Pedantic On Mathematical Software (Middle School Algebra)
This time Maple get it wrong while Wolfram get it flawlessly. Consider the following middle-school level problem:
Solve the following equation on x
\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1
Solution:
The equation is equivalent to \sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3 \right)^2}=1 so that we have
\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1
We divide into three cases:
i) When \sqrt{x-1}< 2 then we have \sqrt{x-1}< 3 so that the above equation becomes
2-\sqrt{x-1}+3-\sqrt{x-1}=1
that is \sqrt{x-1}=2, contradicting \sqrt{x-1}< 2.
ii) When \sqrt{x-1}> 3 then we have \sqrt{x-1}> 2 so that the equation becomes
\sqrt{x-1}-2+\sqrt{x-1}-3=1
that is \sqrt{x-1}=3, contradicting \sqrt{x-1}> 3.
ii) When 2\leq \sqrt{x-1}\leq 3 then we have
\sqrt{x-1}-2+3-\sqrt{x-1}=1
which is equivalent to 1=1, and so all x on this case are the solutions, those are x\in [5,10].
However, if you input the command
solve(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}-1,x)
into Maple 16, it gives you an output
x=x
Other math software like Wolfram Mathematica gives us the correct answer 5\leq x \leq 10.
Using the method of squaring you'll get 1=1 as well.
1 comments :
It will be very interesting if you see the algorithm behind Maple 16 and point out some parts to improve. :)
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