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Math Help - Equation

  1. #1
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    Equation

    \frac{\sqrt{3-2a}+4\sqrt{a}}{\sqrt{1+2a}+4\sqrt{1-a}}=\frac{\sqrt{-2a^{2}+3a}}{\sqrt{-2a^{2}+a+1}}, a is a real number.

    What I did:
    \frac{\sqrt{3-2a}+4\sqrt{a}}{\sqrt{1+2a}+4\sqrt{1-a}}=\frac{\sqrt{-2a^{2}+3a}}{\sqrt{-2a^{2}+a+1}} \Leftrightarrow \frac{\sqrt{3-2a}+4\sqrt{a}}{\sqrt{-2a^{2}+3a}}=\frac{\sqrt{1+2a}+4\sqrt{1-a}}{\sqrt{-2a^{2}+a+1}}
    \Leftrightarrow \frac{\sqrt{1+2(1-a))}+4\sqrt{1-(1-a)}}{\sqrt{[1+2(1-a)][1-(1-a)]}}=\frac{\sqrt{1+2a}+4\sqrt{1-a}}{\sqrt{(1+2a)(1-a)}}<br />

    Let 0, 1)\mapsto \mathbb{R}, f(x)=\frac{\sqrt{1+2x}+4\sqrt{1-x}}{\sqrt{(1+2x)(1-x)}}\Rightarrow \displaystyle f(1-a)=f(a)" alt="\displaystyle f0, 1)\mapsto \mathbb{R}, f(x)=\frac{\sqrt{1+2x}+4\sqrt{1-x}}{\sqrt{(1+2x)(1-x)}}\Rightarrow \displaystyle f(1-a)=f(a)" />. I want to prove that f is injective/ strictly monotone.

    Well, that is my idea, but anything else would be okay. Thanks in advance.
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  2. #2
    Moo
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    Hello,

    If you're trying to prove that there is a unique solution, then your method is good. And it's easy to prove because \displaystyle f(x)=\frac{1}{\sqrt{1-x}}+\frac{4}{1+2x}

    If you want to find a, then that's more complicated I guess can you first confirm what you are looking for ?
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  3. #3
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    I want to find a; if f is injective/ strictly monotone, then 1-a=a \Leftrightarrow a=\frac{1}{2} which verify the equation.
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  4. #4
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    Well, let u = 3a - 2a^2 and v = a + 1 - 2a^2, then:

    v[3 + 14a +8sqrt(u)] = u[17 - 14a +8sqrt(v)]
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  5. #5
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    (\sqrt{v}-\sqrt{u})[8\sqrt{uv}-7(u+v)(\sqrt{u}+\sqrt{v})+10(\sqrt{u}+\sqrt{v})]=0 \Rightarrow
    I. \sqrt{v}-\sqrt{u}=0 \Rightarrow a=\frac{1}{2}
    II. 8\sqrt{uv}-7(u+v)(\sqrt{u}+\sqrt{v})+10(\sqrt{u}+\sqrt{v})=0 \Rightarrow 8\sqrt{uv}+(\sqrt{u}+\sqrt{v})[7(2a-1)^{2}-4]=0
    And now?
    8\sqrt{uv}\geq 0, (\sqrt{u}+\sqrt{v})\geq 0, but 7(2a-1)^{2}-4 \epsilon (-4, 3)
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