Quadratic Equations: Chapter Review

Vocabulary and notation

Quadratic equation, standard form $ax^2 + bx + c = 0$

Every term moved to one side, written in decreasing power. Here $a$ is the leading coefficient and $c$ the constant term. The one unbreakable requirement is $a \neq 0$; $b$ or $c$ may be $0$.

Root (solution)

A value making the standard-form expression equal exactly $0$. A quadratic has two, one, or no real root, so finding one does not finish the problem.

Repeated root

The single root left when the two roots merge into one value, as in $x^2 = 0$.

Factored form

The same equation as a product of factors set to $0$, like $(x - 2)(x + 3) = 0$. It puts the roots on display; standard form hides them.

Monic and non-monic trinomials

Monic means the leading coefficient is exactly $1$, so the squared term is a plain $x^2$. Non-monic means it is not $1$, and that coefficient then splits across both factors.

Irreducible over the integers

A trinomial with no factorization into whole-number coefficients, such as $x^2 + x + 1$. Integer factoring is the wrong tool for it, not a dead end for the equation: $x^2 + x - 1 = 0$ is also irreducible yet has two real roots, while $x^2 + x + 1 = 0$ has no real roots.

Parabola

The U-shaped curve $y = ax^2 + bx + c$. Where it meets the $x$-axis is exactly where the real roots are: it can cross twice, touch once at a repeated root, or miss the axis entirely.

Formulas and theorems

• Zero-product property

  If a product equals $0$, at least one factor is $0$: $AB = 0$ forces $A = 0$ or $B = 0$.

  Use when The product must equal exactly $0$; no other number works. It extends to any number of factors, and a nonzero constant factor out front (the $3$ in $3(x - 4)(x + 2) = 0$) contributes no root.

• Square-root method

  $$\begin{gathered} x^2 = k \ \text{ gives } \ x = \pm\sqrt{k} \\ (x - h)^2 = k \ \text{ gives } \ x = h \pm \sqrt{k} \end{gathered}$$

  Text description

  One parabola cut by a horizontal line above its lowest point (two crossings), through that point (one), and below it (none).

  Use when Needs one squared quantity alone on one side and a constant on the other: either a bare $x^2$, so $b = 0$, or a whole squared bracket already isolated. $k > 0$: two real roots. $k = 0$: one repeated root. $k < 0$: no real root, because no real number squares to a negative. Treat whatever sits inside the square as one quantity, and undo the shift only after taking the root.

  e.g. $(x + 2)^2 = 9$: $x + 2 = \pm 3$, so $x = 1$ or $x = -5$.

• Monic factoring: sum $b$, product $c$

  $$\begin{gathered} (x + p)(x + q) \\ = x^2 + (p + q)x + pq \end{gathered}$$

  Use when Read right to left to factor $x^2 + bx + c$: find integers $p$, $q$ with $p + q = b$ and $pq = c$. Needs a leading coefficient of exactly $1$, so pull out any common numerical factor first. For some trinomials no such pair exists.

  e.g. $(x + 2)(x + 6) = x^2 + 8x + 12$, read backward to factor.

• Reading the signs of $p$ and $q$

  If $c > 0$ the two numbers share a sign, and it is the sign of $b$. If $c < 0$ they have opposite signs, and the one of larger size carries the sign of $b$.

  Use when A monic trinomial factored over the integers; it halves the search before any adding. When $b = 0$ and $c < 0$ the pair are opposites of equal size ($x^2 - 9 = (x + 3)(x - 3)$); when $b = 0$ and $c > 0$ no real pair exists.

• Non-monic factoring: $(px + r)(qx + s)$

  $$\begin{gathered} (px + r)(qx + s) \\ = pq\,x^2 + (ps + qr)\,x + rs \end{gathered}$$

  Text description

  A rectangle split into four cells, the two shaded cross cells adding to the middle term.

  Use when Matching $ax^2 + bx + c$ term by term forces $pq = a$, $rs = c$, $ps + qr = b$. The middle coefficient is a sum of two CROSS products, not a plain sum, so which constant sits beside which $x$ term matters.

  e.g. $(3x + 4)(x + 2) = 3x^2 + 10x + 8$.

• The $a \cdot c$ split

  $$\begin{gathered} (ps) + (qr) = b \\ (ps)(qr) = a \cdot c \end{gathered}$$

  Use when $a \neq 0$. The two cross products are themselves a sum-and-product pair, so the monic search finds them. It aims the trial rather than replacing it: still expand to confirm.

  e.g. $6x^2 + 7x - 3$: $ac = -18$, and $9 + (-2) = 7$, pointing at $(3x - 1)(2x + 3)$.

