TeachLens

Browse lessons
Algebra
Balancing EquationsSimultaneous EquationsDistributive PropertyFactoring QuadraticsCommon FactorDifference of SquaresFactor by GroupingCompleting the SquareQuadratic FormulasCubic FormulasPower Formulas
Geometry
Area of a Triangle
Trigonometry
SAS Triangle AreaSin, Cos, TanHow Sine Is Derived
Circles
How Pi Is DerivedArea of a CircleArea of a SectorPerimeter of a SectorRadians

Quadratic form

Start with x² plus x terms

x² + 6x + 5
x2 + 6x + 5x2half of 6 is 3(x + 3)2subtract 4(x + 3)2 - 4

Completing the square rewrites a quadratic around a perfect square bracket.

Detailed notes

Worked Example

Solve 2x² - 6x - 10 = 0 by changing the quadratic into a perfect square.

Step 1: Divide by the coefficient of x²

Completing the square is easiest when the x² coefficient is 1. Divide every term by 2 first.

2x2 - 6x - 10 = 0
x2 - 3x - 5 = 0

Step 2: Move the constant

Keep the x terms on the left, and move the number term to the right side.

x2 - 3x = 5

Step 3: Add the square of half the x coefficient

Half of -3 is -32. Square it, then add that same value to both sides so the equation stays balanced.

x2 - 3x + (-32)2 = 5 + (-32)2
x2 - 3x + 94 = 5 + 94
x2 - 3x + 94 = 294

Step 4: Factor the left side

The left side is now a perfect square trinomial, so it folds into one squared bracket.

(x - 32)2 = 294

Step 5: Use the square root property

If a square equals a number, the bracket equals the positive or negative square root of that number.

x - 32 = ±294
x - 32 = ±292

Step 6: Solve for x

Add 32 to both sides. The two signs give the two solutions of the quadratic.

x = 32 ± 292
x = 3 ± 292