Equations and inequalities: 1st/2nd degree, systems, absolute value

A first-degree equation $ax+b=0$ ($a \neq 0$) always has a unique solution $x = -b/a$. An inequality is solved the same way, with one detail that changes everything: multiplying or dividing both sides by a negative number reverses the direction of the inequality.

For the second degree $ax^2+bx+c=0$, the discriminant $\Delta = b^2-4ac$ gives the number of real solutions: two if $\Delta>0$, one (double) if $\Delta=0$, none if $\Delta<0$. When they exist, the solutions are $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$. For a second-degree inequality, the sign of the trinomial between the roots (if any) is opposite to the sign of $a$, and outside the roots it matches the sign of $a$.

A system of two linear equations in two unknowns is solved by substitution (isolate a variable in one equation and substitute it into the other) or by elimination (add/subtract multiples of the equations to cancel a variable). Both methods give the same result; the choice is a matter of efficiency given the coefficients.

The absolute value $|x|$ represents a distance from zero: $|x|=a$ ($a>0$) is equivalent to $x=a$ OR $x=-a$. For an inequality, $|x|<a$ is equivalent to $-a<x<a$, whereas $|x|>a$ is equivalent to $x<-a$ OR $x>a$ — never confuse these two structures.

Illustration: sign chart for a rational inequality

Calculation: solving a rational inequality

A 3-step method: (1) move everything to one side to compare against 0; (2) locate the critical values (roots of the numerator and of the denominator); (3) test the sign of the expression on each interval bounded by those values.

Example: solve $\dfrac{x-2}{x+1} \geq 0$. Critical values: $x=2$ (numerator zero) and $x=-1$ (denominator zero, forbidden value). Testing each interval: for $x=-2$ (< −1): $\dfrac{-4}{-1}=4>0$; for $x=0$ (between −1 and 2): $\dfrac{-2}{1}=-2<0$; for $x=3$ (> 2): $\dfrac{1}{4}>0$. The solution is $x \in (-\infty,-1) \cup [2,+\infty)$ — note that $x=-1$ stays excluded even though the inequality is non-strict, since it makes the expression undefined, while $x=2$ is included because it only zeroes out the numerator.

Classic IMAT trap: in an inequality of the form $\dfrac{P(x)}{Q(x)} \geq k$, never multiply directly by $Q(x)$ without knowing its sign — the inequality flips if $Q(x)<0$. Always reduce to a comparison against 0 first instead. Same logic for a quadratic inequality like $x^2-5x+6<0$: the roots $2$ and $3$ split the line into 3 intervals, and since the parabola opens upward, the expression is negative only between its roots — solution $x \in (2,3)$.

• Discriminant and number of real roots
• Sign of a second-degree trinomial
• Substitution vs elimination for a system
• $|x|<a$ (interval) vs $|x|>a$ (union of two half-lines)

• $\Delta = b^2-4ac$
• $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$
• Sum of roots: $-b/a$ — Product of roots: $c/a$
• $|x|<a$ ($a>0$) $\iff$ $-a<x<a$

The discriminant "announces", before any calculation, how many times the parabola $y=ax^2+bx+c$ crosses the x-axis: twice, once (tangent), or never. Visualizing the parabola avoids a lot of sign errors on second-degree inequalities.

Forgetting to reverse the direction of an inequality when dividing by a negative number is mistake number one. For the second degree, a common error is assuming the trinomial is always positive between the roots — it's the opposite when $a<0$. For absolute values, mixing up the "interval" form and the "union" form is very common under time pressure.