no solution :?
    No because \sqrt{x} is always greater than or equal to 0.

    Besides, 16+4\not= 12 when you plug it back through the equation.
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    (Original post by Pixel Warrior)
    I guess that since sqrt(16) gives +/- 4 you are correct, but you should ask a teacher or something as its been a while since i've done this.

    And no probs, happy to help
    \sqrt{16} \not= \pm 4

    \Rightarrow \sqrt{16}=4
 
 
 
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