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Thread: Change subject

  1. #1
    Junior Member
    Jun 2009

    Change subject

    Make p the subject of the formula, given that c is an odd number.

    (z^b)(p^25c)=[2(z^3b)(p^25c)+ (z^2a) -(z^4b)]^0.5
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  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, yeoky!

    This is an ugly problem!

    Solve for $\displaystyle p$, given that $\displaystyle c$ is an odd number.

    . . $\displaystyle z^bp^{25c} \;=\; \bigg[2z^{3b}p^{25c} + z^{2a} - z^{4b}\bigg]^{\frac{1}{2}}$
    Let $\displaystyle X \:=\:p^{25c}$

    The equation becomes: .$\displaystyle z^bX \;=\;\bigg[2z^{3b}X + z^{2a} - z^{4b}\bigg]^{\frac{1}{2}} $

    Square both sides: .$\displaystyle z^{2b}X^2 \;=\;2z^{3b}X + z^{2a} - z^{4b} $

    And we have: .$\displaystyle z^{2b}X^2 - 2z^{3b}X + z^{4b} - z^{2a} \;=\;0 $

    Quadratic Formula:

    . . $\displaystyle X \;=\;\frac{2z^{3b} \pm \sqrt{(2z^{3b})^2 - 4(z^{2b})(z^{4b}-z^{2a})}}{2z^{2b}} \;=\;\frac{2z^{3b} \pm\sqrt{4z^{6b} - 4z^{6b} + 4z^{2x+2b}}}{2z^{2b}} $

    . . . . $\displaystyle = \;\frac{2z^{3b} \pm \sqrt{4z^{2a+2b}}}{2z^{2b}} \;=\; \frac{2z^{3b} \pm 2z^{a+b}}{2z^{2b}} \;=\;\frac{z^{3b} \pm z^{a+b}}{z^{2b}} \;=\;z^b \pm z^{a-b}$

    Therefore: .$\displaystyle p^{25c} \;=\;z^b \pm z^{a-b} \quad\Rightarrow\quad p \;=\;\left[z^b \pm z^{a-b}\right]^{\frac{1}{25c}} $

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  3. #3
    MHF Contributor
    Oct 2009
    Well done! And the condition that c is odd is probably used to justify that [...]^(1/25c), as a root of an odd degree, exists whether [...] is positive or negative.

    I am wondering about the expression "Make p the subject of the formula" in the original post. Is it the same as "Find the root p of the equation"? I just have not seen it before.

    (Puts "necessary/sufficient Nazi" hat on)
    Since $\displaystyle x = y$ implies $\displaystyle x^2=y^2$, but not necessarily the other way around, this solution proves only the necessary condition: If a solution to the original equation exists, then it is equal to... Squaring may lead to acquiring extra roots, so after finding the solutions for the squared equation one generally has to prove that they are also roots of the original equation, e.g., by substituting the solutions and computing both sides of the equation.

    It seems to me here that after substitution we get a full square under the square root, so it is always defined. On the other hand, the left-hand side $\displaystyle z^b p^{25c} = z^b(z^b \pm z^{a-b}) = z^{2b}\pm z^a$ may be positive or negative depending on $\displaystyle z$, $\displaystyle a$, and $\displaystyle b$. I am not sure if this can be simplified, so the complete solution would be something like $\displaystyle p=[z^b\pm z^{a-b}]^{1/25c}$ provided that $\displaystyle z^{2b}\pm z^a \ge 0$.
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  4. #4
    Junior Member
    Jun 2009
    Thanks. Soroba! It's indeed an ugly problem.
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