Mathematics · Algebra

pq Formula

Q: How does the pq formula work step by step?

First bring the equation into the normalized form $x^{2} + px + q = 0$; nothing may stand in front of $x^{2}$. Then read off p and q with their signs and insert them into x₁,₂ = −p/2 ± √((p/2)² − q). Compute $-\frac{p}{2}$ first, then the expression under the root, take the root and form the two solutions with plus and minus. Example: $x^{2} - 6x + 8 = 0$ has $p = -6$ and $q = 8$, so x₁,₂ = 3 ± √(9 − 8) = 3 ± 1, giving $x_{1} = 4$ and $x_{2} = 2$. Finally check by substituting into the original equation; it takes seconds and catches sign errors.

Q: What is the difference between the pq formula and the general quadratic formula?

Both solve quadratic equations and give identical results, but they require different starting forms. The pq formula needs the normalized form $x^{2} + px + q = 0$ with leading coefficient 1; if a factor a stands in front of $x^{2}$, you must divide by a first. The general quadratic formula works directly with $ax^{2} + bx + c = 0$ and saves the normalization, but is slightly longer. Which one you use is a matter of taste and curriculum: many German schools teach the pq formula as the standard. What matters is consistency: choose the formula, produce its form, read off the coefficients with signs. For $a = 1$ both routes are equally fast.

Q: What do I do if there is a factor in front of $x^{2}$?

Divide the complete equation by that factor before reading off p and q. From $2x^{2} - 8x + 6 = 0$, division by 2 gives the normalized form $x^{2} - 4x + 3 = 0$ with $p = -4$ and $q = 3$; the pq formula yields x₁,₂ = 2 ± √(4 − 3) = 2 ± 1, that is 3 and 1. Important: divide every single term, including the constant, and the right side (0/2 = 0 stays 0). Dividing does not change the solution set, because both sides are treated equally. The most common mistake is reading p and q from the unnormalized equation; then all subsequent results are wrong. Alternatively use the general quadratic formula, which handles the factor a directly.

Q: What does the expression $(\frac{p}{2})^{2} - q$ under the root tell you?

This expression is the discriminant D of the normalized form and decides solvability before you finish calculating. If D is positive, the root gives a genuine value and there are two distinct real solutions. If D is exactly zero, the root vanishes and both solutions merge into the double root $x = -\frac{p}{2}$; the parabola only touches the x-axis. If D is negative, no real root can be taken, the equation has no real solution and the parabola lies entirely above or below the x-axis. In exams it pays to compute D first: it answers questions about the number of roots without the full formula.

Q: How do I check my solutions with Vieta formulas?

Vieta formulas provide a lightning check: for $x^{2} + px + q = 0$, $x_{1} + x_{2} = -p$ and $x_{1}\cdot{}x_{2} = q$. So once you have computed two solutions, add and multiply them and compare with −p and q. Example: for $x^{2} + 4x - 5 = 0$ the solutions are $x_{1} = 1$ and $x_{2} = -5$; their sum is $-4 = -p$ ✓ and their product $-5 = q$ ✓. If both values match, your solutions are almost certainly correct; if one does not, you have nearly always flipped a sign. Vieta also works backwards: for integer solutions you often find them by cleverly guessing two numbers with the right sum and product.

The pq formula solves any quadratic equation in normalized form $x^{2} + px + q = 0$ , that is with leading coefficient 1.

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Formula

x_{1, 2} = - \frac{p}{2} \pm (\frac{p}{2})^{2} - q

Dimensionless (algebra)

Variables & units

Variables & units
Symbol	Meaning	Unit
$p$	Coefficient of x in the normalized form	dimensionless
$q$	Constant term of the normalized form	dimensionless
x₁, x₂	Solutions of the equation	dimensionless
$D$	Discriminant $D = (\frac{p}{2})^{2} - q$	dimensionless

LaTeX

x_{1,2} = -\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^{2} - q}

QUANTA STUDYquanta-study.de

An upward-opening standard parabola has a dashed axis of symmetry at −p/2, from which two arrows of length √D lead to the roots x₂ and x₁. — The roots x₁ and x₂ lie symmetrically about the vertex position −p/2, each at distance √D.

Derivation & background

The pq formula is the German school form of the solution formula for quadratic equations in normalized form. It follows from completing the square. The expression $D = (\frac{p}{2})^{2} - q$ under the root is called the discriminant: $D > 0$ means two solutions, $D = 0$ one double root, $D < 0$ no real solution. Compared with the general quadratic formula it saves reading off a, but requires normalizing to leading coefficient 1 first.

Exam blueprint

Validity range

Use normalized form $x^2+px+q=0$ with leading coefficient 1; otherwise normalize first. Real roots require $(\frac{p}{2})^2-q\ge0$ .

Derivation steps

Completing the square turns $x^{2} + px + q$ into a shifted square.

