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Solving Quadratics: Factoring and Completing the Square

A free Algebra I lesson from the “Quadratic Foundations” unit, with a worked example and practice problems including step-by-step solutions.

Factoring rewrites a quadratic without changing its value. If a product equals zero, at least one factor must be zero; this is why (x - 2)(x - 3) = 0 gives x = 2 or x = 3. The zero-product rule does not apply to a product equal to 6 or to a sum equal to zero. Completing the square provides another route when a quadratic does not factor conveniently. The identity (x + p)^2 = x^2 + 2px + p^2 shows why x^2 + bx needs (b/2)^2 to become a perfect square. First divide by the coefficient of x^2 if it is not 1, then move the constant to the other side. Add the same square to both sides to preserve equality. Once (x + p)^2 = k, use x + p = sqrt(k) or x + p = -sqrt(k) when k is positive. A zero right side yields one root; a negative right side has no real root. Dividing an equation by an expression containing x can lose a root when that expression is zero, so factor instead or consider that case separately. Check solutions in the original equation.

What you'll learn

Why it matters: Completing the square reveals the turning point of an area or height model and solves equations whose roots are not whole numbers.

Worked example

Problem. Solve 2x^2 - 12x + 10 = 0 by completing the square, explaining why equality is preserved.

  1. Divide every term by 2: x^2 - 6x + 5 = 0. Division by the nonzero number 2 preserves the solution set.
  2. Subtract 5: x^2 - 6x = -5.
  3. Half of -6 is -3, and (-3)^2 = 9. Add 9 to both sides: x^2 - 6x + 9 = 4.
  4. Use the perfect-square identity to write (x - 3)^2 = 4.
  5. Both 2 and -2 square to 4, so x - 3 = 2 or x - 3 = -2. Hence x = 5 or x = 1.
  6. Check: at x = 1, 2 - 12 + 10 = 0; at x = 5, 50 - 60 + 10 = 0.

Answer: x = 1 or x = 5.

Lesson preview: this page shows 5 practice items. The interactive activity contains 32 practice items and 17 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.

Practice problems

1. What number must be added to x^2 + 10x to make a perfect square trinomial?

Show solution
  1. Half of 10 is 5; its square is 25.
  2. x^2 + 10x + 25 = (x + 5)^2.
  3. When solving an equation, also add 25 to its other side.

Answer: 25

2. Which equation follows from x^2 + 4x = 3 by correctly completing the square?

Choices: (x + 2)^2 = 7 · (x + 2)^2 = 3 · (x + 4)^2 = 19 · (x - 2)^2 = 7

Show solution
  1. Half of the x coefficient 4 is 2, so add 2^2 = 4 to both sides.
  2. x^2 + 4x + 4 = 3 + 4 becomes (x + 2)^2 = 7.

Answer: (x + 2)^2 = 7

3. A student solves x^2 = 5x by dividing both sides by x and reports only x = 5. What solution was lost?

Choices: x = 0 · x = -5 · x = 1 · No solution was lost.

Show solution
  1. Move all terms to one side: x^2 - 5x = x(x - 5) = 0.
  2. The factors give x = 0 or x = 5.
  3. Division by x assumes x != 0 and therefore omits the valid zero root.

Answer: x = 0

4. Solve x^2 - 8x = -7 by completing the square. Enter the smaller root.

Show solution
  1. Add (-8/2)^2 = 16 to both sides: x^2 - 8x + 16 = 9.
  2. Then (x - 4)^2 = 9, so x - 4 = 3 or -3.
  3. The roots are 7 and 1; the smaller is 1.

Answer: 1

5. Solve (x - 4)(x + 2) = 0.

Choices: 4 and -2 · -4 and 2 · 4 and 2 · -4 and -2

Show solution
  1. Set each factor equal to zero.

Answer: 4 and -2

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Explain a method, then repair an error

Solve x^2 + 2x - 4 = 0 by completing the square. Explain why the same number is added to both sides and why there are two roots. A learner writes x^2 + 2x = 4, then (x + 1)^2 = 4. Identify and correct the error.

Review rubric and solution
  • Add 1 to both sides to obtain (x + 1)^2 = 5.
  • Explain that adding only to one side changes the equation.
  • Use both signs to obtain -1 + sqrt(5) and -1 - sqrt(5).
  • Verify by expanding (x + 1)^2 = 5 back to the original equation.
  1. From x^2 + 2x = 4, add (2/2)^2 = 1 to both sides, giving x^2 + 2x + 1 = 5.
  2. Thus (x + 1)^2 = 5. The proposed right side 4 is wrong because it omitted the added 1.
  3. Both sqrt(5) and -sqrt(5) square to 5, so x = -1 + sqrt(5) or x = -1 - sqrt(5).
  4. Expanding the completed square gives x^2 + 2x + 1 = 5, equivalent to x^2 + 2x - 4 = 0.

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