• Root read off a linear factor

  $$px + r = 0 \quad \text{gives} \quad x = -\frac{r}{p}$$

  Use when $p \neq 0$. The root comes out a fraction whenever $p$ does not divide $r$: routine for non-monic quadratics, and not an error.

  e.g. $2x - 3 = 0$ gives $x = \tfrac{3}{2}$.

• Quadratic with no constant term ($c = 0$)

  $$ax^2 + bx = x(ax + b)$$

  Use when Applies exactly when $c = 0$. The roots are $x = 0$ and $x = -\tfrac{b}{a}$, so $x = 0$ is always one of them; when $b = 0$ as well the two coincide and $0$ is a repeated root.

  e.g. $3x^2 + 12x = 3x(x + 4)$, with roots $0$ and $-4$.

• Sum and product of the roots (Vieta's formulas)

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

  Use when $a \neq 0$, read from standard form; both totals divide by $a$. When $a = 1$ they collapse to $r + s = -b$ and $rs = c$, and the monic factoring numbers $p$, $q$ are the negatives of the roots: same product, opposite sum. They hold for every quadratic with $a \neq 0$, whether or not its roots are real.

  e.g. $5x^2 - 3x - 2$: sum $\tfrac{3}{5}$, product $-\tfrac{2}{5}$.

• Monic quadratic from its roots

  $$x^2 - (r + s)x + rs = 0$$

  Use when Builds the monic quadratic with roots $r$ and $s$; the constant keeps the product's own sign. Scaling the equation by any nonzero number changes the coefficients but never the roots, which is how fractions clear.

  e.g. Roots $-3$ and $5$: sum $2$, product $-15$, so $x^2 - 2x - 15 = 0$.

Problem types, step by step

Exam traps

• Trap Splitting a product that does not equal $0$: from $(x - 2)(x + 3) = 14$, writing $x - 2 = 14$ or $x + 3 = 14$, so $x = 16$ or $x = 11$.

  Fix The property needs a bare $0$ on one side. Expand and rearrange to $x^2 + x - 20 = 0$, factor to $(x + 5)(x - 4) = 0$, and the roots are $x = -5$ or $x = 4$.

• Trap Dividing both sides by $x$: from $x^2 = 5x$, cancelling leaves only $x = 5$.

  Fix Dividing by an expression that can be zero deletes the root $x = 0$. Move everything to one side and factor: $x(x - 5) = 0$ keeps both.

• Trap Reading a root straight off its factor, so $(x + 4)$ looks like the root $4$.

  Fix A factor and the root it produces have opposite signs: $x + 4 = 0$ gives $x = -4$.

• Trap Answering $(x + 1)^2 = 25$ with $x = \pm 5$.

  Fix The $\pm 5$ is the value of $x + 1$, not of $x$. Finish both branches: $x = 4$ or $x = -6$.

• Trap Treating $(2x + 3)(x + 1)$ and $(2x + 1)(x + 3)$ as the same factorization.

  Fix Swapping the constants is free only when the two $x$-coefficients match, as they always do in a monic trinomial. Here they are $2$ and $1$, so the $2$ doubles whichever constant sits beside it: the first expands to $2x^2 + 5x + 3$, the second to $2x^2 + 7x + 3$.

• Trap Stopping at $2x^2 + 6x + 4 = (2x + 2)(x + 2)$.

  Fix True but incomplete: $2x + 2$ still hides a factor of $2$. Pull common factors all the way out front, giving $2(x + 1)(x + 2)$.

• Trap Using $-b$ and $c$ on a non-monic quadratic: reading $2x^2 - 9x + 4$ as sum $9$, product $4$.

  Fix Both totals divide by $a$: the sum is $-\tfrac{b}{a} = \tfrac{9}{2}$ and the product is $\tfrac{c}{a} = 2$. Dividing only one of the two is the usual half-done version.

• Trap Saying the roots of $x^2 - 7x + 10$ sum to $-7$.

  Fix The sum is $-b$, and $b$ is already $-7$, so the roots sum to $+7$. That minus is never optional, and it never touches the product.

• Trap Building from roots as $x^2 + Sx + P$.

  Fix The coefficient of $x$ is the negative of the sum, so the template is $x^2 - Sx + P$. The constant keeps the product's own sign.