1Complete $x^{2} + px$ to $(x +\frac{ p}{2})^{2} - (\frac{p}{2})^{2}$ and move q to the other side.
2Taking the root and solving for x gives $x = -\frac{p}{2} \pm{} \sqrt{((\frac{p}{2})^{2} - q)}$ .

Rearrangements

Determine discriminant

D = \left(\frac{p}{2}\right)^{2} - q

$D > 0$ : two solutions, $D = 0$ : one double root, $D < 0$ : no real solution.

Vieta formulas

x_{1} + x_{2} = -p, \quad x_{1} \cdot x_{2} = q

Quick check: sum and product of the solutions must give −p and q.

Vertex of the parabola

x_{S} = -\frac{p}{2}

The solutions lie symmetrically around the vertex $x = -\frac{p}{2}$ .

Task variant

Solve $x^{2} + 4x - 5 = 0$ .

$p = 4$ , $q = -5$ . x₁,₂ = −2 ± √(4 + 5) = −2 ± 3, so $x_{1} = 1$ , $x_{2} = -5$ .
Check with Vieta: $1 + (-5) = -4 = -p$ ✓ and $1\cdot{}(-5) = -5 = q$ ✓.

Solve $2x^{2} - 8x + 6 = 0$ with the pq formula.

Normalize first: $x^{2} - 4x + 3 = 0$ , so $p = -4$ , $q = 3$ . x₁,₂ = 2 ± √(4 − 3) = 2 ± 1, so $x_{1} = 3$ , $x_{2} = 1$ .

Common mistakes

A factor $a \ne{} 1$ sits in front of $x^{2}$ , yet p and q are read off directly.

First divide the whole equation by a; only then is it in normalized form.

Copying the sign of p wrongly: in $x^{2} - 6x + 8$ , $p = -6$ , not 6.

$-\frac{p}{2}$ then becomes +3; carry signs through consistently.

Forgetting the ± and giving only one solution.

With a positive discriminant there are always two solutions.

Computing $(\frac{p}{2})^{2}$ as $\frac{p^{2}}{2}$ .

Halve first, then square: $(\frac{6}{2})^{2} = 9$ , not $\frac{36}{2} = 18$ .

Exam context

Standard tool for roots, intersections and substitution tasks in exams.

Formula cluster

Quadratic equations

pq formula, general quadratic formula and binomial formulas solve the same class of equations.

x = (−b ± √D) / 2a

Quadratic Formula

(a ± b)² = a² ± 2ab + b²

Binomial Formulas

d/dx(xⁿ) = n·xⁿ⁻¹

Power Rule of Differentiation

Worked example

$x^{2} - 6x + 8 = 0$ : $p = -6$ , $q = 8$ . $x_{1,2} = 3 \pm \sqrt{9-8} = 3 \pm 1$ , so $x_{1} = 4$ , $x_{2} = 2$ . Check: $4^{2} - 6\cdot{}4 + 8 = 0$ ✓ and $2^{2} - 6\cdot{}2 + 8 = 0$ ✓.

Applications

Roots of parabolas, intersections of function graphs, projectile and braking-distance problems, substitution for biquadratic equations

How do you calculate with pq Formula?

Here is how to work through a typical pq Formula (x₁,₂ = −p/2 ± √((p/2)² − q)) task step by step:

1
Task
Solve $x^{2} + 4x - 5 = 0$ .
Solution path
1. $p = 4$ , $q = -5$ . x₁,₂ = −2 ± √(4 + 5) = −2 ± 3, so $x_{1} = 1$ , $x_{2} = -5$ .
2. Check with Vieta: $1 + (-5) = -4 = -p$ ✓ and $1\cdot{}(-5) = -5 = q$ ✓.
2
Task
Solve $2x^{2} - 8x + 6 = 0$ with the pq formula.
Solution path
Normalize first: $x^{2} - 4x + 3 = 0$ , so $p = -4$ , $q = 3$ . x₁,₂ = 2 ± √(4 − 3) = 2 ± 1, so $x_{1} = 3$ , $x_{2} = 1$ .

Frequently asked questions about pq Formula

How does the pq formula work step by step?+

First bring the equation into the normalized form $x^{2} + px + q = 0$ ; nothing may stand in front of $x^{2}$ . Then read off p and q with their signs and insert them into x₁,₂ = −p/2 ± √((p/2)² − q). Compute $-\frac{p}{2}$ first, then the expression under the root, take the root and form the two solutions with plus and minus. Example: $x^{2} - 6x + 8 = 0$ has $p = -6$ and $q = 8$ , so x₁,₂ = 3 ± √(9 − 8) = 3 ± 1, giving $x_{1} = 4$ and $x_{2} = 2$ . Finally check by substituting into the original equation; it takes seconds and catches sign errors.

