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

TEKS A.8A lists four methods: 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 5x equals 24 must first become x squared plus 5x minus 24 equals 0; factoring while one side is 24 tells you nothing, because 24 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.

Why do you reject some solutions in a quadratic word problem?

A quadratic usually has two solutions, but the context often makes one impossible. A negative time before launch, a negative length or width, or a negative count cannot occur in reality, so you discard that root and report only the meaningful solution with units. Rejecting the unrealistic root is part of the interpretive credit STAAR awards.

TexasMaths

STAAR Algebra I: a complete guide to solving quadratic equations

A deep-dive STAAR Algebra I guide to solving quadratic equations, the solving side of the Quadratic Functions and Equations reporting category (about 25 percent of the test). Covers factoring and the zero-product property, the square-root property and completing the square, the quadratic formula and the discriminant, and real-world applications like projectile motion and area.

Generated by Claude Opus 4.816 min readA.6, A.8Updated 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
Applications and interpretation
How this category is examined
Check your knowledge

What this category demands

This guide covers the solving side of the Quadratic Functions and Equations reporting category (TEKS A.6, A.8), about 25 percent of the STAAR Algebra I test. The function side (graphing, transformations, domain and range, writing) is in the companion module. Solving quadratics is where the Meets and Masters standards are usually decided. 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, and quadratic applications.

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, and STAAR'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.

Applications and interpretation

Real-world quadratics model projectile motion ( $h(t) = -16t^2 + v_0 t + h_0$ , with the ground at $h = 0$ and the peak at the vertex) and area (length times width equals the given area). Solve, then interpret: reject negative times, lengths, or counts, and state the answer with units. The vertex gives a maximum or minimum; the zeros give start and end points.

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; $h(1.25) = -16(1.5625) + 40(1.25) + 6 = -25 + 50 + 6 = 31$ feet.

step 2 Set height to zero for landing

$-16t^2 + 40t + 6 = 0$ . Use the quadratic formula 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

$t = \frac{-40 - 44.5}{-32} \approx 2.6$ s (the positive landing time); the negative root is rejected. The ball reaches 31 feet and lands at about 2.6 seconds.

How this category is examined

Multiple choice and multiselect. Solve factorable quadratics, count solutions from the discriminant, 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.
Inline choice. State the number of solutions or whether the graph crosses the axis.

Check your knowledge

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

Solve $x^2 - 7x + 12 = 0$ by factoring. (1 point)
Solve $x^2 - 2x - 15 = 0$ by factoring. (1 point)
Solve $(x - 4)^2 = 25$ . (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)
A rectangle is 5 m longer than wide with area 36. Find the width. (2 points)

Solutions

Step 1: Q1. Factor and apply the zero-product property

Find two numbers that multiply to $12$ and add to $-7$ : $-3$ and $-4$ . Set each factor to zero:

(x - 3)(x - 4) = 0 \Rightarrow x = 3 \text{ or } x = 4

Step 2: Q2. Factor and solve

Find two numbers that multiply to $-15$ and add to $-2$ : $-5$ and $3$ . Set each factor to zero:

(x - 5)(x + 3) = 0 \Rightarrow x = 5 \text{ or } x = -3

Step 3: Q3. Use the square-root property

The equation is already in squared form, so isolate the square and take the square root of both sides, remembering the $\pm$ :

x - 4 = \pm 5 \Rightarrow x = 9 \text{ or } x = -1

Step 4: Q4. Complete the square

Move the constant, then add $\left(\frac{8}{2}\right)^2 = 16$ to both sides to create a perfect-square trinomial:

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

Step 5: Q5. Apply the quadratic formula

Identify $a = 2$ , $b = 3$ , $c = -2$ , substitute into the formula, and simplify the radical:

x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4} \Rightarrow x = \frac{1}{2} \text{ or } x = -2

Step 6: Q6. Use the discriminant to count solutions

Compute $b^2 - 4ac$ with $a = 1$ , $b = 1$ , $c = 4$ . A negative discriminant means no real solutions:

b^2 - 4ac = 1 - 16 = -15 < 0 \Rightarrow \text{no real solutions}

Step 7: Q7. Set height to zero and factor

Set $h(t) = 0$ and factor out the common factor $-16t$ . The zero $t = 0$ is the launch; the positive zero is the landing:

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

Step 8: Q8. Set up and solve a quadratic area problem

Let $w$ be the width and $w + 5$ the length. Area equals length times width, so set up the equation, rearrange to standard form, and factor. Reject the negative root because width cannot be negative:

w(w + 5) = 36 \Rightarrow w^2 + 5w - 36 = 0 \Rightarrow (w + 9)(w - 4) = 0 \Rightarrow \text{width } 4 \text{ m}

Sources & how we know this

STAAR Algebra I Reference Materials — Texas Education Agency (2024)
STAAR Algebra I Assessed Curriculum — Texas Education Agency (2024)