Complete the square step by step

• Choose the square-completing term by halving and squaring the x coefficient.
• Preserve an equation by adding the same term to both sides before factoring.
• Use both square-root branches and verify the resulting solutions.

Build the idea

Worked example

1. Step 1

   Add sixteen to both sides so the variable terms remain on the left and the constant to be balanced is on the right.

   $$x^2+6x=16$$

2. Step 2

   Half of six is three, and three squared is nine. Add nine to both sides, then factor the completed trinomial.

   $$x^2+6x+9=25\Rightarrow(x+3)^2=25$$

3. Step 3

   Take both square-root branches. Subtracting three from five gives two, while subtracting three from negative five gives negative eight.

   $$x=2\quad\text{or}\quad x=-8$$

• Adding nine only to the left side because it completes the visible square.

  The square model explains the left expression, but equality requires adding the same nine to the right side as well.

• Taking only the positive square root of twenty-five.

  Both five and negative five square to twenty-five, so both branches must be solved unless later restrictions remove one.

Solve x squared minus eight x plus seven equals zero by completing the square and check both values.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Lesson overview

   Completing the square adds the exact area needed to turn x squared plus a linear term into a square binomial, while preserving equality by making the same addition on both sides. A perfect square trinomial has the form x squared plus two a x plus a squared. Its middle coefficient is twice the number inside the binomial, so halving that coefficient identifies the missing side length. Choose the square-completing term by halving and squaring the x coefficient.

2. Part 2

   Visual model and equations

   Adding the missing square changes an expression, but an equation must remain balanced. The same amount is therefore added to both sides before the left side is rewritten as one squared binomial. Arrange one x-squared tile and six x-tiles into an almost-square with three x-tiles along two sides. A three-by-three corner of nine unit tiles completes the square. x squared plus six x plus nine equals the quantity x plus three, squared. x plus three equals positive or negative five.

   • x squared plus six x plus nine equals the quantity x plus three, squared.
   • x plus three equals positive or negative five.
3. Part 3

   Worked example

   Solve x squared plus six x minus sixteen equals zero by completing the square, and verify both solutions. Move the constant, add the square of half the linear coefficient to both sides, factor the perfect square, and solve both square-root cases. Add sixteen to both sides so the variable terms remain on the left and the constant to be balanced is on the right. Half of six is three, and three squared is nine. Add nine to both sides, then factor the completed trinomial. Take both square-root branches. Subtracting three from five gives two, while subtracting three from negative five gives negative eight. The equation has two solutions: x equals two and x equals negative eight. Substitution gives four plus twelve minus sixteen equals zero for two, and sixty-four minus forty-eight minus sixteen equals zero for negative eight.

   • x squared plus six x equals sixteen.
   • x squared plus six x plus nine equals twenty-five, so the quantity x plus three, squared, equals twenty-five.
   • x equals two or x equals negative eight.
4. Part 4

   Checks, practice, and scope

   Once the square is isolated, taking square roots creates two cases unless the right side is zero. Keeping the plus-or-minus symbol prevents the common loss of one valid solution. Adding nine only to the left side because it completes the visible square. The square model explains the left expression, but equality requires adding the same nine to the right side as well. Taking only the positive square root of twenty-five. Both five and negative five square to twenty-five, so both branches must be solved unless later restrictions remove one. Solve x squared minus eight x plus seven equals zero by completing the square and check both values. Subtract seven from both sides, then add sixteen because half of negative eight is negative four. The completed equation is the quantity x minus four, squared, equals nine, giving x equals one or seven. Half the linear coefficient determines the side length that completes the square. Every added square-completing term must appear on both sides of an equation. A positive squared value usually produces two square-root branches. This lesson uses a monic quadratic; a different leading coefficient should be factored out before the same pattern is applied. Complex solutions are outside this lesson when the completed square equals a negative real number.

Scope and limitations

• This lesson uses a monic quadratic; a different leading coefficient should be factored out before the same pattern is applied.
• Complex solutions are outside this lesson when the completed square equals a negative real number.