Completing the square and the quadratic formula
Not every quadratic equation factors neatly. Completing the square and the quadratic formula solve every quadratic equation, including those with irrational answers.
Quadratic equations
A quadratic equation has the form \(ax^2 + bx + c = 0\) with \(a \neq 0\). It can have two, one, or no real solutions. Factoring is quickest when it works, but \(x^2 + 4x - 1 = 0\) has no nice integer factors, so we need more general methods.
Square root property
If \(x^2 = k\) with \(k \ge 0\), then \(x = \pm\sqrt{k}\). So \(x^2 = 20\) gives \(x = \pm 2\sqrt{5}\), and \((x - 3)^2 = 16\) gives \(x - 3 = \pm 4\), so x = 7 or x = -1. Forgetting the negative root is a very common error.
Completing the square
To solve \(x^2 + 6x - 7 = 0\), move the constant: \(x^2 + 6x = 7\). Add the square of half the x-coefficient, \((6/2)^2 = 9\), to both sides: \(x^2 + 6x + 9 = 16\). The left side is now \((x + 3)^2\), so \(x + 3 = \pm 4\), giving x = 1 or x = -7.
When a is not 1
Divide by a first. For \(2x^2 + 8x - 3 = 0\): \(x^2 + 4x = \tfrac{3}{2}\), add 4 to get \((x + 2)^2 = \tfrac{11}{2}\), so \(x = -2 \pm \sqrt{\tfrac{11}{2}}\), about 0.35 or -4.35. Completing the square is also how the vertex form of a parabola is found.
The quadratic formula
Completing the square on \(ax^2 + bx + c = 0\) in general gives the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It always works. Write the equation in standard form first and identify a, b, and c carefully, including their signs.
Using the formula
For \(x^2 + 4x - 1 = 0\): a = 1, b = 4, c = -1, so \(x = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}\). That is about 0.236 and -4.236. For \(2x^2 - 7x + 3 = 0\): \(x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}\), so x = 3 or x = 1/2.
Choosing a method
Try factoring first if the numbers look friendly. Use the square root property when there is no x term. Use the quadratic formula for everything else; it never fails. Completing the square is worth mastering because it explains where the formula comes from and is needed for graphing.
Worked example
A rectangular poster has area 60 cm² and its length is 4 cm more than its width: \(w(w + 4) = 60\), so \(w^2 + 4w - 60 = 0\). The formula gives \(w = \frac{-4 \pm \sqrt{16 + 240}}{2} = \frac{-4 \pm 16}{2}\), so w = 6 or w = -10. The width is 6 cm and the length 10 cm.
Key takeaways
Quadratic equations, ax squared + bx + c = 0, can have two, one, or no real solutions. The square root property solves x squared = k as x = plus or minus the square root of k; never forget the negative root. Completing the square adds (b/2) squared to both sides after making a = 1. The quadratic formula, x = (-b plus or minus the square root of b squared - 4ac) over 2a, solves every quadratic; write the equation in standard form and watch the signs. Factor when easy, otherwise use the formula.
Flashcard
ax^2 + bx + c = 0, a not zero
Check your understanding
Solve \(x^2 + 6x - 7 = 0\).