Absolute Value (Modulus) Function
The absolute value of a real number $x$, denoted as $|x|$, is defined as the non-negative value of $x$ without regard to its sign. Mathematically, $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$. In equations involving $|x - a|$, the critical point is $x = a$, where the function changes its algebraic definition. When multiple absolute values are present, the number line is divided into intervals by all critical points. Each interval must be solved separately by replacing the modulus terms with their appropriate algebraic signs based on that interval.
Quadratic Equations and Discriminant
A quadratic equation of the form $ax^2 + bx + c = 0$ has its roots determined by the formula $x = \frac{-b \pm \sqrt{D}}{2a}$, where $D = b^2 - 4ac$ is the discriminant. The nature of roots depends on $D$: if $D > 0$, there are two distinct real roots; if $D = 0$, there is one repeated real root; and if $D < 0$, there are no real roots (the roots are complex). In interval-based solving, once roots are found, they must be checked against the interval boundary to ensure validity.
Interval Boundary Conditions
When solving modulus equations using the case method, the choice of including or excluding boundary points in specific intervals (e.g., using $\le$ vs $<$) is mathematically flexible as long as every point on the real number line is covered exactly once. Since absolute value functions are continuous, $|x-a|$ equals zero at the boundary $x=a$, making both the positive and negative definitions yield the same result. It is conventional to include the boundary point in the interval where the modulus expression is non-negative.
Extraneous vs. Valid Solutions
In algebra, an extraneous solution is a root of a simplified version of an equation that does not satisfy the original constraints. In the context of modulus equations, a solution found for a specific case (like $x < -4$) is considered valid only if its numerical value actually lies within that range. For example, if Case 1 yields $x = 0.53$, but the condition for Case 1 is $x < -4$, then $0.53$ is extraneous and must be discarded. Only valid solutions contribute to the final count.
Q
What are critical points?
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Critical points are the values of x that make the expression inside the modulus zero. They are where the function's definition changes.
Q
Why did Case 2 have no real solutions?
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Because the resulting quadratic equation had a negative discriminant ($D = -15$), which means it has no real roots.
Q
Can I solve this graphically?
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Yes, you can sketch the graph of $y = f(x)$ and count the intersections with the x-axis ($y=0$).
Q
How to handle |x + a| + |x + b| = c?
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Use the same interval method. Define cases for $x < \min(-a,-b)$, $\min < x < \max$, and $x > \max$.
Q
What is the discriminant formula?
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$D = b^2 - 4ac$. It tells you if real roots exist.
Q
Why check the range $x < -4$ specifically?
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Because it's the interval where both expressions inside the modulus are negative.
Q
What is $\sqrt{65}$ approximately?
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It is approximately 8.06, since $8^2 = 64$ and $9^2 = 81$.
Q
Can the total solution count be zero?
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Yes, if every case results in either an extraneous solution or no real roots at all.
Q
Is x = -4 a solution?
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Plug it in: $-4|0| + 3|-2| + 10 = 0 + 6 + 10 = 16 \ne 0$. So, no.
Q
What if the equation had |x + 4|²?
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Since $|a|^2 = a^2$, the modulus sign would just disappear for that term.