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Behind You

07/08/2017 at 11:59
Answers
2
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Solve : \(\sqrt{2x^2+1}+\sqrt{x^2-3x+\dfrac{17}{2}}=\sqrt[3]{45+29\sqrt{2}}-\sqrt[3]{40+14\sqrt{2}}\)




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  • ...
    WhySoSerious 07/08/2017 at 13:42

    Shorten right side : \(\sqrt[3]{45+29\sqrt{2}}-\sqrt[3]{40+14\sqrt{2}}\)

    \(=\sqrt[3]{2\sqrt{2}+3\left(\sqrt{2}\right)^2\cdot3+3\sqrt{2}\cdot9+27}-\sqrt[3]{2\sqrt{2}+3\left(\sqrt{2}\right)^2\cdot2+3\sqrt{2}\cdot4+8}\)

    \(=\sqrt[3]{\left(\sqrt{2}+3\right)^3}-\sqrt[3]{\left(\sqrt{2}+2\right)^3}=\sqrt{2}+3-\left(\sqrt{2}+2\right)=1\)

    So the expression is equal : \(\sqrt{2x^2+1}+\sqrt{x^2-3x+\dfrac{17}{2}}=1\)

    \(\Leftrightarrow\sqrt{2x^2+1}+\sqrt{\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}}=1\)

    The expression has no solution cause the left side \(>\sqrt{2x^2+1}\ge1\).

    Behind You selected this answer.
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    Behind You 08/08/2017 at 14:51

    Thank you !


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