Re: Algebric problem

On 06/26/2024 08:30 AM, Jim Burns wrote:

On 6/26/2024 10:17 AM, Richard Hachel wrote:
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One commenter posed the problem:
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sqrt(3x+7)+sqrt(x+2)=1
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On the French forum,
mathematicians had some problems with this.
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However,
the solution is very simple,
and requires three lines of calculations
on a piece of paper.

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Is the problem that
some of the possible solutions presented by
three lines of calculations
do not produce equality in the original equation?
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Squaring can introduce more possible solutions.
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for each x such that  √(3x+7)+√(x+2)=1
also  √(3x+7)=1-√(x+2)
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for each x such that  √(3x+7)=1-√(x+2)
also  3x+7=1-2√(x+2)+x+2
but also
for each x such that  -√(3x+7)=1-√(x+2)
also  3x+7=1-2√(x+2)+x+2
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for each x such that
  √(3x+7)=1-√(x+2) or
  -√(3x+7)=1-√(x+2)
also
  3x+7=1-2√(x+2)+x+2  and
  x+2=-√(x+2)
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for each x such that
  √(3x+7)=1-√(x+2) or
  -√(3x+7)=1-√(x+2) or
  x+2=-√(x+2) or
  x+2=√(x+2)
also
  (x+2)²=x+2
  (x+1)(x+2)=0
  x=-1 or x=-2
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Only -1 and -2 _can be_ a solution.
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√(3⋅-1+7)+√(-1+2)=√4+√1≠1
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√(3⋅-2+7)+√(-2+2)=√1+√0=1
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Only -2 _is_ a solution.
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R.H.

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y = 3x+7
z = x+2
root y = 1 - root z && root z = 1 - root y -> 1 >= y, z >= 0
y <= 1 -> 3x+7 <= 1 -> 3x <= 8 -> x <= 8/3
z <= 1 -> x+2 <= 1 -> x < 3 -> x <= 3
y >= 0 -> 3x+7 >= 0 -> 3x >= -7 -> x >= -7/3
z >= 0 -> x+2 >= 0 -> x > -2
-> -2 <= x <= 8/3