How do you solve a quadratic equation on the GRE?

First try factoring: find two numbers that multiply to the constant and add to the middle coefficient, then set each factor to zero. If it does not factor cleanly, use the quadratic formula x = (-b ± √(b² - 4ac)) / (2a), which works for every quadratic. Always keep both roots.

What does the discriminant tell you?

The discriminant b² - 4ac tells you how many real solutions a quadratic has without solving it. Positive means two distinct real roots, zero means one repeated real root, and negative means no real solutions. The GRE never asks about complex roots.

What is the difference-of-squares shortcut?

Any expression of the form A² - B² factors into (A - B)(A + B). This lets you solve equations like 9x² - 25 = 0 instantly and simplify products such as (a - b)(a + b) = a² - b², which appears frequently in GRE algebra traps.

How do you find the vertex of a parabola?

The x-coordinate of the vertex is x = -b/(2a); substitute it back to get the y-coordinate. If the equation is already in vertex form a(x - h)² + k, the vertex is simply (h, k). The sign of a tells you whether it is a minimum (a > 0) or maximum (a < 0).

CramClub • Quadratic Equations and Expressions — GRE® General Test Guide & Practice

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x^2 - y^2 = (x - y)(x + y)$

$(x + y)^2 = x^2 + 2xy + y^2$

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

Why Quadratics Reward the Prepared

A quadratic is any expression of the form $ax^2 + bx + c$ with $a \ne 0$ . On the GRE Quantitative section, quadratics show up as pure algebra ("solve for $x$ "), as Quantitative Comparisons that hinge on a hidden second root, and inside word problems about area, projectile height, or revenue. The exam rarely rewards brute force here. Instead it rewards recognizing structure — a factorable trinomial, a difference of squares, a perfect-square pattern — so you can skip the heavy algebra entirely.

Expanding: The Special Products

Before solving, you must fluently expand. Three patterns repay memorization because the GRE hides them constantly:

$(x + y)^2 = x^2 + 2xy + y^2$ $(x - y)^2 = x^2 - 2xy + y^2$ $(x + y)(x - y) = x^2 - y^2$

The middle terms are where students slip. A frequent GRE trap gives you $x + y$ and $xy$ and asks for $x^2 + y^2$ . Since $(x + y)^2 = x^2 + 2xy + y^2$ , you get $x^2 + y^2 = (x + y)^2 - 2xy$ — no need to find $x$ and $y$ individually.

Solving by Factoring

If a quadratic factors, factoring is the fastest route to its roots, thanks to the zero-product property: if a product equals zero, at least one factor must be zero.

Solve $x^2 - 7x + 10 = 0$ . Find two numbers that multiply to $10$ and add to $-7$ : those are $-2$ and $-5$ .

$x^2 - 7x + 10 = (x - 2)(x - 5) = 0$

Set each factor to zero: $x = 2$ or $x = 5$ . Those values are the roots (also called zeros or $x$ -intercepts). Critically, a quadratic usually has two solutions — forgetting the second is the single most expensive GRE error, especially in Quantitative Comparison.

Difference of Squares and Completing the Square

The difference-of-squares pattern factors instantly: $x^2 - 36 = (x - 6)(x + 6)$ , giving $x = \pm 6$ . When a trinomial does not factor over the integers, completing the square rewrites it so you can isolate $x$ :

$x^2 + 6x - 4 = 0 \;\Rightarrow\; x^2 + 6x = 4 \;\Rightarrow\; (x + 3)^2 = 13 \;\Rightarrow\; x = -3 \pm \sqrt{13}.$

Take half of the middle coefficient ( $6 \div 2 = 3$ ), square it ( $9$ ), and add it to both sides to build the perfect square $(x+3)^2$ .

