Quadratic Equations

A1.J.1Solve by factoring01/05

Solve a quadratic by factoring.

Solve for x

x^{2} - 5 x + 6 = 0

Step01of 05

The whole method rests on one rule: the zero product property.

If A \cdot B = 0, then A = 0 or B = 0.

The only way two things multiply to zero is if at least one of them IS zero. This is why factoring + setting equal to zero is the recipe.

A1.J.1 — Solve a quadratic by factoring.

Solve for x

x^{2} - 5 x + 6 = 0

Step 01 of 05

The whole method rests on one rule: the zero product property.

If A \cdot B = 0, then A = 0 or B = 0.

The only way two things multiply to zero is if at least one of them IS zero. This is why factoring + setting equal to zero is the recipe.

Step 02 of 05

Step 1 — make sure one side is zero. Already done here. If the equation reads $x^{2} - 5 x = - 6$ , move everything to one side first.

Step 2 — factor. We need two numbers that multiply to $6$ and add to $- 5$ . Both negative: $- 2$ and $- 3$ .

x^{2} - 5 x + 6 = (x - 2) (x - 3) = 0

Step 03 of 05

Step 3 — set each factor equal to zero.

x - 2 = 0 or x - 3 = 0

Two simple linear equations replaced one quadratic — that's the magic of factoring.

Step 04 of 05

Step 4 — solve each.

x = 2 or x = 3

Two solutions. Check both: $(2)^{2} - 5 (2) + 6 = 0 ✓$ and $(3)^{2} - 5 (3) + 6 = 0 ✓$ .

Step 05 of 05

Common pitfall — leaving a GCF behind. If the quadratic isn't monic, factor the GCF first, but don't drop it.

2 x^{2} - 10 x + 12 = 0

2 (x^{2} - 5 x + 6) = 0

The $2$ on the outside doesn't change the solutions: dividing both sides by 2 gives the same equation as before. Solutions: $x = 2$ or $3$ .

Key insight

Get one side to zero, factor, set each factor to zero, solve each. The quadratic gives at most TWO solutions because there are at most two factors that can equal zero independently.

A1.J.2 — Solve by completing the square.

Solve for x

x^{2} + 6 x - 7 = 0

Step 01 of 05

When a quadratic doesn't factor nicely (or you need the vertex form),completing the square turns one side into a perfect square — and perfect squares are easy to undo with a square root.

The trick: take HALF the coefficient of x, square it, add it to both sides. That makes the left side fold up into $(x + p)^{2}$ .

Step 02 of 05

Step 1 — move the constant to the other side.

x^{2} + 6 x - 7 = 0

x^{2} + 6 x = 7

The variable terms now sit alone on the left, ready to receive the "completing" piece.

Step 03 of 05

Step 2 — add $(b /2)^{2}$ to BOTH sides. Here $b = 6$ , so half of it is $3$ , squared is $9$ .

x^{2} + 6 x + 9 = 7 + 9

x^{2} + 6 x + 9 = 16

Step 04 of 05

Step 3 — factor the left as a perfect square. The "p" inside is exactly half of $b$ :

(x + 3)^{2} = 16

Verify the pattern: $(x + 3)^{2} = x^{2} + 6 x + 9$ $✓$ — the very thing we built.

Step 05 of 05

Step 4 — square root both sides, remembering the $\pm$ .

x + 3 = \pm 4

x = - 3 + 4 = 1 or x = - 3 - 4 = - 7

Key insight

Five-beat recipe: isolate the variable terms, add

(b /2)^{2}

to both sides, factor as

(x + b /2)^{2}

, square-root both sides with

\pm

, solve the two linear equations. Works for every quadratic — also gives you vertex form for graphing.

A1.J.3 — The quadratic formula and the discriminant.

The quadratic formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Step 01 of 05

The quadratic formula solves $a x^{2} + b x + c = 0$ for ANY $a$ , $b$ , $c$ — even when factoring fails. Memorize it; it's derived by completing the square in full generality.

The expression under the radical, $b^{2} - 4 a c$ , is called the discriminant. Its sign tells you HOW MANY real solutions exist before you finish solving.

