Completing the Square

A square with a corner missing

Draw $x^2+4x$x² + 4x as an area. The $x^2$x² is a square with sides x. Split the 4x into two equal strips, each 2 by x, and put one along the right side of the square and one along the bottom.

The shape is nearly a square with sides x + 2. Only one piece is missing: the small corner, which is 2 by 2. Adding that corner would complete the square.

$x^2+4x$x² + 4x arranged as a square with sides x + 2. The corner is missing, and its sides are both 2.

Add the corner, then take it away

The corner is 2 × 2 = 4. With it, the shape is a full square: $x^2+4x+4={\left(x+2\right)}^2$x² + 4x + 4 = (x + 2)². But $x^2+4x$x² + 4x does not have that extra 4, so subtract it again: $x^2+4x={\left(x+2\right)}^2-4$x² + 4x = (x + 2)² − 4.

Check by expanding: ${\left(x+2\right)}^2-4=x^2+4x+4-4=x^2+4x$(x + 2)² − 4 = x² + 4x + 4 − 4 = x² + 4x.

The 2 is half of 4, the coefficient of x, and this is always so. ${\left(x+p\right)}^2=x^2+2px+p^2$(x + p)² = x² + 2px + p², so the middle term is 2p times x. To make a middle term of 4x, p must be 2, which is half of 4. The corner is then $p^2$p², the square of that half.

Drag the right edge to change b. The two strips are each half of bx, and the dashed corner is the square of that half. Set it up for $x^2+6x$x² + 6x.

With a number term

Complete the square on $x^2+6x-1$x² + 6x − 1. Half of 6 is 3, and ${\left(x+3\right)}^2=x^2+6x+9$(x + 3)² = x² + 6x + 9, so $x^2+6x={\left(x+3\right)}^2-9$x² + 6x = (x + 3)² − 9. Put the −1 back: $x^2+6x-1={\left(x+3\right)}^2-9-1={\left(x+3\right)}^2-10$x² + 6x − 1 = (x + 3)² − 9 − 1 = (x + 3)² − 10.

The coefficient of x can be negative. For $x^2-6x+5$x² − 6x + 5, half of −6 is −3, and ${\left(x-3\right)}^2=x^2-6x+9$(x − 3)² = x² − 6x + 9. So $x^2-6x+5={\left(x-3\right)}^2-9+5={\left(x-3\right)}^2-4$x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4.

Reading the lowest point

The completed square shows the least value at a glance. In $x^2+4x={\left(x+2\right)}^2-4$x² + 4x = (x + 2)² − 4, the square ${\left(x+2\right)}^2$(x + 2)² is never below 0, because no square is negative. So ${\left(x+2\right)}^2-4$(x + 2)² − 4 is never below −4.

It equals −4 when the square is 0, which happens when x + 2 = 0, at x = −2. So the least value of $x^2+4x$x² + 4x is −4, at x = −2, and the lowest point of the curve $y=x^2+4x$y = x² + 4x is (−2, −4). This lowest point is the turning point of the curve, also called its vertex.

The curve $y={\left(x+2\right)}^2-4$y = (x + 2)² − 4 turns at (−2, −4), where the square is 0. It meets the x-axis at −4 and 0.

Any completed square

Read ${\left(x-2\right)}^2-3$(x − 2)² − 3 the same way. The bracket is 0 when x = 2, and the number after it is −3, so the lowest point is (2, −3).

Take care with the signs. The x-coordinate is the value that makes the bracket 0, so (x − 2) gives 2 and (x + 2) gives −2. The y-coordinate is the number after the bracket, with its own sign.

The curve $y={\left(x-2\right)}^2-3$y = (x − 2)² − 3 turns at (2, −3).

Solving an equation this way

A completed square also solves equations. Solve $x^2+8x+3=0$x² + 8x + 3 = 0. Half of 8 is 4, and ${\left(x+4\right)}^2=x^2+8x+16$(x + 4)² = x² + 8x + 16, so the equation is ${\left(x+4\right)}^2-16+3=0$(x + 4)² − 16 + 3 = 0, which is ${\left(x+4\right)}^2-13=0$(x + 4)² − 13 = 0.

Add 13 to both sides: ${\left(x+4\right)}^2=13$(x + 4)² = 13. A number whose square is 13 is $\sqrt{13}$√13 or $-\sqrt{13}$−√13, so $x+4=\sqrt{13}$x + 4 = √13 or $x+4=-\sqrt{13}$x + 4 = −√13. Subtract 4: $x=-4+\sqrt{13}$x = −4 + √13 or $x=-4-\sqrt{13}$x = −4 − √13, written $x=-4\pm \sqrt{13}$x = −4 ± √13. The quadratic formula gives the same answer.

The usual mistakes

Not halving. $x^2+6x$x² + 6x is not ${\left(x+6\right)}^2$(x + 6)² minus something: ${\left(x+6\right)}^2$(x + 6)² has 12x in the middle. The bracket takes half of 6, which is 3.

Forgetting to subtract the corner. ${\left(x+3\right)}^2$(x + 3)² is $x^2+6x+9$x² + 6x + 9, so $x^2+6x-1$x² + 6x − 1 is ${\left(x+3\right)}^2-10$(x + 3)² − 10, not ${\left(x+3\right)}^2-1$(x + 3)² − 1.

Mixing up where and how low. In ${\left(x+2\right)}^2-4$(x + 2)² − 4, the least value is −4, and it happens at x = −2. The number inside the bracket gives the x-coordinate, with its sign changed; the number outside gives the value.

When $x^2$x² has a coefficient

Take out the coefficient of $x^2$x² as a factor first, then complete the square inside the bracket. For $-2x^2+12x$−2x² + 12x, take out −2: $-2x^2+12x=-2\left(x^2-6x\right)$−2x² + 12x = −2(x² − 6x). Half of −6 is −3, and $x^2-6x={\left(x-3\right)}^2-9$x² − 6x = (x − 3)² − 9, so $-2x^2+12x=-2\left[{\left(x-3\right)}^2-9\right]=18-2{\left(x-3\right)}^2$−2x² + 12x = −2[(x − 3)² − 9] = 18 − 2(x − 3)².

Check by expanding: $18-2\left(x^2-6x+9\right)=18-2x^2+12x-18=-2x^2+12x$18 − 2(x² − 6x + 9) = 18 − 2x² + 12x − 18 = −2x² + 12x. Now read it. The square ${\left(x-3\right)}^2$(x − 3)² is never negative, so $18-2{\left(x-3\right)}^2$18 − 2(x − 3)² is never more than 18. The greatest value is 18, at x = 3. With a negative coefficient of $x^2$x², the completed square gives a highest point instead of a lowest one.