The Quadratic Formula Always Works

When nothing factors cleanly, the quadratic formula solves any $ax^2 + bx + c = 0$ :

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Solve $2x^2 + 3x - 4 = 0$ with $a = 2$ , $b = 3$ , $c = -4$ :

x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)} \\\\ x &= \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4} \end{aligned}$$ Irrational roots are perfectly normal on the GRE. Note the sign care: because $c = -4$ is negative, $-4ac = -4(2)(-4) = +32$. ## The Discriminant: Counting Real Roots The quantity under the radical, $b^2 - 4ac$, is the **discriminant**. Its sign tells you how many real solutions exist *without solving*: | Discriminant $b^2 - 4ac$ | Real solutions | |---|---| | Positive | **Two** distinct real roots | | Zero | **One** (repeated) real root | | Negative | **No real solutions** | The GRE tests only real solutions in this domain. If the discriminant is negative, the correct GRE answer is simply "no real solutions" — you are never asked about complex numbers. ## The Parabola: Direction and Vertex The graph of $y = ax^2 + bx + c$ is a **parabola**. It opens **upward** when $a > 0$ (so its vertex is a *minimum*) and **downward** when $a < 0$ (vertex is a *maximum*). The $x$-coordinate of the vertex — also the axis of symmetry — is: $$x = -\frac{b}{2a}$$ Substitute that back to get the vertex's $y$-value. In **vertex form** $y = a(x - h)^2 + k$, the vertex $(h, k)$ is visible directly; watch the sign, since $(x - h)$ means the $x$-coordinate is $+h$. ## A Reliable Game Plan - **Two integers multiply to $c$, add to $b$?** Factor and use the zero-product property. - **Looks like $x^2 - k^2$?** Difference of squares: $(x-k)(x+k)$. - **Given $x+y$ and $xy$?** Use $(x+y)^2 = x^2 + 2xy + y^2$ — do not solve for each variable. - **Nothing factors?** Quadratic formula, and keep both $\pm$ roots. - **Only need the number of solutions?** Compute the discriminant. - **Need a max or min?** Vertex at $x = -\frac{b}{2a}$; the sign of $a$ says max vs. min. ## Common Mistakes to Avoid - Reporting only one root when a quadratic has two — fatal in Quantitative Comparison. - Sign errors in $-4ac$ when $c$ is negative (a double negative becomes addition). - Reading the vertex of $a(x-h)^2 + k$ as $(-h, k)$ instead of $(h, k)$. - Assuming a negative discriminant is an error, rather than "no real solutions."

Solve $x^2 + 2x - 15 = 0$ .

Solve $9x^2 - 25 = 0$ .

Solve $3x^2 - 5x - 1 = 0$ using the quadratic formula.

If $x + y = 9$ and $xy = 14$ , what is the value of $x^2 + y^2$ ?

Solve $x^2 - 8x + 5 = 0$ by completing the square.

For which values of $k$ does $x^2 + kx + 16 = 0$ have exactly one real solution?

Reporting only one root. Solving $x^2 = 49$ as just $x = 7$ drops the negative root and wrecks Quantitative-Comparison answers.

A quadratic normally has two roots. $x^2 = 49$ gives $x = 7$ OR $x = -7$ . Always ask whether a second solution changes the comparison.

Sign errors in the discriminant when $c$ is negative — treating $-4(3)(-1)$ as $-12$ instead of $+12$ .

Multiply the signs deliberately: $-4ac$ with a negative $c$ becomes positive, which enlarges $b^2 - 4ac$ .

Misreading vertex form: taking $y = 2(x - 5)^2 - 3$ to have vertex $(-5, -3)$ .

Vertex form $a(x - h)^2 + k$ has vertex $(h, k)$ . Because it subtracts $h$ , $(x - 5)$ means $h = +5$ : the vertex is $(5, -3)$ .

Completing the square while adding the correction term to only one side of the equation.

Whatever you add to build the perfect square, add the identical amount to the other side so the equation stays balanced.

What is the sum of all solutions to $x^2 - 3x - 28 = 0$ ?

If $a - b = 6$ and $a^2 - b^2 = 54$ , what is the value of $a + b$ ?

How many distinct real solutions does $2x^2 + 3x + 5 = 0$ have?

A rectangular garden is $3$ meters longer than it is wide, and its area is $54$ square meters. What is the width, in meters?

For what value of $c$ does $x^2 - 12x + c = 0$ have exactly one real solution?

The parabola $y = -2(x + 3)^2 + 8$ is written in vertex form. Which statement about it is true?