Problem 11

Find all numbers $x$ for which: (i) \[ \begin{align*} |x - 3| &= 8 \\ \sqrt{(x - 3)^2} &= 8 \\ \pm(x - 3) &= 8 \\ \\ -x_1 + 3 &= 8 \\ x_1 &= -5 \\ \\ x_2 - 3 &= 8 \\ x_2 &= 11 \\ \end{align*} \]

(ii) \[ \begin{align*} |x - 3| &< 8 \\ \sqrt{(x - 3)^2} &< 8 \\ \pm(x - 3) &< 8 \\ \\ -x_1 + 3 &< 8 \\ x_1 &> -5 \\ \\ x_2 - 3 &< 8 \\ x_2 &< 11 \\ \end{align*} \]

(iii)* \[ \begin{align*} |x + 4| &< 2 \\ \pm(x + 4) &< 2 \\ \\ -x_1 - 4 &< 2 \\ x_1 &> -6 \\ \\ x_2 + 4 &< 2 \\ x_2 &< -2 \\ \end{align*} \]

(iv) \[ \begin{align*} |x - 1| + |x + 1| &< 2 \\ \pm(x - 1) + \pm(x + 1) &< 2 \\ \\ -x_1 + 1 - x_1 + 2 &> 1 \\ -2x_1 + 3 &> 1 \\ 2x_1 &< 2 \\ x_1 &< 1 \\ \\ -x_2 + 1 + x_2 - 2 &> 1 \\ -1 &> 1 && \text{| contradiction} \\ \\ x_3 - 1 - x_3 + 2 &> 1 \\ 1 &> 1 && \text{| contradiction} \\ \\ x_4 - 1 + x_4 - 2 &> 1 \\ 2x_4 - 3 &> 1 \\ x_4 &> 2 \\ \end{align*} \] $x_2$ and $x_3$ are an impossibility and therefore only $x_1$ and $x_4$ are left.

(v) \[ \begin{align*} |x - 1| + |x + 1| &< 2 \\ \pm(x - 1) + \pm(x + 1) &< 2 \\ \\ -x_1 + 1 - x_1 - 1 &< 2 \\ x_1 &> -1 \\ \\ x_2 - 1 + x_2 + 1 &< 2 \\ x_2 &< 1 \\ \end{align*} \]

(vi) \[ \begin{align*} |x - 1| + |x + 1| &< 1 \\ \pm(x - 1) + \pm(x + 1) &< 1 \\ \\ -x_1 + 1 - x_1 - 1 &< 1 \\ x_1 &> -\frac{1}{2} \\ \\ x_2 - 1 + x_2 + 1 &< 1 \\ x_2 &< \frac{1}{2} \\ \end{align*} \]

(vii) \[ \begin{align*} |x - 1| \cdot |x + 1| &= 0 \\ \pm(x - 1) \cdot \pm(x + 1) &= 0 \\ (x - 1)(x + 1) &= 0 && \text{| the signs can be removed by multiplication with } -1 \\ \\ x_1 - 1 &= 0 \\ x_1 &= 1 \\ \\ x_2 &= -1 \\ \end{align*} \]

(viii) \[ \begin{align*} |x - 1| \cdot |x + 2| &= 3 \\ \pm(x - 1) \cdot \pm(x + 2) &= 3 \\ \\ (-1)(x_1 - 1) \cdot (-1)(x_1 + 2) &= 3 \\ (x_1 - 1)(x_1 + 2) &= 3 \\ x_1^2 + x_1 - 2 &= 3 \\ x_1^2 + x_1 - 5 &= 0 && \text{| quadratic equation} \\ x_{1_{12}} &= \frac{-1 \pm\sqrt{1 - 4 \cdot 1 \cdot -5}}{2} \\ x_{1_1} &= \frac{-1 -\sqrt{21}}{2} \\ x_{1_2} &= \frac{-1 +\sqrt{21}}{2} \\ \\ (-1)(x_2 - 1) \cdot (x_2 + 2) &= 3 \\ (-x_2 + 1)(x_2 + 2) &= 3 \\ -x_2^2 - x_2 + 2 &= 3 \\ -x_2^2 - x_2 - 1 &= 0 && \text{| quadratic equation} \\ x_{2_1} &= \frac{-1 + \sqrt{-3}}{2} \\ x_{2_2} &= \frac{-1 - \sqrt{-3}}{2} \\ \\ (x_3 - 1) \cdot (-1)(x_3 + 2) &= 3 \\ (x_3 - 1)(-x_3 - 2) &= 3 \\ -x_3^2 - x_3 + 2 &= 3 && \text{| identical to } x_2 \\ \\ (x_4 - 1) \cdot (x_4 + 2) &= 3 \\ x_4^2 + x_4 - 2 &= 3 && \text{| identical to } x_1 \\ \end{align*} \]

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