Grade 9

Quadratic equations

Equations of the form ax²+bx+c=0, solved by factoring, completing the square or the quadratic formula.

IntuitionA curved path, not a straight one

Throw a ball, and it does not travel in a straight line — gravity curves its path into a parabola. Anywhere a quantity depends on the square of another (area of a square field, distance fallen under gravity, compound growth over two steps), the relationship $y=ax^2+bx+c$ shows up, and asking "when does y hit a target value?" means solving $ax^2+bx+c=0$ . Unlike a straight line, a parabola can cross a fixed height zero, one, or two times — so a quadratic equation can have two solutions, one, or none.

The parabola

y=x^2-2x-3

: drag

c

and

d

to watch the two crossings of

y=0

move together, merge into one, or disappear entirely.

SchoolStandard form and the discriminant

Definition: Quadratic equation and discriminant

A quadratic equation in $x$ is one that can be written in the standard form $ax^2+bx+c=0$ with $a\neq0$ . Its discriminant is the number $\Delta=b^2-4ac$ , built entirely from the three coefficients, and — as the theorems below show — its sign alone decides how many real solutions the equation has.

ax^2+bx+c=0

Three techniques solve a quadratic: factoring when the left side splits into two simple linear factors (fast, but only works for nice numbers), completing the square (always works, and is the engine behind the next technique), and the quadratic formula $x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$ , which always works and needs no guessing.

x=\dfrac{-b\pm\sqrt{\Delta}}{2a}

What the discriminant tells you
Sign of Δ	Number of real roots	Formula for the root(s)
$\Delta>0$	Two, distinct	$x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$
$\Delta=0$	One (double root)	$x=-\dfrac{b}{2a}$
$\Delta<0$	None (real)	—

UndergraduateTwo key theorems

Theorem: Deriving the quadratic formula

For $a\neq0$ , every real solution of $ax^2+bx+c=0$ satisfies $x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$ , where $\Delta=b^2-4ac$ ; conversely, whenever $\Delta\ge0$ this formula produces genuine solutions.

Why is it true?

Factoring only works when we get lucky with nice numbers; completing the square turns any quadratic into a perfect square equalling a number, which we can then "undo" with a square root — this gives one formula that solves every quadratic equation, no guessing required.

Proof

Step 1 (Normalize). Since $a\neq0$ , divide every term of $ax^2+bx+c=0$ by $a$ : this gives $x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$ , an equivalent equation where the leading coefficient is $1$ .

Step 2 (Complete the square). The first two terms $x^2+\dfrac{b}{a}x$ are the start of the perfect square $\left(x+\dfrac{b}{2a}\right)^2=x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}$ . Adding and subtracting $\dfrac{b^2}{4a^2}$ inside $x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$ lets us rewrite it as $\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0$ .

Step 3 (Isolate the square). Move the constant terms to the right side and combine them over the common denominator $4a^2$ : $-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=\dfrac{-b^2+4ac}{4a^2}=-\dfrac{b^2-4ac}{4a^2}=-\dfrac{\Delta}{4a^2}$ . This gives $\left(x+\dfrac{b}{2a}\right)^2=\dfrac{\Delta}{4a^2}$ .

Step 4 (Take the square root). When $\Delta\ge0$ , the right side is a nonnegative real number, so both sides have real square roots: $x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{\Delta}}{2a}$ (the $\pm$ already accounts for both signs, whichever sign $a$ itself has).

Step 5 (Isolate x). Subtract $\dfrac{b}{2a}$ from both sides: $x=-\dfrac{b}{2a}\pm\dfrac{\sqrt{\Delta}}{2a}=\dfrac{-b\pm\sqrt{\Delta}}{2a}$ , which is exactly $x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$ . Every algebraic step used (dividing by nonzero $a$ , adding/subtracting the same quantity, taking square roots of equal nonnegative numbers) is reversible, so this formula is both necessary and sufficient whenever $\Delta\ge0$ .

Theorem: Discriminant sign determines the root count

For $a\neq0$ , the equation $ax^2+bx+c=0$ has exactly two distinct real roots if $\Delta>0$ , exactly one real root $x=-\dfrac{b}{2a}$ if $\Delta=0$ , and no real root if $\Delta<0$ .

