$|x-1|\cdot|x+1|=0$ and $|x-1|\cdot|x+2|=3$

Question

Can you help me with these equations? $$|x-1|\cdot |x+1|=0$$ $$|x-1|\cdot |x+2|=3$$ I don't know how to solve them.

Answer 1

Hint:

For ii), I would use this:

• $\lvert A\rvert \lvert B\rvert= \lvert A B\rvert$,
• $\lvert A\rvert=a \;(\ge0)\iff (A =a\;\text{or}\;A=-a)$.

Answer 2

(i) $|x-1|=0$ or $|x+1|=0$

Then $x=\pm1$

(ii)

1)If $x<-2$ then $|x-1|\cdot|x+2|=-(x-1)\cdot(-(x+2))$. Then $$(x-1)(x+2)=3$$ $$x^2+x-5=0$$

2) $-2\le x \le 1$. Similarly

3) $x> 1$. Similarly

Answer 3

i) $|x-1|\cdot |x+1|=0$ iff $|x-1|=0$ or $|x+1|=0$ that is $x\in\{1,-1\}$

ii) $|x-1|\cdot |x+2|=|(x-1)(x+2)|=3$ which is equivalent to solve the quadratic equations $(x-1)(x+2)=3$ or $(x-1)(x+2)=-3$. The first equation has two solutions, namely $(-1\pm \sqrt{21})/2$ and the second one none. So the solutions of ii) are $(-1\pm \sqrt{21})/2$.