5.3 Quadratic Equations, Roots & Factoring

A quadratic equation in standard form is $ax^2 + bx + c = 0$ with $a \ne 0$. On GMAT Quant the coefficients are almost always integers, and more than nine times in ten the trinomial factors over the integers. When it does not, the quadratic formula and completing the square still produce exact roots — fractions or simplified radicals. Decimal approximations of $\sqrt{2}$ or $\sqrt{5}$ are not what Problem Solving is asking for, and you have no calculator to produce them anyway.

This is Official Guide Math Review 3.2. The same algebra appears inside word problems (area, consecutive integers, revenue) once you have translated the sentence into $ax^2 + bx + c = 0$.

Factoring: the AC method

For $ax^2 + bx + c = 0$:

1. Compute the product $ac$.
2. Find integers $m$ and $n$ with $m \cdot n = ac$ and $m + n = b$.
3. Split $bx$ into $mx + nx$ and factor by grouping.
4. Set each factor equal to $0$.

Worked example. Solve $2x^2 - 7x - 15 = 0$.

Here $ac = 2 \times (-15) = -30$. The pair that multiplies to $-30$ and adds to $-7$ is $-10$ and $3$. Split:

$2x^2 - 10x + 3x - 15 = 0 \implies 2x(x - 5) + 3(x - 5) = 0 \implies (2x + 3)(x - 5) = 0$

Roots: $x = -3/2$ or $x = 5$. Leave $-3/2$ as a fraction.

Worked example (monic). Solve $x^2 - 5x - 24 = 0$.

Integers that multiply to $-24$ and add to $-5$ are $-8$ and $3$: $(x - 8)(x + 3) = 0$, so $x = 8$ or $x = -3$.

If no integer pair works, stop factoring and move to the discriminant. Forcing a factoring guess wastes the 128-second budget.

The quadratic formula and the four discriminant cases

For $ax^2 + bx + c = 0$,

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

The discriminant $\Delta = b^2 - 4ac$ sits under the radical and classifies the real roots. On Quant, split the positive case by whether $\Delta$ is a perfect square. That tells you whether the roots are rational (nice fractions or integers) or irrational (keep a simplified radical).

Worked example ($\Delta = 0$). $x^2 - 6x + 9 = 0$ has $\Delta = 36 - 36 = 0$. Repeated root $x = 3$. In fact $(x - 3)^2 = 0$.

Worked example (perfect-square $\Delta$). $x^2 - 5x + 6 = 0$ has $\Delta = 25 - 24 = 1 = 1^2$. Roots $(5 \pm 1)/2$, namely $3$ and $2$, matching $(x - 3)(x - 2) = 0$.

Worked example (non-square $\Delta$). $x^2 - 4x - 1 = 0$ has $\Delta = 16 + 4 = 20 = 4 \cdot 5$. Roots

$x = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$

Do not write $4.236$ and $-0.236$. Those are calculator values, and they are not the GMAT answer form.

Worked example (negative $\Delta$). $x^2 + x + 1 = 0$ has $\Delta = 1 - 4 = -3 < 0$. No real solution. (Complex roots are not tested on Quant.)

If a stem asks how many real solutions an equation has, compute $\Delta$ and stop. You do not need the roots.

Completing the square

Completing the square is both a solving method and the derivation of the quadratic formula. It is also the fastest way to read the vertex of $y = ax^2 + bx + c$.

For $x^2 + bx + c = 0$ (leading coefficient $1$):

1. Move $c$: $x^2 + bx = -c$.
2. Add $(b/2)^2$ to both sides.
3. The left side is $(x + b/2)^2$.
4. Take square roots, remembering $\pm$.

Worked example. Solve $x^2 - 6x - 7 = 0$ by completing the square.

$x^2 - 6x = 7$ $x^2 - 6x + 9 = 16$ $(x - 3)^2 = 16$ $x - 3 = \pm 4$

So $x = 7$ or $x = -1$. Same roots as $(x - 7)(x + 1) = 0$.

For $ax^2 + bx + c = 0$ with $a \ne 1$, divide through by $a$ first, or factor $a$ out of the $x$ terms:

$a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c = 0$

The vertex is at $x = -b/(2a)$. Completing the square on $2x^2 - 8x + 3$ begins by factoring $2$ from the first two terms: $2(x^2 - 4x) + 3 = 2(x - 2)^2 - 8 + 3 = 2(x - 2)^2 - 5$.

Vieta: sum and product of roots

If $r$ and $s$ are the roots of $ax^2 + bx + c = 0$,

$r + s = -\frac{b}{a}, \qquad rs = \frac{c}{a}$

The sign on the sum is the trap. For $x^2 - 6x + 8 = 0$, $b = -6$, so $r + s = 6$, not $-6$. A quick check: the factored form is $(x - r)(x - s) = x^2 - (r + s)x + rs$, matching $x^2 - (\text{sum})x + (\text{product})$.

Symmetric expressions rewrite in $r + s$ and $rs$:

• $r^2 + s^2 = (r + s)^2 - 2rs$
• $1/r + 1/s = (r + s)/rs$, provided $rs \ne 0$
• $r/s + s/r = (r^2 + s^2)/rs = \bigl((r + s)^2 - 2rs\bigr)/rs$
• $|r - s| = \sqrt{(r + s)^2 - 4rs} = \sqrt{\Delta}/|a|$

Worked example. For $3x^2 - 12x + 5 = 0$, $r + s = 12/3 = 4$ and $rs = 5/3$. Then

$r^2 + s^2 = 16 - 2\cdot\frac{5}{3} = \frac{48 - 10}{3} = \frac{38}{3}$

You never computed $r$ or $s$. The discriminant is $144 - 60 = 84$, not a perfect square, so the individual roots are irrational anyway.

Quadratic in form, and the divide-by-zero trap

Rational disguise. $x + 12/x = 7$, with $x \ne 0$. Multiply through by $x$:

$x^2 + 12 = 7x \implies x^2 - 7x + 12 = 0 \implies (x - 3)(x - 4) = 0$

Both $3$ and $4$ satisfy the original.

Radical disguise. $x - 3\sqrt{x} - 10 = 0$. Let $u = \sqrt{x}$ with $u \ge 0$. Then $u^2 - 3u - 10 = 0$, $(u - 5)(u + 2) = 0$, so $u = 5$ (discard $u = -2$). Thus $x = 25$.

Exponential disguise. $2^{2x} - 10 \cdot 2^x + 16 = 0$. Let $u = 2^x > 0$. Then $(u - 8)(u - 2) = 0$, so $2^x = 8$ or $2^x = 2$, hence $x = 3$ or $x = 1$.

The lost-root trap. Never divide $x^2 = 6x$ by $x$. That assumes $x \ne 0$ and drops the genuine root $x = 0$. Always rearrange to standard form:

$x^2 - 6x = 0 \implies x(x - 6) = 0 \implies x = 0 \text{ or } x = 6$

The same warning applies to $x^3 = 4x$: bring to one side, factor $x(x^2 - 4) = 0$, and keep $x = 0, 2, -2$.

Traps

• Writing decimal approximations instead of $2 \pm \sqrt{5}$.
• Using $r + s = b/a$ instead of $-b/a$.
• Treating $\Delta > 0$ as “two integer roots.” Only a perfect-square discriminant (with rational coefficients) guarantees rational roots.
• Forgetting that $\Delta < 0$ means no real solution, not “two solutions you cannot compute without a calculator.”
• Cancelling $x$ from both sides and losing $x = 0$.