Which method should you use to solve a quadratic?

Factor when it factors easily over the integers, that is fastest. Take square roots when there is a squared term but no linear x term, like x squared equals 49 or (x minus 3) squared equals 16. Use the quadratic formula when factoring is hard or impossible; it always works and is on the reference sheet. Completing the square is mainly for producing vertex form and for understanding where the formula comes from.

What is the zero product property?

If a product of factors equals zero, then at least one of the factors must be zero, because the only way to multiply real numbers to zero is to include a zero. So once a quadratic is factored and set equal to zero, set each factor to zero and solve. This is why you must move everything to one side first: the property needs a zero on the other side.

What does the discriminant tell you?

The discriminant is b squared minus 4ac, the expression under the square root in the quadratic formula. If it is positive there are two real solutions, if it is zero there is one (a double root), and if it is negative there are no real solutions. You can answer how many real solutions an equation has just by computing the discriminant.

What constant completes the square for x squared plus bx?

Take half of the coefficient b and square it: add (b over 2) squared. For x squared plus 8x, half of 8 is 4 and 4 squared is 16, so add 16 to get (x plus 4) squared. The number inside the binomial is always half of b. When solving, add that constant to both sides to keep the equation balanced.

How do you find the vertex and axis of symmetry of a parabola?

The axis of symmetry is the vertical line x equals negative b over 2a. The vertex sits on it, so its x-coordinate is negative b over 2a, and you find its y-coordinate by substituting that x into the function. The sign of a tells you the opening: positive opens up with a minimum, negative opens down with a maximum. The axis formula and vertex form are not on the reference sheet.

How do you interpret a projectile-height quadratic?

For a downward parabola like h(t) equals negative 16 t squared plus v-naught t plus h-naught, the vertex is the maximum height and when it occurs, the y-intercept is the starting height, and the positive zero is when it lands. Solve, then discard non-viable answers such as a negative time, because the domain of a height model starts at zero.

LouisianaMaths

Louisiana LEAP 2025 Algebra I: a complete guide to quadratics (A-REI, A-SSE, F-IF)

A deep-dive Louisiana LEAP 2025 Algebra I guide to quadratics: solving by factoring (A-REI.B.4), by square roots and completing the square, with the reference-sheet quadratic formula and the discriminant, graphing parabolas with vertex and axis of symmetry (F-IF.C.7), and quadratic applications such as projectile height.

Generated by Claude Opus 4.815 min readA1: A-REI.B.4, A-SSE.B.3, F-IF.C.7Updated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this module covers
Solving by factoring
Square roots and completing the square
The quadratic formula and the discriminant
Graphing quadratics
Quadratic applications
How this module is examined
Check your knowledge

What this module covers

This guide covers quadratics on the Louisiana LEAP 2025 Algebra I test, a core part of the Major Content category: solving by factoring (A-REI.B.4), by square roots and completing the square, with the reference-sheet quadratic formula and the discriminant, graphing parabolas with vertex and axis of symmetry (F-IF.C.7), and applications such as projectile height (A-CED.A.1, F-IF.B.4). Each dot-point page has its own practice: solving quadratics by factoring, solving by square roots and completing the square, the quadratic formula and the discriminant, graphing quadratic functions, and quadratic applications.

Solving by factoring

Put the quadratic in standard form $ax^2 + bx + c = 0$ , factor, then use the zero product property: set each factor to zero. The solutions are the zeros (the $x$ -intercepts).

Square roots and completing the square

Take square roots when there is no linear term: $(x - 3)^2 = 16$ gives $x - 3 = \pm 4$ . Complete the square by adding $\left(\frac{b}{2}\right)^2$ , which also gives vertex form.

The quadratic formula and the discriminant

The reference-sheet formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ solves any quadratic. The discriminant $b^2 - 4ac$ counts real solutions: positive gives two, zero gives one, negative gives none.

Graphing quadratics

A quadratic graphs as a parabola. The sign of $a$ sets the opening (up for $a > 0$ , down for $a < 0$ ). The axis of symmetry is $x = \frac{-b}{2a}$ , the vertex sits on it, the $y$ -intercept is $c$ , and the $x$ -intercepts are the zeros.

