Solving an equation using standard form

Solving f(x) = 0 using the standard form

Starting from a(x - alpha)^2 + beta = 0, we isolate the square term:

(x - alpha)^2 = -beta/a

There are three cases depending on the sign of -beta/a:

Case	Consequence
-beta/a < 0	no real solution
-beta/a = 0	a single solution x = alpha
-beta/a > 0	two solutions x = alpha - sqrt(-beta/a) and x = alpha + sqrt(-beta/a)

Example

Let f(x) = 2(x - 2)² - 3. We solve f(x) = 0:

2(x - 2)² - 3 = 0 (x - 2)² = 3/2 x – 2 = √(3/2) or x – 2 = –√(3/2) x = 2 + √(3/2) or x = 2 – √(3/2)

A simpler case

Let g(x) = (x - 1)^2 - 9. We solve (x - 1)^2 = 9, so x - 1 = 3 or x - 1 = -3, i.e. x = 4 or x = -2.

A classic pitfall

When converting from (x - alpha)^2 = k to x - alpha = ± sqrt(k), one must never forget the negative solution: a square root has only one positive value, but the equation retains two solutions that are symmetrical about alpha.