Math101learn.math101.caCompleting the Square
Completing the square rewrites a quadratic as a perfect square plus a constant, revealing its vertex and solutions.
Completing the square turns a quadratic expression into a form whose geometry and algebra are easier to read.
The perfect-square pattern
Expanding a squared binomial gives
Therefore an expression $x^2+bx$ becomes part of a perfect square when we add
Half the coefficient of $x$, then square it. This single pattern drives the method.
Rewriting a monic quadratic
To rewrite $x^2+bx+c$:
- group $x^2+bx$;
- add and subtract $(b/2)^2$ so the value does not change;
- factor the resulting perfect-square trinomial;
- combine the constants.
For example,
The expression is unchanged, but the vertex structure is now visible.
Worked example: convert to vertex form
When the leading coefficient is not one
If $a\ne1$, first factor $a$ from the quadratic and linear terms only.
Rewrite $2x^2+12x+7$:
Complete the square inside the brackets:
Forgetting that the added square is inside a factor of $2$ changes the expression.
Solving a quadratic by completing the square
The method can isolate a square and then use square roots. Solve
Move the constant and complete the square:
Take both square-root branches:
The $\pm$ is necessary because both a positive and negative number can have the same square.
Why it reveals the vertex
In vertex form $y=a(x-h)^2+k$, the square is never negative. If $a>0$, its smallest value occurs when $x=h$, so the minimum point is $(h,k)$. If $a<0$, the vertex is a maximum.
Completing the square therefore changes an expanded formula into a form that directly displays the turning point, axis of symmetry, opening direction, and vertical stretch.
Connection to the quadratic formula
The quadratic formula can be derived by completing the square on
After dividing by $a$, moving the constant, and completing the square, solving for $x$ produces
This explains why the discriminant appears: it is the expression left under the square root.
Choosing among quadratic methods
Factoring is fastest when factors are obvious. The quadratic formula works uniformly. Completing the square is especially useful when vertex form or geometric features are required, or when a quadratic lacks easy integer factors.
All three methods agree because they transform the same equation without changing its solutions.
Common mistakes
Adding a number without subtracting it. Rewriting an expression must preserve its value.
Using half of $b$ instead of its square. Add $(b/2)^2$.
Factoring $a$ from the constant too. Initially factor it only from the $x^2$ and $x$ terms unless you deliberately factor the entire expression.
Forgetting the outside multiplier. A square added inside brackets is also multiplied by $a$.
Taking only one square root. Solving $(x-h)^2=q$ requires $x-h=\pm\sqrt q$ when $q>0$.
Quick self-check
- Is the coefficient of $x^2$ inside the working brackets equal to $1$?
- Did I half the linear coefficient and square the result?
- Did I preserve equality by balancing what was added?
- Does expanding the final form reproduce the original?
- If solving, did I include both square-root branches?
Related topics
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Which expression is equivalent to x² + 8x + 3?
- x² + 8x + 3 = (x² + 8x + 16) − 16 + 3
- The trinomial is (x + 4)².
- So the result is (x + 4)² − 13.
End of lesson
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