Quadratic applications

Projectile height $h(t) = -16t^2 + v_0 t + h_0$ is a downward parabola: the vertex is the maximum height and when it occurs, the $y$ -intercept is the start, and the positive zero is landing. Discard non-viable (negative) times.

How this module is examined

Equation response. Solve by any method, or find a vertex, axis, or intercept.
Type III modeling. Use a quadratic model and interpret a feature with units.
Multiple choice. Identify the number of solutions, the opening direction, or the completing-the-square constant.
Graphing items. Plot a parabola or its vertex.

Check your knowledge

Work these as you would for credit on the online test.

Solve $x^2 + 2x - 15 = 0$ by factoring. (2 points)
Solve $x^2 - 9x = 0$ . (2 points)
Solve $(x - 3)^2 = 16$ . (2 points)
What constant completes the square for $x^2 + 8x$ ? (1 point)
Solve $x^2 + 6x - 7 = 0$ by completing the square. (2 points)
Solve $2x^2 + 4x - 3 = 0$ with the quadratic formula. (2 points)
How many real solutions does $x^2 + 2x + 5 = 0$ have? (1 point)
For $y = x^2 - 6x + 5$ , find the vertex. (2 points)
Which way does $y = -2x^2 + 3$ open, and is the vertex a max or min? (1 point)
For $h(t) = -16t^2 + 32t + 48$ , when does it hit the ground? (3 points)

Solutions

Step 1: Q1 - solving by factoring

Factor the trinomial, then set each factor equal to zero using the zero-product property:

$(x + 5)(x - 3) = 0$ , so $x = -5$ or $x = 3$ .

Step 2: Q2 - factoring out the GCF

Factor out $x$ (the GCF), then set each factor equal to zero:

$x(x - 9) = 0$ , so $x = 0$ or $x = 9$ .

Step 3: Q3 - solving by square roots

When the equation has the form $(x - k)^2 = c$ with no linear term, take the square root of both sides and remember both the positive and negative roots:

x - 3 = \pm 4, \text{ so } x = 7 \text{ or } x = -1

Step 4: Q4 - completing the square

To complete the square for $x^2 + 8x$ , take half the coefficient of $x$ , then square it. Half of $8$ is $4$ , and $4^2 = 16$ .

Answer: add $16$ .

Step 5: Q5 - solving by completing the square

Rewrite the left side as a perfect square, take the square root of both sides (keeping both signs), then solve for $x$ :

(x + 3)^2 = 16 \implies x + 3 = \pm 4 \implies x = 1 \text{ or } x = -7

Step 6: Q6 - applying the quadratic formula

Identify $a = 2$ , $b = 4$ , $c = -1$ , substitute into $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , then simplify:

x = \frac{-4 \pm \sqrt{16 + 8}}{4} = \frac{-4 \pm \sqrt{40}}{4} = \frac{-2 \pm \sqrt{10}}{2}

Step 7: Q7 - using the discriminant

The discriminant $b^2 - 4ac$ tells how many real solutions exist. A negative discriminant means no real solutions:

b^2 - 4ac = 4 - 20 = -16 < 0: \text{ no real solutions}

Step 8: Q8 - finding the vertex

The $x$ -coordinate of the vertex is $x = -\dfrac{b}{2a}$ . Substitute back to find the $y$ -coordinate:

Axis $x = \dfrac{-(-6)}{2(1)} = 3$ ; $y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$ . Vertex $(3, -4)$ .

Step 9: Q9 - direction of opening

The sign of the leading coefficient $a$ determines whether the parabola opens up or down. When $a < 0$ the parabola opens downward and the vertex is a maximum:

$a = -2 < 0$ : opens down, vertex is a maximum.

Step 10: Q10 - projectile problem

Set $h(t) = 0$ to find when the object lands, solve the quadratic, and discard any negative time value as non-viable:

$-16t^2 + 32t + 48 = 0 \implies t^2 - 2t - 3 = 0 \implies (t - 3)(t + 1) = 0$ ; viable $t = 3\ \text{s}$ .

Sources & how we know this

Louisiana Student Standards for Mathematics — Louisiana Department of Education (2025)
LEAP 2025 Assessment Guide for Algebra I — Louisiana Department of Education (2025)