What are the four ways to solve a quadratic on TNReady?

Factoring (using the zero-product property), taking square roots, completing the square, and applying the quadratic formula. Factoring is fastest when the quadratic factors with integers; square roots suit equations with no linear term or in squared form; completing the square works on any quadratic and builds vertex form; and the quadratic formula, on the reference sheet, solves every quadratic.

Why must a quadratic be set to zero before factoring?

Because the zero-product property is the engine of factoring, and it only works when a product equals zero. If A times B equals zero, then A equals zero or B equals zero. So x squared plus 4x equals 21 must first become x squared plus 4x minus 21 equals 0; factoring while one side is 21 tells you nothing, because 21 is not zero.

What does the discriminant tell you?

The discriminant is b squared minus 4ac, the part under the radical in the quadratic formula. A positive discriminant gives two real solutions (the parabola crosses the x-axis twice), a zero discriminant gives one solution (a double root, the vertex touches the axis), and a negative discriminant gives no real solutions (the parabola misses the axis). It counts solutions without solving.

How do you complete the square?

Move the constant to the other side, then add the square of half the linear coefficient, that is (b over 2) squared, to both sides. This makes the left side a perfect-square trinomial, which factors as a binomial squared. Then take the square root of both sides, remembering the plus or minus, and solve. If the leading coefficient is not 1, divide every term by it first.

In a word problem, when do you use the vertex versus the zeros?

The zeros (where the quadratic equals zero) answer when something is at ground level or breaks even, such as the landing time of a projectile. The vertex answers the extreme value, such as the maximum height or maximum revenue. Words like how high, maximum, or least point to the vertex, while when does it land or hits the ground point to a zero.

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, and the vertex sits on it. Substitute that x-value back into the function to get the vertex's y-coordinate. The sign of the leading coefficient a tells you whether the parabola opens up (minimum) or down (maximum).

TennesseeMaths

TNReady Algebra I: a complete guide to quadratic equations

A deep-dive TNReady Algebra I guide to quadratic equations, part of the Equations and Inequalities and Functions reporting categories. Covers solving by factoring and the zero-product property, the square-root property and completing the square, the quadratic formula and the discriminant, graphing parabolas and their key features, and real-world applications such as projectile motion and area.

Generated by Claude Opus 4.816 min readA1.A.REI.B.4, A1.F.IF.D.7aUpdated 2026-06-02

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

Jump to a section

What this category demands
Factoring and the zero-product property
Square roots and completing the square
The quadratic formula and the discriminant
Graphing and applications
How this category is examined
Check your knowledge

What this category demands

This guide covers quadratic equations and functions (TN domains A1.A.REI.B.4 and A1.F.IF.D.7a), a high-value block that spans the Equations and Inequalities and Functions reporting categories. Solving and graphing quadratics is where the Met and Exceeded standards are often decided. Each dot-point page has its own practice: solving by factoring, square roots and completing the square, the quadratic formula and discriminant, quadratic applications, and graphing quadratic functions.

Factoring and the zero-product property

To solve by factoring, set the equation to zero first, factor, then apply the zero-product property: if a product is zero, at least one factor is zero. For $(x - 5)(x + 2) = 0$ , $x = 5$ or $x = -2$ , each the opposite sign of its factor's constant. The solutions are the zeros (x-intercepts) of the related parabola. Factoring is fastest when the quadratic factors with integers, which TNReady's multiple-choice quadratics usually do.

Square roots and completing the square

The square-root property: if $(x - h)^2 = k$ , then $x = h \pm \sqrt{k}$ , with the $\pm$ giving both solutions. Use it when there is no linear term or the equation is in squared form. Completing the square turns any quadratic into squared form: move the constant, add $\left(\frac{b}{2}\right)^2$ to both sides, factor the perfect square, then take square roots. It also converts standard form to vertex form.

The quadratic formula and the discriminant

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

Graphing and applications

To graph a parabola, find the direction (sign of $a$ ), the axis $x = \frac{-b}{2a}$ , the vertex on it, and the intercepts. The vertex is a minimum (opens up) or maximum (opens down). For applications, projectile motion uses $h(t) = -16t^2 + v_0 t + h_0$ (ground at $h = 0$ , peak at the vertex) and area uses length times width. Solve, then interpret: reject negative times, lengths, or counts, and state units.

