Mathematics · Algebra

pq Formula

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

First bring the equation into the normalized form x² + px + q = 0; nothing may stand in front of x². Then read off p and q with their signs and insert them into x₁,₂ = −p/2 ± √((p/2)² − q). Compute −p/2 first, then the expression under the root, take the root and form the two solutions with plus and minus. Example: x² − 6x + 8 = 0 has p = −6 and q = 8, so x₁,₂ = 3 ± √(9 − 8) = 3 ± 1, giving x₁ = 4 and x₂ = 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 x² + px + q = 0 with leading coefficient 1; if a factor a stands in front of x², you must divide by a first. The general quadratic formula works directly with ax² + 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 x²?

Divide the complete equation by that factor before reading off p and q. From 2x² − 8x + 6 = 0, division by 2 gives the normalized form x² − 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 (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 = −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 x² + px + q = 0, x₁ + x₂ = −p and x₁·x₂ = q. So once you have computed two solutions, add and multiply them and compare with −p and q. Example: for x² + 4x − 5 = 0 the solutions are x₁ = 1 and x₂ = −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 x² + 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

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

Dimensionless (algebra)

Variables & units – pq Formula

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 = (p/2)² − q	dimensionless

Derivation & background – pq Formula

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 = (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

Applies only to quadratic equations in normalized form x² + px + q = 0; if there is a factor in front of x², divide by it first. Real solutions exist only for (p/2)² − q ≥ 0.

Derivation steps

Completing the square turns x² + px + q into a shifted square.

1Complete x² + px to (x + p/2)² − (p/2)² and move q to the other side.
2Taking the root and solving for x gives x = −p/2 ± √((p/2)² − q).

Rearrangements

Determine discriminant

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

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

Vieta formulas

x_{1} + x_{2} = - p, 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 = −p/2.

Task variant

Solve x² + 4x − 5 = 0.

p = 4, q = −5. x₁,₂ = −2 ± √(4 + 5) = −2 ± 3, so x₁ = 1, x₂ = −5. Check with Vieta: 1 + (−5) = −4 = −p ✓ and 1·(−5) = −5 = q ✓.

Solve 2x² − 8x + 6 = 0 with the pq formula.

Normalize first: x² − 4x + 3 = 0, so p = −4, q = 3. x₁,₂ = 2 ± √(4 − 3) = 2 ± 1, so x₁ = 3, x₂ = 1.

Common mistakes

A factor a ≠ 1 sits in front of x², 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 x² − 6x + 8, p = −6, not 6.

−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 (p/2)² as p²/2.

Halve first, then square: (6/2)² = 9, not 36/2 = 18.

Exam context

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Formula cluster

Quadratic equations

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

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

Quadratic Formula

(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}, (a + b) (a - b) = a^{2} - b^{2}

Binomial Formulas

\frac{d}{d x} (x^{n}) = n \cdot x^{n - 1}

Power Rule of Differentiation

Worked example

x² − 6x + 8 = 0: p = −6, q = 8. x₁,₂ = 3 ± √(9 − 8) = 3 ± 1, so x₁ = 4, x₂ = 2. Check: 4² − 6·4 + 8 = 0 ✓ and 2² − 6·2 + 8 = 0 ✓.

Applications

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

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Which formula describes pq Formula?

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How do you rearrange x₁,₂ = −p/2 ± √((p/2)² − q) for Determine discriminant?

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Which common mistake happens with pq Formula?

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Scientific sources

[1]Bronstein, I.N. et al. (2020). Taschenbuch der Mathematik. Springer.

Common notations & search queries

x=-p/2±sqrt((p/2)^2-q)x^2+px+q=0pq Formelp q Formelminus p halbe plus minus wurzelPQ-Formel Rechnerquadratische Gleichung Normalformpq formulaMitternachtsformel Unterschied

Related formulas

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

Quadratic Formula

(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}, (a + b) (a - b) = a^{2} - b^{2}

Binomial Formulas

\frac{d}{d x} (x^{n}) = n \cdot x^{n - 1}

Power Rule of Differentiation

More Mathematics formulas

a^{2} + b^{2} = c^{2}

Pythagorean Theorem

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

Quadratic Formula

e^{i φ} = cos φ + i \cdot sin φ

Euler's Formula

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

Product Rule of Differentiation

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Frequently asked questions about pq Formula

How does the pq formula work step by step?+

First bring the equation into the normalized form x² + px + q = 0; nothing may stand in front of x². Then read off p and q with their signs and insert them into x₁,₂ = −p/2 ± √((p/2)² − q). Compute −p/2 first, then the expression under the root, take the root and form the two solutions with plus and minus. Example: x² − 6x + 8 = 0 has p = −6 and q = 8, so x₁,₂ = 3 ± √(9 − 8) = 3 ± 1, giving x₁ = 4 and x₂ = 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 x² + px + q = 0 with leading coefficient 1; if a factor a stands in front of x², you must divide by a first. The general quadratic formula works directly with ax² + 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 x²?+

Divide the complete equation by that factor before reading off p and q. From 2x² − 8x + 6 = 0, division by 2 gives the normalized form x² − 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 (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 = −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 x² + px + q = 0, x₁ + x₂ = −p and x₁·x₂ = q. So once you have computed two solutions, add and multiply them and compare with −p and q. Example: for x² + 4x − 5 = 0 the solutions are x₁ = 1 and x₂ = −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.

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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 x² + 4x − 5 = 0.
Solution path
p = 4, q = −5. x₁,₂ = −2 ± √(4 + 5) = −2 ± 3, so x₁ = 1, x₂ = −5. Check with Vieta: 1 + (−5) = −4 = −p ✓ and 1·(−5) = −5 = q ✓.
2
Task
Solve 2x² − 8x + 6 = 0 with the pq formula.
Solution path
Normalize first: x² − 4x + 3 = 0, so p = −4, q = 3. x₁,₂ = 2 ± √(4 − 3) = 2 ± 1, so x₁ = 3, x₂ = 1.