QuickCalculators rewrites a quadratic by completing the square, producing vertex form and optional roots from the completed square. Enter coefficients for ax² + bx + c, then read the half-coefficient steps, the squared binomial, and the vertex coordinates.
Completing the square is a method, not only an answer. The work panel shows each add-and-subtract stage so the algebra stays checkable by hand.
Complete the square of a quadratic
Completing the square turns ax² + bx + c into a form a(x − h)² + k (or (x − h)² + k when a = 1). QuickCalculators scales to a leading coefficient of 1 when needed, then builds the perfect-square trinomial from the linear term.
Vertex form makes the vertex (h, k) readable at a glance. The same rewrite supports solving by isolating the squared binomial and taking square roots.
Follow the completing the square steps
For x² + bx + c with a = 1, take half of b, square that half, and add and subtract the square so a perfect-square trinomial appears. QuickCalculators shows half of b, the square, and the regrouping into (x + b/2)² plus a constant.
Example outline for x² + 6x + 5: half of 6 is 3, and 3² is 9. Rewrite as x² + 6x + 9 − 9 + 5 = (x + 3)² − 4. When a ≠ 1, factor a from the x terms first, then complete the square inside.
Find the vertex from vertex form
In vertex form a(x − h)² + k, the vertex is (h, k). If a > 0 the parabola opens upward and (h, k) is a minimum. If a < 0 it opens downward and (h, k) is a maximum. QuickCalculators reports those coordinates from the completed square.
For (x + 3)² − 4, h = −3 and k = −4, so the vertex is (−3, −4). Sign carefulness matters: (x + 3) means h = −3, not +3.
Solve the quadratic by completing the square
After completing the square, isolate the squared binomial and take square roots on both sides. The two roots (when they exist over the reals) come from the ± step. QuickCalculators can continue from vertex form to those solutions.
From (x + 3)² − 4 = 0, move the constant: (x + 3)² = 4, so x + 3 = ±2, hence x = −1 or x = −5.
Checking in the original quadratic confirms both roots.
Complete the square of x² + 6x + 5
The fixture completes the square for x² + 6x + 5 on QuickCalculators.
- Half of 6 is 3; square to get 9.
- Write x² + 6x + 9 − 9 + 5.
- Regroup as (x + 3)² − 4.
- Read vertex (−3, −4). Solving (x + 3)² = 4 yields x = −1 and x = −5.
Avoid this common misconception
A frequent error is forgetting to balance the constant when adding (b/2)², which changes the expression's value. Completing the square adds and subtracts the same square, or distributes carefully when a ≠ 1. Dropping the compensating subtraction invents a different quadratic. QuickCalculators keeps both the added and subtracted square visible.
Frequently asked questions
What does completing the square mean?
Completing the square rewrites a quadratic so a perfect-square binomial appears, usually producing vertex form a(x − h)² + k.
How do you complete the square for x² + 6x + 5?
Half of 6 is 3, and 3² is 9, so the form becomes (x + 3)² − 4. Entering those coefficients returns that fixture.
How do you find the vertex after completing the square?
Read (h, k) from a(x − h)² + k. For (x + 3)² − 4, the vertex is (−3, −4).
Can completing the square solve any quadratic?
Completing the square solves any quadratic with real or complex roots. Real square-root steps require a nonnegative right-hand side after isolating the square; otherwise the roots are complex.
How does completing the square relate to the quadratic formula?
The quadratic formula is derived by completing the square on ax² + bx + c in general form. Both methods share the same algebraic core.
What if the leading coefficient is not 1?
Factor a from the x² and x terms first, complete the square inside the parentheses, then distribute a back across the squared binomial and constant as needed.
Summary
QuickCalculators rewrites quadratics by completing the square into vertex form and can continue to roots. Half of b, square that half, and balance the constant. The fixture x² + 6x + 5 becomes (x + 3)² − 4 with vertex (−3, −4) and roots −1 and −5.
Always add and subtract the same square so the value does not change.