Polynomials & quadratics
Complete the square
ALEKS placement
Add the square of half the linear coefficient to build a perfect-square trinomial.
What this covers
x2+bx+(2b)2=(x+2b)2- Divide through so the leading coefficient is 1.
- Move the constant to the other side.
- Halve the x-coefficient, square it, and add it to both sides.
- Why half of b, squared: the two numbers that multiply to (b/2)² and add to b are b/2 and b/2, so the trinomial is a perfect square.
- Write the left side as a squared binomial: x, the sign of the x-term, then the half you had before squaring.
- Take square roots of both sides and keep both signs.
Use this when the question asks for vertex form, the centre of a circle, or specifically says to complete the square. For roots, try the factoring method you already know first. If it will not factor and the x-coefficient is even, completing the square keeps the numbers small; otherwise the next quadratic method handles it.
Worked example
Solve x² + 6x − 7 = 0 by completing the square
- x2+6x=7Move the constant across.
- (6÷2)2=9Halve the linear coefficient, then square that half. This is the missing value that makes a squared group.
- x2+6x+9=7+9Add the missing value to both sides so the equation stays balanced.
- (x+3)2=16The left side is now a perfect square.
- x+3=4orx+3=−4Positive four and negative four both square to sixteen. Each gives a possible solution.
- x=1orx=−7Subtract three in each case. Both values satisfy the original equation.
Rewrite x² − 6x + 11 in vertex form.
- x2−6x+9−9+11Half of −6 is −3, and (−3)² = 9. Adding 9 and subtracting 9 adds zero, so the value is unchanged.
- (x−3)2+2x² − 6x + 9 is the square (x − 3)². The leftover numbers give −9 + 11 = 2.
A common mistake
From x2−4x=5
“Adding (−4 ÷ 2)² = 4 completes the square on the left, so 4 has to be added to the right as well: 5 + 4 = 9. Adding it to one side only changes the equation.”
Try one
Rewrite as a squared binomial plus or minus a constant.
x2+12xPractice complete square free
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