Loading [MathJax]/jax/output/HTML-CSS/jax.js

Saturday, 24 November 2012


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. :)

Reply