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Find the number of integers x satisfying \(\sqrt[3]{24 + x} + \sqrt{12 - x} \le 6\)

 Jun 15, 2020
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Find the number of integers x satisfying \(\sqrt[3]{24 + x} + \sqrt{12 - x} \le 6 \)

 

Hello Guest!

 

  \(\sqrt[3]{24 + x} + \sqrt{12 - x} \le 6 \)

 

    216   192       36    -24      6  

    125    101       25   -13      5 

    64      40        16    -4        4

    8       -16          9    -3        3

    1       -23           4     8       2

    0       -24          1    11       1

                             0   12        0

 

\(\sqrt[3]{24 + x} + \sqrt{12 - x} \le 6 \)   

 

\(\large x=-24\)

   

\(\sqrt[3]{24 +(-24)} + \sqrt{12 - (-24)} \le 6\\ \sqrt[3]{0} + \sqrt{36} \le 6 \)   

laugh  !

 Jun 15, 2020
edited by asinus  Jun 15, 2020

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