🔢 Quant

Quadratics, Functions & Graphs

Factoring identities, the discriminant, vertex max/min, function notation, and lines in the coordinate plane.

~6h

to master

Quadratics: factor first, formula last

A quadratic equation in standard form is $ax^2 + bx + c = 0$ with $a \neq 0$ . The GMAT's favourite solving route is factoring — move everything to one side, factor, and use the zero-product rule: a product is 0 exactly when some factor is 0.

📐Core Rule

The three identities that unlock most factoring: $a^2 - b^2 = (a+b)(a-b)$ · $a^2 + 2ab + b^2 = (a+b)^2$ · $a^2 - 2ab + b^2 = (a-b)^2$ . Spot them in disguise: $x^2 - 49$ , $4y^2 - 12y + 9 = (2y-3)^2$ , $\frac{x^2 - 1}{x - 1} = x + 1$ (for $x \neq 1$ ).

When factoring stalls, the quadratic formula always works: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . The discriminant $b^2 - 4ac$ tells you how many real roots exist without solving:

$b^2 - 4ac$	Real roots
positive	two distinct
zero	exactly one
negative	none

Count roots, don't find them: .

⚠️GMAT Trap

Never divide both sides by a variable expression. $x^2 = 5x$ → dividing by $x$ silently discards the solution $x = 0$ . Factor instead: $x(x - 5) = 0$ , so $x = 0$ or $x = 5$ . The "lost root" is a stock wrong answer.

Max/min: the vertex in one line

For

f(x) = ax^2 + bx + c

: if

a > 0

the parabola opens up (unique minimum); if

a < 0

it opens down (unique maximum). The extreme value occurs at

x = -\dfrac{b}{2a}

and equals

c - \dfrac{b^2}{4a}

. Revenue/area/projectile problems are this formula wearing a costume — try .

Functions: machine in, machine out

$f(x) = x^2 - 3x$ defines a machine: input $x$ , output $f(x)$ . Evaluate by substitution everywhere the variable appears: $f(-2) = 4 + 6 = 10$ .

Domain = allowed inputs (exclude division by zero, negative numbers under square roots).
Range = achievable outputs (a quadratic's range starts/ends at its vertex value).
One input gives at most one output, but different inputs may share an output.
For compound machines, work inside-out: $f(g(2))$ means run $g$ first.

Graphs and lines

Every non-vertical line is $y = mx + b$ : slope $m$ , $y$ -intercept $b$ .

📐Core Rule

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$ — subtract coordinates in the same order. Positive slope rises left-to-right, negative falls, zero is horizontal; vertical lines ( $x = c$ ) have undefined slope. Parallel lines share $m$ ; perpendicular slopes multiply to $-1$ .

$x$ -intercept: set $y = 0$ . $y$ -intercept: set $x = 0$ . For any function, the $x$ -intercepts are the solutions of $f(x) = 0$ — algebra and geometry are the same fact.
Two distinct lines intersect at the unique shared solution of their equations; parallel ⇔ no solution; same line ⇔ infinitely many. This is the graphical face of the linear-systems trichotomy.

💡Exam Tip

A point lies on a graph exactly when its coordinates satisfy the equation — checking membership is substitution, never plotting. For "which quadrant" questions, just track the two signs: $(+,-)$ → IV.

Checklist

One side $= 0$ before factoring; never divide away a variable
Discriminant for how many roots; formula for which
Vertex $x = -b/2a$ for any max/min wording
Domain: ban zero denominators and negative radicands
Slope subtraction in consistent order

Sample Questions

22 practice questions

Medium

The cost $C$ , in dollars, to remove $p$ percent of a certain contaminant from a reservoir is estimated by using the formula $C = \frac{20,000p}{100 - p}$ . According to this estimate, how much more would it cost to remove 98 percent of the contaminant from the reservoir than it would cost to remove 75 percent of the contaminant?

Hard

At her regular hourly rate, Mira had estimated the labour cost of a repair job as $180, and she was paid that amount. However, the job took 6 hours longer than she had estimated and, consequently, she earned $5 per hour less than her regular hourly rate. What was the time Mira had estimated for the job, in hours?

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