What is the difference between the pq formula and the general quadratic formula?+

Both solve quadratic equations and give identical results, but they require different starting forms. The pq formula needs the normalized form $x^{2} + px + q = 0$ with leading coefficient 1; if a factor a stands in front of $x^{2}$ , you must divide by a first. The general quadratic formula works directly with $ax^{2} + bx + c = 0$ and saves the normalization, but is slightly longer. Which one you use is a matter of taste and curriculum: many German schools teach the pq formula as the standard. What matters is consistency: choose the formula, produce its form, read off the coefficients with signs. For $a = 1$ both routes are equally fast.

What do I do if there is a factor in front of

x^{2}

Divide the complete equation by that factor before reading off p and q. From $2x^{2} - 8x + 6 = 0$ , division by 2 gives the normalized form $x^{2} - 4x + 3 = 0$ with $p = -4$ and $q = 3$ ; the pq formula yields x₁,₂ = 2 ± √(4 − 3) = 2 ± 1, that is 3 and 1. Important: divide every single term, including the constant, and the right side (0/2 = 0 stays 0). Dividing does not change the solution set, because both sides are treated equally. The most common mistake is reading p and q from the unnormalized equation; then all subsequent results are wrong. Alternatively use the general quadratic formula, which handles the factor a directly.

What does the expression

(\frac{p}{2})^{2} - q

under the root tell you?+

This expression is the discriminant D of the normalized form and decides solvability before you finish calculating. If D is positive, the root gives a genuine value and there are two distinct real solutions. If D is exactly zero, the root vanishes and both solutions merge into the double root $x = -\frac{p}{2}$ ; the parabola only touches the x-axis. If D is negative, no real root can be taken, the equation has no real solution and the parabola lies entirely above or below the x-axis. In exams it pays to compute D first: it answers questions about the number of roots without the full formula.

How do I check my solutions with Vieta formulas?+

Vieta formulas provide a lightning check: for $x^{2} + px + q = 0$ , $x_{1} + x_{2} = -p$ and $x_{1}\cdot{}x_{2} = q$ . So once you have computed two solutions, add and multiply them and compare with −p and q. Example: for $x^{2} + 4x - 5 = 0$ the solutions are $x_{1} = 1$ and $x_{2} = -5$ ; their sum is $-4 = -p$ ✓ and their product $-5 = q$ ✓. If both values match, your solutions are almost certainly correct; if one does not, you have nearly always flipped a sign. Vieta also works backwards: for integer solutions you often find them by cleverly guessing two numbers with the right sum and product.

Quanta exam set

Curated exam set for "pq Formula":

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Which formula describes pq Formula?

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How do you rearrange pq Formula for Determine discriminant?

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Which common mistake happens with pq Formula?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, College Algebra 2e, §2.5: Quadratic Equations

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pq Formula

The pq formula solves any quadratic equation in normalized form

x^{2} + px + q = 0

, that is with leading coefficient 1.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

p

Coefficient of x in the normalized form

dimensionless

q

Constant term of the normalized form

dimensionless

x₁, x₂

Solutions of the equation

dimensionless

D

Discriminant

D = (\frac{p}{2})^{2} - q

dimensionless

Derivation & background

Exam blueprint

Validity range

Use normalized form $x^2+px+q=0$ with leading coefficient 1; otherwise normalize first. Real roots require $(\frac{p}{2})^2-q\ge0$ .

Derivation steps

Completing the square turns $x^{2} + px + q$ into a shifted square.

1Complete $x^{2} + px$ to $(x +\frac{ p}{2})^{2} - (\frac{p}{2})^{2}$ and move q to the other side.
2Taking the root and solving for x gives $x = -\frac{p}{2} \pm{} \sqrt{((\frac{p}{2})^{2} - q)}$ .

How do you calculate with pq Formula?

Here is how to work through a typical pq Formula (x₁,₂ = −p/2 ± √((p/2)² − q)) task step by step:

Task

Solve $x^{2} + 4x - 5 = 0$ .

Solution path

$p = 4$ , $q = -5$ . x₁,₂ = −2 ± √(4 + 5) = −2 ± 3, so $x_{1} = 1$ , $x_{2} = -5$ .
Check with Vieta: $1 + (-5) = -4 = -p$ ✓ and $1\cdot{}(-5) = -5 = q$ ✓.

Task

Solve $2x^{2} - 8x + 6 = 0$ with the pq formula.

Solution path

Normalize first: $x^{2} - 4x + 3 = 0$ , so $p = -4$ , $q = 3$ . x₁,₂ = 2 ± √(4 − 3) = 2 ± 1, so $x_{1} = 3$ , $x_{2} = 1$ .

Frequently asked questions about pq Formula

How does the pq formula work step by step?+

What is the difference between the pq formula and the general quadratic formula?+

What do I do if there is a factor in front of

x^{2}

What does the expression

(\frac{p}{2})^{2} - q

under the root tell you?+

How do I check my solutions with Vieta formulas?+

Retain pq Formula for exams

Create a curated FSRS exam set for x₁,₂ = −p/2 ± √((p/2)² − q): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential