Completing the Square in a Quadratic Equation

Completing the square turns a quadratic expression into a squared binomial plus or minus a constant. It is useful when factoring is difficult and it explains the structure behind the quadratic formula. The key pattern is x² + bx + (b/2)² = (x + b/2)². Half the coefficient of x, then square that half.

A complete example

Solve x² + 6x + 5 = 0. Move the constant to the other side: x² + 6x = −5. Half of 6 is 3, and 3² = 9. Add 9 to both sides to preserve equality: x² + 6x + 9 = 4. The left factors as (x + 3)², so (x + 3)² = 4. Take both square roots: x + 3 = ±2. Therefore x = −1 or x = −5. Substituting either value into the original equation gives zero.

The added number is not arbitrary. Expanding (x + 3)² gives x² + 6x + 9, which explains why 9 completes the square. In general, the first two terms x² + bx match the expansion of (x + b/2)² except for the final square term.

When the equation has a leading coefficient

If the coefficient of x² is not 1, first make it 1 or factor it from the x² and x terms. For 2x² + 8x + 3 = 0, divide the entire equation by 2: x² + 4x + 3/2 = 0. Move the constant: x² + 4x = −3/2. Half of 4 is 2, and its square is 4. Add 4 to both sides: (x + 2)² = 5/2. Then x + 2 = ±√(5/2), giving x = −2 ± √(5/2). You can rationalize the denominator if a problem requests a particular form, but the exact radical is already a valid solution.

Another route is to factor 2 from the quadratic and linear terms: 2(x² + 4x) + 3 = 0. Completing the square inside the parentheses requires adding 4 inside, which adds 2 × 4 = 8 to the left side overall. Balance the equation accordingly: 2(x² + 4x + 4) + 3 = 8, so 2(x + 2)² = 5. Divide by 2 to reach the same result. The factor outside the parentheses changes how much is added to the equation.

Build a perfect-square trinomial

For x² − 10x, take half of −10, which is −5, then square it: 25. Thus x² − 10x + 25 = (x − 5)². The sign inside the binomial follows the original linear coefficient, while the added square is nonnegative. For x² + 10x, the completed expression is x² + 10x + 25 = (x + 5)².

If the linear coefficient is odd, the added value may be a fraction. For x² + 3x, half of 3 is 3/2 and its square is 9/4. Then x² + 3x + 9/4 = (x + 3/2)². Fractions are expected; avoid rounding because it can change the exact equation.

Use the method to create vertex form

Completing the square also rewrites a quadratic function in vertex form. For y = x² + 6x + 2, group the x-terms and add and subtract 9: y = (x² + 6x + 9) + 2 − 9 = (x + 3)² − 7. The vertex is (−3, −7). The added 9 creates a perfect square, and subtracting 9 preserves the original value.

For a function ax² + bx + c, first factor a from the x² and x terms: a(x² + (b/a)x) + c. Complete the square inside using half of b/a, then compensate outside. This procedure is easier to organize in separate lines, especially when fractions appear.

Common mistakes

• Adding the full coefficient of x instead of half its value squared.
• Adding a number to one side only when solving an equation.
• Forgetting that a factor outside parentheses also multiplies the added square.
• Using a negative value for the square term; a real square is nonnegative.
• Taking only the positive square root after a squared binomial is isolated.
• Changing the original expression when rewriting a function without compensating for the added constant.

A check is to expand the binomial you created and verify it matches the original quadratic terms. Then substitute the solutions into the original equation, not only the transformed one. Equivalent algebra steps should preserve the solution set.

Make the coefficient of the squared variable one

Completing the square is easiest when the coefficient on x² is 1. For 2x² + 8x − 3 = 0, first divide every term by 2: x² + 4x − 1.5 = 0. Move the constant: x² + 4x = 1.5. Take half of 4, square it to get 4, and add 4 to both sides: (x + 2)² = 5.5. Then x = −2 ± √5.5.

If the coefficient is not one, factor it from the x² and x terms before balancing. Do not factor it from the constant unless the entire expression is being factored. Multiplying back or substituting an approximate root into the original equation provides a useful check.

Exam takeaway

Make the coefficient of x² equal to 1, take half the x coefficient, and square it. Add that value to both sides or compensate for it elsewhere in a function. Factor the resulting perfect-square trinomial, take both square roots when solving, and check the result in the original equation.

Keep the equation balanced and track both roots

For x² + 6x + 2 = 0, move the constant first: x² + 6x = −2. Half the x-coefficient, 6, and square the result: (6 ÷ 2)² = 9. Add 9 to both sides to preserve equality: x² + 6x + 9 = 7, so (x + 3)² = 7. Taking square roots gives x + 3 = ±√7, and the two solutions are x = −3 ± √7. The ± is essential because both a positive and a negative number can square to 7.

If the coefficient of x² is not 1, divide every term by that coefficient before completing the square. For 2x² + 8x + 3 = 0, divide by 2 to obtain x² + 4x + 3/2 = 0, then isolate the constant and add (4 ÷ 2)² = 4 to both sides. Fractions are expected and should be kept exact until the final step. Check a solution by substituting it into the original equation or by comparing the sum and product of the roots with the equivalent factored or standard form.