Questions: Solving Quadratic Equations by Completing the Square

Question 1

To complete the square for x² + 10x = 3, what value must be added to both sides?

• A. 5, giving (x + 5)² = 8
• B. 25, giving (x + 5)² = 28
• C. 10, giving (x + 10)² = 13
• D. 100, giving (x + 10)² = 103

Answer: (B)

The completing term is (b/2)², where b is the coefficient of x. Here b = 10, so (10/2)² = 5² = 25. Adding 25 to both sides gives x² + 10x + 25 = 28, which factors as (x + 5)² = 28. Option A uses b/2 = 5 instead of (b/2)² = 25 — the most common error. Option C uses b itself. Option D squares b rather than b/2. The critical step is halving the coefficient *before* squaring.

Question 2

After completing the square on a quadratic equation, a student obtains (x − 4)² = −9. How many real solutions does the original equation have?

• A. Two real solutions: x = 4 + 9 = 13 and x = 4 − 9 = −5
• B. One real solution: x = 4, since the negative sign cancels the square
• C. Zero real solutions: the square root of a negative number is not real
• D. Two real solutions obtained by taking ±√9 = ±3, giving x = 7 and x = 1

Answer: (C)

The equation (x − 4)² = −9 requires taking the square root of −9, which is not a real number. No real value of x can satisfy this — the square of any real number is non-negative. The solutions are complex: x = 4 ± 3i. Geometrically, this means the parabola sits entirely above or below the x-axis with no real x-intercepts. The sign of k in (x − h)² = k tells you everything: k > 0 gives two real solutions, k = 0 gives one repeated root, k < 0 gives two complex solutions.

Question 3

When completing the square for a quadratic where the leading coefficient is not 1, you can skip dividing by that coefficient first and still arrive at the correct answer.

• T. True
• F. False

Answer: False

If the leading coefficient a ≠ 1, dividing through by a first is essential. Without this step, the left side after adding the completing term does not factor as a perfect square trinomial. For 2x² + 8x = 6: dividing by 2 gives x² + 4x = 3, then adding (4/2)² = 4 yields (x + 2)² = 7. Attempting to complete the square directly on 2x² + 8x without dividing first produces a different and incorrect factorization. The technique only works cleanly when the coefficient of x² is 1.

Question 4

When solving (x + 3)² = 16 by taking the square root of both sides, there are two solutions: x = 1 and x = −7.

• T. True
• F. False

Answer: True

Taking the square root of both sides gives x + 3 = ±4. The positive case: x + 3 = 4, so x = 1. The negative case: x + 3 = −4, so x = −7. Both check out: (1 + 3)² = 16 ✓ and (−7 + 3)² = (−4)² = 16 ✓. The ± is the structural source of two solutions — a very common error is taking only the positive root and missing the second solution, especially when the completed square has a large or positive h value.

Question 5

Why must the completing term (b/2)² be added to both sides of the equation, rather than only to the left side?

Model answer: Adding a value to only one side of an equation changes the equality — it creates a different equation with different solutions. To maintain equivalence (to produce a new equation with the same solutions as the original), whatever is added to the left side must also be added to the right. The completing term is added to create a perfect square trinomial on the left; adding it to the right preserves the original equation's solution set.

This is the fundamental property of equality: if A = B, then A + c = B + c for any constant c. Forgetting to add the completing term to the right side is one of the most common errors in completing the square. The resulting 'solved' equation is inequivalent to the original and its solutions are wrong. Checking by substituting back into the original equation immediately exposes this error, which is why checking solutions is important in algebraic manipulation.