Step 02 of 05

Worked example. Solve $2 x^{2} - 7 x + 3 = 0$ . Here $a = 2$ , $b = - 7$ , $c = 3$ .

x = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( 2 ) ( 3 )}{2 ( 2 )} = \frac{7 \pm 49 - 24}{4}

= \frac{7 \pm 25}{4} = \frac{7 \pm 5}{4}

x = 3 or x = \frac{1}{2}

Step 03 of 05

The discriminant $b^{2} - 4 a c$ tells you the type of solutions before you compute them.

$b^{2} - 4 a c > 0$	→two distinct real solutions
$b^{2} - 4 a c = 0$	→one real solution (double root)
$b^{2} - 4 a c < 0$	→zero real solutions (the parabola doesn't cross the x-axis)

Geometrically, this counts how many times the parabola $y = a x^{2} + b x + c$ crosses the x-axis.

Step 04 of 05

The three discriminant cases — visualized. Each parabola below is a quadratic with the same $a = 1$ , but the constant term shifts the x-axis crossings.

Forest:

b^{2} - 4 a c > 0

(2 crossings). Blue:

= 0

(touches x-axis at vertex). Teal:

< 0

(no crossing).

Step 05 of 05

When to reach for the formula.

Quadratic factors over integers	→factoring is faster
Coefficients are messy / large	→quadratic formula
You only need the discriminant	→compute b² − 4ac, stop there
Need exact (irrational) answers	→quadratic formula

Key insight

The discriminant is the most efficient tool in the toolbox: one calculation tells you whether to expect 2, 1, or 0 real solutions before you commit to grinding through the full formula. Use it whenever the question asks "how many solutions?" — there's no need to compute them.

A1.J.4 — Quadratic word problems — projectile, area.

The problem

A ball is launched from a 80-foot platform at 64 ft/s upward. Its height after $t$ seconds is $h (t) = - 16 t^{2} + 64 t + 80$ . (a) When does it hit the ground? (b) What's its maximum height?

Step 01 of 05

The $- 16 t^{2}$ is gravity (32 ft/s² acceleration, half of which is the $16$ ). The $64 t$ is the upward launch. The $+ 80$ is the starting platform height.

Translate each question into a math operation on $h (t)$ :

"When does it hit the ground?"	→solve $h (t) = 0$
"What is its max height?"	→find the vertex of $h$
"How high after 2 seconds?"	→compute $h (2)$
"When is it 100 ft high?"	→solve $h (t) = 100$

Step 02 of 05

(a) When does it hit the ground? Set $h (t) = 0$ and solve.

- 16 t^{2} + 64 t + 80 = 0

Divide both sides by $- 16$ to simplify (signs flip but RHS is 0):

t^{2} - 4 t - 5 = 0 \Rightarrow (t - 5) (t + 1) = 0

t = 5 or t = - 1

Reject $t = - 1$ (negative time isn't physical). Answer: 5 seconds.

Step 03 of 05

(b) Maximum height? The max of a downward-opening parabola is at the vertex, $t = - b / (2 a)$ .

t = - \frac{64}{2 ( - 16 )} = - \frac{64}{- 32} = 2

So the peak occurs at $t = 2$ . Plug in:

h (2) = - 16 (4) + 64 (2) + 80 = - 64 + 128 + 80 = 144

Max height: 144 feet, reached 2 seconds after launch.

Step 04 of 05

The trajectory in one picture. The launch ( $t = 0$ ) at height 80, peak at $(2, 144)$ , ground impact at $(5, 0)$ .

Time on the x-axis (seconds), height on the y-axis (feet).

Step 05 of 05

Area problems work the same way. Set up the area as a quadratic in one variable, then optimize or solve.

Example: a rectangular pen with 60 ft of fencing. If width is $x$ , length is $30 - x$ , area is $A = x (30 - x) = - x^{2} + 30 x$ . Max area is at the vertex: $x = 15$ , giving a 15×15 square with $A = 225$ sq ft.

Key insight

Translate the question first ("when zero?" → solve, "max?" → vertex, "after t?" → evaluate), then run the corresponding mechanic. Reject any negative time / negative dimension answer that the algebra spits out — physics doesn't care about extraneous roots.

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