Recipe

Complete the square when the leading coefficients aren't 1 (Step 4 continued)

Use when: Step 4 of the general-form recipe needs to complete the square, but A and/or C aren't both 1

Goal: a fully factored, standard-form equation, ready for Step 5's center, vertex, and focus read-off

Taught in Translated conics and completing the square · extends Analyze a conic from general form

1. 1

   Group the x-terms and the y-terms separately, then factor A out of the x-group and C out of the y-group.

   Example

   2x^2-8x+3y^2+12y=0 groups and factors to 2(x^2-4x)+3(y^2+4y)=0.

   Why: completing the square only works on a group shaped like x^2+bx. A leading coefficient other than 1 has to come out first, or the "half the middle coefficient, squared" step reads the wrong number.

2. 2

   Take half the inner linear coefficient, square it, and add it inside each group. Since that addition sits inside a factored group, add the SAME amount times the factored-out coefficient to the other side.

   Example

   Half of -4 squared is 4; half of 4 squared is 4. Add 4 inside each group, and 2\times4=8 plus 3\times4=12 to the right: 2(x^2-4x+4)+3(y^2+4y+4)=8+12.

   Why: the parentheses are multiplied by the factored-out coefficient, so whatever number is tucked inside is really added that many times over. Skipping this scaling is the single most common error in this step.

3. 3

   Factor each completed-square group and simplify the constant on the right; the equation is now in the family's standard form.

   Example

   2(x^2-4x+4)+3(y^2+4y+4)=20 factors to 2(x-2)^2+3(y+2)^2=20.

   Why: this is exactly the standard form Step 5 of the general-form recipe expects. The leading coefficients ride along instead of vanishing, since they weren't 1 to begin with.