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MathematicsGrade 10· U.S. National — Common Core & NGSS
Aligned to:Common Core State Standards (Math)

Deriving the Quadratic Formula by Completing the Square

Students transform a general quadratic equation by completing the square to derive the quadratic formula and explain how it identifies the equation’s solutions.

Deriving the Quadratic Formula by Completing the Square

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Reviewing Perfect-Square Trinomials

A perfect-square trinomial results from squaring a binomial. The patterns are (x + m)² = x² + 2mx + m² and (x − m)² = x² − 2mx + m². The first and last terms are squares, and the middle term is twice the product of x and m. For example, x² + 6x + 9 is a perfect-square trinomial because 9 = 3² and 6x = 2(x)(3). Therefore, x² + 6x + 9 = (x + 3)². This pattern also shows how to complete an incomplete square. For x² + 8x, half of 8 is 4, so adding 4², or 16, creates x² + 8x + 16 = (x + 4)². The same idea will be applied to a general quadratic equation.

A square area model shows x squared plus 6x plus 9 forming a binomial square, alongside the completion of x squared plus 8x.
A square area model shows x squared plus 6x plus 9 forming a binomial square, alongside the completion of x squared plus 8x.Source: Illustrated for this lesson

Writing the General Quadratic Equation

Every quadratic equation in one variable can be written in the general form ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. The condition a ≠ 0 matters because without the x² term, the equation would be linear rather than quadratic. The coefficient a controls the squared term, b controls the linear term, and c is the constant. For example, 2x² − 7x + 3 = 0 has a = 2, b = −7, and c = 3. Writing an equation in general form makes its coefficients easy to identify and prepares it for a consistent solution process. To derive a formula that works for every quadratic equation, we begin with ax² + bx + c = 0 and use reversible operations that preserve its solutions.

The general quadratic equation is color-coded to connect each coefficient with its corresponding term and a numerical example.
The general quadratic equation is color-coded to connect each coefficient with its corresponding term and a numerical example.Source: Illustrated for this lesson

Completing the Square Symbolically

Begin with ax² + bx + c = 0. First, divide every term by a to make the coefficient of x² equal to 1: x² + (b/a)x + c/a = 0. Move the constant term to the right: x² + (b/a)x = −c/a. Half the coefficient of x is b/(2a), so add its square, b²/(4a²), to both sides. The left side becomes the perfect square (x + b/(2a))². Combining the fractions on the right gives (x + b/(2a))² = (b² − 4ac)/(4a²). These operations keep both sides equal. For example, 2x² + 8x − 10 = 0 becomes x² + 4x = 5 after dividing by 2 and moving the constant. Adding 4 gives (x + 2)² = 9.

A balanced sequence of equations shows division by a, movement of the constant, equal additions, and the resulting completed square.
A balanced sequence of equations shows division by a, movement of the constant, equal additions, and the resulting completed square.Source: Illustrated for this lesson

Deriving the Quadratic Formula

From the completed-square equation, (x + b/(2a))² = (b² − 4ac)/(4a²), multiply both sides by 4a² to obtain (2ax + b)² = b² − 4ac. Taking square roots requires both possible signs: 2ax + b = ±√(b² − 4ac). Subtract b and divide by 2a to isolate x. The result is x = (−b ± √(b² − 4ac))/(2a), the quadratic formula. The plus-or-minus symbol represents the two possible square roots and can produce two solutions. For example, applying the formula to x² − 5x + 6 = 0 gives x = (5 ± √25 − 24)/2 = (5 ± 1)/2. Thus, the solutions are x = 3 and x = 2.

The completed-square equation transforms step by step into the quadratic formula and then produces two roots in an example.
The completed-square equation transforms step by step into the quadratic formula and then produces two roots in an example.Source: Illustrated for this lesson

Checking the Formula with an Example

Consider 2x² + 3x − 2 = 0, where a = 2, b = 3, and c = −2. Substitute these values into the quadratic formula: x = (−3 ± √(3² − 4(2)(−2)))/(2(2)). The expression under the radical is 9 + 16 = 25, so x = (−3 ± 5)/4. Using the plus sign gives x = 2/4 = 1/2. Using the minus sign gives x = −8/4 = −2. Check each solution in the original equation. For x = 1/2, 2(1/2)² + 3(1/2) − 2 = 0. For x = −2, 2(−2)² + 3(−2) − 2 = 0. Both substitutions produce true equations, confirming that the formula identifies the equation’s solutions.

A complete worked example shows coefficient substitution, two calculated roots, and both roots checked in the original equation.
A complete worked example shows coefficient substitution, two calculated roots, and both roots checked in the original equation.Source: Illustrated for this lesson