A full projectile problem

A ball's height is $h(t) = -16t^2 + 40t + 6$ feet. Find when it lands (to the nearest tenth) and its maximum height.

step 1 Maximum height at the vertex

$t = \frac{-40}{2(-16)} = 1.25$ s; $h(1.25) = -16(1.5625) + 40(1.25) + 6 = 31$ feet.

step 2 Set height to zero for landing

$-16t^2 + 40t + 6 = 0$ , with $a = -16$ , $b = 40$ , $c = 6$ .

step 3 Apply the formula

$t = \frac{-40 \pm \sqrt{1600 + 384}}{-32} = \frac{-40 \pm \sqrt{1984}}{-32}$ , and $\sqrt{1984} \approx 44.5$ .

step 4 Choose the realistic root

$t \approx 2.6$ s (the positive landing time); the negative root is rejected. The ball reaches $31$ feet and lands at about $2.6$ s.

How this category is examined

Multiple choice and multiple select. Solve factorable quadratics, count solutions from the discriminant, identify a vertex or direction, or select all true statements. Sign-reversal and "positive root only" distractors are standard.
Equation editor and number entry. Solve by any method and enter solutions in simplest radical form, or set up and solve an application.
Graphing. Plot the vertex and intercepts of a parabola.

Check your knowledge

Work these as you would for credit on the EOC.

Solve $x^2 - 7x + 12 = 0$ by factoring. (1 point)
Solve $x^2 + 3x = 10$ . (1 point)
Solve $(x - 2)^2 = 25$ . (1 point)
Solve $x^2 + 8x + 3 = 0$ by completing the square (simplest radical form). (2 points)
Solve $2x^2 + 3x - 2 = 0$ using the quadratic formula. (2 points)
How many real solutions does $x^2 + x + 4 = 0$ have? (1 point)
A ball's height is $h(t) = -16t^2 + 48t$ . When does it land? (2 points)
For $f(x) = x^2 - 6x + 5$ , find the vertex. (2 points)

Solutions to the check-your-knowledge questions

Step 1: Q1. Solve by factoring

Find two numbers that multiply to $12$ and add to $-7$ : they are $-3$ and $-4$ . Apply the zero-product property:

(x - 3)(x - 4) = 0, \quad x = 3 \text{ or } x = 4.

Step 2: Q2. Rearrange then factor

The equation must equal zero before factoring. Move $10$ to the left, then find factors of $-10$ that add to $3$ :

x^2 + 3x - 10 = 0 \Rightarrow (x + 5)(x - 2) = 0, \quad x = -5 \text{ or } x = 2.

Step 3: Q3. Use the square-root property

The equation is already in squared form, so isolate the square and take both roots. The $\pm$ gives both solutions:

x - 2 = \pm 5, \quad x = 7 \text{ or } x = -3.

Step 4: Q4. Complete the square

Move the constant to the right, add $\left(\frac{8}{2}\right)^2 = 16$ to both sides, then take the square root:

x^2 + 8x = -3 \;\Rightarrow\; x^2 + 8x + 16 = 13 \;\Rightarrow\; (x + 4)^2 = 13 \;\Rightarrow\; x = -4 \pm \sqrt{13}.

Step 5: Q5. Apply the quadratic formula

With $a = 2$ , $b = 3$ , $c = -2$ , compute the discriminant $9 + 16 = 25$ , then simplify:

x = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4}, \quad x = \frac{1}{2} \text{ or } x = -2.

Step 6: Q6. Count solutions using the discriminant

Identify $a = 1$ , $b = 1$ , $c = 4$ , then compute $b^2 - 4ac = 1 - 16 = -15$ . A negative discriminant means no real solutions.

Step 7: Q7. Find the landing time

Factor out $-16t$ to expose both roots. The ball starts at $t = 0$ (the launch time), so the landing is at the other root:

-16t(t - 3) = 0, \quad t = 0 \text{ or } t = 3; \text{ it lands at } t = 3 \text{ s.}

Step 8: Q8. Find the vertex

The axis of symmetry is $x = \frac{-b}{2a} = \frac{6}{2} = 3$ . Substitute to get the $y$ -coordinate:

f(3) = 9 - 18 + 5 = -4, \quad \text{vertex } (3, -4).

Final answer: Q1: $x = 3$ or $4$ . Q2: $x = -5$ or $2$ . Q3: $x = 7$ or $-3$ . Q4: $x = -4 \pm \sqrt{13}$ . Q5: $x = \tfrac{1}{2}$ or $-2$ . Q6: no real solutions. Q7: $t = 3$ s. Q8: vertex $(3, -4)$ .

Sources & how we know this

Math EOC Reference Sheet — Tennessee Department of Education (2024)
Tennessee Academic Standards for Mathematics — Tennessee Department of Education (2024)