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## Homework Statement

Find solutions to the given equality

**2. Relevant equation**

$$ x^2 +3|x-1|=1 $$

## The Attempt at a Solution

The above can be rewritten as:

$$ |x-1| = {1-x^2\over 3}$$

If my understanding of absolute values is correct, the above simply means that:

$$ x-1 = {1-x^2\over 3} \lor x-1 = -({1-x^2\over 3}) $$

For the first equation:

$$ x-1 = {1-x^2\over 3} = 3x-3=1-x^2 \iff x^2+3x-4=0 \iff x \in \{-4, 1\}$$

For the second equation:

$$ x-1 = -({1-x^2\over 3}) \iff 1-x={1-x^2\over 3} \iff x^2-3x+2 =0 \iff x \in \{1, 2\} $$

So apparently, any ## x \in \{-4, 1, 2\}## fulfills at least one of the ##\lor## stipulations for the absolute value, so I would expect either of these three values to fulfill the original equation. However, only ##x=1## is correct - the intersection of the solutions to the two absolute value equations.

What am I missing?

As an example, if we were solving ##|x-1|=5##, this would mean that either ##x-1=5 \lor x-1=-5##. This means that ##x=6 \lor x=-4## and I expect either of these solutions to solve the original equation, not just the intersection of the two solutions, which in this case would be ##\emptyset##.

I'm probably missing something extremely obvious here.

EDIT:

Is this because, in the original equation ## |x-1| = {1-x^2\over 3}##, the domain is restricted to values where ##{1-x^2\over 3} \geq 0##? This would restrict the domain to ##-1 \geq x \leq 1##, which explains the result.

Thanks

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