Split equal absolute values into the two signed linear equations. Use ordered roots and interval signs to solve a cubic inequality. Introduce a new variable, then undo the substitution completely.
Notes and key methods
For real U,V, \(|U|=|V|\) is equivalent to \(\displaystyle U=V\) or \(\displaystyle U=-V\). Solve both equations and remove duplicate roots. Squaring is also valid for this equality, but can hide the useful linear factors.
When solving an inequality, split at the zeros of the expressions inside the moduli or compare squared nonnegative sides. Test an interval, rather than carrying an equality argument over without checking the sign.
For three distinct simple real roots, a cubic changes sign at each root. The leading coefficient fixes the sign to the far right. A test value in each interval can confirm the sign chart.
A non-strict inequality includes its zero boundaries. For \(\displaystyle y=|P(x)|\), reflect only the intervals where the cubic is negative; these roots remain the meeting points of the reflected pieces.
An equation involving \(\displaystyle x^{4}\) and \(\displaystyle x^{2}\) can become a quadratic in \(\displaystyle u=x^{2}\). Solve for u before returning to x. Positive u gives two real square roots, zero gives one, and negative u gives no real x.
Other useful substitutions include \(\displaystyle u=a^{x}\), with u>0. Domain information belongs to the substituted variable and must be checked before recovering the original one.
Check the conditions
Both expressions are real.
An inequality requires sign analysis after finding its boundaries.
The three roots in these problems are distinct and simple.
Include or exclude roots according to the inequality symbol.
For a real \(\displaystyle x, x^{2}\ge 0\).
Recover every permitted square-root branch.
Three worked examples
Example 1 · Two modulus expressions
Write \(\displaystyle U=\)\(3 x + 2\) and \(\displaystyle V=\)\(5 x + 5\). Consider \(\left|{3 x + 2}\right| = \left|{5 x + 5}\right|\).
Solve the modulus equation completely.
Solve \(\displaystyle |U|\le |V|\) for these two expressions.
Find the distance between the two solutions.
Find the midpoint of the two solutions.
Show the reasoning
(a)
\(3 x + 2 = 5 x + 5\) and \(3 x + 2 = - 5 x - 5\) give the displayed two roots.
Solve \(x^{4} - 89 x^{2} + 1600 = 0\). For a substitution, write \(\displaystyle u=x^{2}\).
Solve the equation over the real numbers.
Find the larger positive real root.
Find the product of all four real roots.
Find the discriminant of the quadratic in \(\displaystyle u=x^{2}\).
Show the reasoning
(a)
\(\left(x^{2} - 64\right) \left(x^{2} - 25\right) = 0\) gives \(\displaystyle x^{2}=a^{2}\) or \(\displaystyle b^{2}\); each is positive and gives two roots.
Answer: \(\left\{-8, -5, 5, 8\right\}\)
(b)
Take the positive square root of the larger u-root.
\(8\)
Answer: \(8\).
(c)
Pair each positive root with its negative.
\(1600\)
Answer: \(1600\).
(d)
Use \(\displaystyle u^{2}-(a^{2}+b^{2})u+a^{2}b^{2}=0\).
\(1521\)
Answer: \(1521\).
Common mistakes
Solving only \(\displaystyle U=V\) loses a branch.
Do not keep a duplicated solution twice.
Do not assume a quadratic sign pattern for a cubic.
Multiple roots do not necessarily reverse the sign.
The substituted roots are not automatically x-roots.
Do not discard the negative x branch of a positive square.
