Given equation x + 3 − 2 x + 2 + x + 27 − 10 x + 2 = 4 , find its solution

Question

Given equation       x    +    3    −    2          x      +      2        +      x    +    27    −    10          x      +      2        =  4 , find its solution

upornompe · Accepted Answer

Step 1Solving the equation via substitution   t  =      x    +    2   we get:      x    +    3    −    2    t    +      x    +    27    −    10    t    =  4  x  +  3  −  2  t  =  16  −  8      x    +    27    −    10    t    +  x  +  27  −  10  tHere, you have to have      x    +    3    −    2    t    =  4  −      x    +    27    −    10    t        ≥    0        t    2    −  10  t  +  25  =  x  +  27  −  10  tHere, you have to have      x    +    27    −    10    t    =  5  −  t      ≥    0  By the way, the equation can be written as      (          x      +      2        −    1          )      2        +      (          x      +      2        −    5          )      2        =  4  , ,i.e.      |        x    +    2    −  1      |    +      |        x    +    2    −  5      |    =  4which should be easy to deal with.

Emmy Dillon · Accepted Answer

Step 1Put   t  =      x    +    2   , so we require   t  ≥  0. Then       x    +    3    −    2    t    =            t      2        +    1    −    2    t    =      (    t    −    1          )      2        =      |    t  −  1      |   . Similarly,       x    +    27    −    10    t    =      (    t    −    5          )      2        =      |    t  −  5      |   . So we have       |    t  −  1      |    +      |    t  −  5      |    =  4 . That holds for   t  ∈  [  1  ,  5  ] .So we must have   1  ≤      x    +    2    ≤  5 and hence   1  ≤  x  +  2  ≤  25 , so   −  1  ≤  x  ≤  23 .

Given equation <msqrt> x + 3 − 2 <msqrt> x

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