pq Formula

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.

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.

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.

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.

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.