8.3 Equations Involving Absolute Value
Recall that \(\left |a\right |\) is the distance from \(a\) to \(0\), so \(\left |a\right |=b\) with \(b>0\) says that \(a\) is at distance \(b\) from the origin — and there are two such points.
Theorem 8.9. For \(b>0\), \[\left |a\right |=b \iff a=b \ \text { or } \ a=-b .\] If \(b=0\) the only solution is \(a=0\), and if \(b<0\) there is no solution.
Solution. By Theorem 8.9 the equation splits into two: \[2x-5=7 \quad \implies \quad x=6, \qquad \text {or}\qquad 2x-5=-7 \quad \implies \quad x=-1 .\] Both satisfy the original, so the solutions are \(x=6\) and \(x=-1\).
Solution. The right side is not a constant, so it may be negative and the check matters. Splitting, \[x-1=2x+3 \quad \implies \quad x=-4, \qquad \text {or}\qquad x-1=-(2x+3) \quad \implies \quad 3x=-2 \quad \implies \quad x=-\tfrac {2}{3}.\] Checking: at \(x=-4\) the left side is \(\left |-5\right |=5\) and the right is \(-8+3=-5\); these differ, so \(x=-4\) is extraneous. At \(x=-\frac {2}{3}\) the left side is \(\left |-\frac {5}{3}\right |=\frac {5}{3}\) and the right is \(-\frac {4}{3}+3=\frac {5}{3}\); genuine.
The only solution is \(x=-\frac {2}{3}\).
Note 8.12. This is the same hazard as squaring, in a different costume. An absolute value is never negative, so any solution must also satisfy \(2x+3\geq 0\), that is \(x\geq -\frac {3}{2}\) — which rules out \(x=-4\) without any substitution. Splitting the modulus, like squaring, produces candidates; the original equation decides.
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