Why is it true?

This turns "how many times does the parabola cross the x-axis" — a question about a picture — into "check the sign of one number" — a question you can answer without drawing anything.

Proof

Every step of this proof reuses the identity $\left(x+\dfrac{b}{2a}\right)^2=\dfrac{\Delta}{4a^2}$ established in the previous theorem, which holds for any real $a,b,c$ with $a\neq0$ regardless of the sign of $\Delta$ — only the final square-root step depended on $\Delta\ge0$ .

**Case $\Delta>0$ .** The right side $\dfrac{\Delta}{4a^2}$ is a positive number (a positive numerator over a positive denominator $4a^2$ ), so it has two distinct square roots, $+\dfrac{\sqrt{\Delta}}{2a}$ and $-\dfrac{\sqrt{\Delta}}{2a}$ , which differ because $\sqrt{\Delta}\neq0$ . Each gives a different value of $x+\dfrac{b}{2a}$ , hence a different value of $x$ : exactly two distinct real roots.

**Case $\Delta=0$ .** The right side becomes $0$ , and the only real number whose square is $0$ is $0$ itself — there is no " $\pm$ " ambiguity left. So $x+\dfrac{b}{2a}=0$ , giving the single root $x=-\dfrac{b}{2a}$ . This matches factoring the left side directly as $a\left(x+\dfrac{b}{2a}\right)^2=0$ , a perfect square touching zero exactly once.

**Case $\Delta<0$ .** The right side $\dfrac{\Delta}{4a^2}$ is now negative. But the left side $\left(x+\dfrac{b}{2a}\right)^2$ is a square of a real number, and the square of any real number is always $\ge0$ — it can never equal a negative number. So no real $x$ can satisfy $\left(x+\dfrac{b}{2a}\right)^2=\dfrac{\Delta}{4a^2}$ in this case, meaning the original equation has no real solution at all.

UndergraduateReal-World Applications and Worked Examples

Quadratics appear whenever motion under constant acceleration, area, or products of two changing quantities are involved: a thrown object's height over time, the dimensions of a rectangular plot given a fixed perimeter and target area, or a firm's profit as a quadratic function of price. Solving "when is the quantity zero (or some target value)" is exactly solving a quadratic equation.

Example: Physics — when does a thrown object land?

A ball is thrown so that its height in meters after $t$ seconds is $h(t)=-t^2+4t+5$ . Find the time at which the ball hits the ground (height $0$ ).

Solution

Step 1 (Set height to zero and standardize). Landing means $-t^2+4t+5=0$ . Multiply both sides by $-1$ to get a positive leading coefficient: $t^2-4t-5=0$ .

Step 2 (Factor and interpret). Since $-5\times1=-5$ and $-5+1=-4$ , this factors as $(t-5)(t+1)=0$ , giving $t=5$ or $t=-1$ . A negative time makes no physical sense here (the ball is thrown at $t=0$ ), so we discard it, leaving $t=5$ seconds as the only meaningful answer.

Example: Business — zero-profit prices

A firm's monthly profit (in thousands of dollars) as a function of price level $x$ is $P(x)=-x^2+10x-16$ . Find the two price levels at which the firm exactly breaks even (profit $=0$ ).

Solution

Step 1 (Set profit to zero and standardize). Break-even means $-x^2+10x-16=0$ . Multiplying both sides by $-1$ gives $x^2-10x+16=0$ .

Step 2 (Factor and interpret both roots). Since $-2\times-8=16$ and $-2+-8=-10$ , this factors as $(x-2)(x-8)=0$ , giving $x=2$ and $x=8$ . Both are physically meaningful price levels: profit is exactly zero at $x=2$ and $x=8$ , and (since the profit parabola opens downward) positive in between — the firm should price somewhere between these two break-even points to be profitable.

Compute the discriminant of $2x^2-4x-6=0$ .

If $\Delta<0$ , how many real roots does the quadratic have?

Solve $x^2-5x+6=0$ by factoring.

A ball's height is $h(t)=-t^2+4t+5$ . At what positive time does it land?