How to Solve a Quadratic Equation by Completing the Square?
Solving a Quadratic Equation by Completing the Square
Completing the square rewrites \(x^2 + bx + c\) as a perfect square plus a constant, \((x + \tfrac{b}{2})^2 + k\). It reveals the vertex instantly and lets you solve any quadratic with a square root. We’ll build the method carefully, with a solver and a worksheet maker a tap away.
Solve a Quadratic Equation by Completing the Square: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Read the formFactored, standard, and vertex forms reveal different features.
- Choose the methodFactor when friendly, complete the square for structure, or use the formula when needed.
- Connect to the graphRoots are x-intercepts and the vertex is the minimum or maximum point.
Worked examples
Factor and solve
- Factor into (x – 3)(x – 4).
- Set each factor equal to zero.
- Solve both small equations.
Find the axis
- Use x = -b/(2a).
- Here a = 2 and b = -8.
- Compute 8/4.
Try one before moving on
Solve a Quadratic Equation by Completing the Square: pop-up practice
Completing the square is the technique that turns a messy quadratic into a tidy perfect square, \((x + \tfrac{b}{2})^2\) plus a leftover constant. It’s worth learning for two reasons: it solves any quadratic with a square root, and it hands you the vertex for free. It’s also where the quadratic formula comes from.
In short: rewrite \(x^2 + bx + c\) as \((x + \tfrac{b}{2})^2 + (c – (\tfrac{b}{2})^2)\), then solve by isolating the square and taking the square root. For \(x^2 + 6x + 5 = 0\), this gives \(x = -1\) or \(-5\).
Make a Perfect Square
A perfect-square trinomial like \(x^2 + 6x + 9\) factors neatly as \((x+3)^2\). The middle coefficient \(6\) tells you the \(3\): it’s half of \(6\). Completing the square forces this pattern by adding and subtracting \((\tfrac{b}{2})^2\), so you can fold the first part into a square.
How to solve by completing the square (steps):
- Move the constant to the right side.
- Add \((\tfrac{b}{2})^2\) to both sides.
- Write the left side as \((x + \tfrac{b}{2})^2\).
- Take the square root of both sides (keep \(\pm\)) and solve for \(x\).
Worked Examples
The completed square reveals the vertex; the graph confirms where the curve sits and crosses.
Example A, Build the square
Complete the square for \(x^2 + 6x + 5\).
- Half the middle coefficient: \(\tfrac{6}{2} = 3\); square it: \(3^2 = 9\).
- Add and subtract 9: \(x^2 + 6x + 9 – 9 + 5\).
- Fold the square: \((x+3)^2 – 4\), vertex \((-3,-4)\).
Answer: \((x+3)^2 – 4\)
Example B, Solve it
Solve \(x^2 + 6x + 5 = 0\).
- Use the completed form: \((x+3)^2 – 4 = 0\).
- Isolate the square: \((x+3)^2 = 4\), so \(x + 3 = \pm 2\).
- Solve: \(x = -1\) or \(x = -5\), the x-intercepts.
Answer: \(x = -1\) or \(-5\)
Example C, An irrational answer
Solve \(x^2 – 4x – 1 = 0\).
- Complete the square: \((x-2)^2 – 5 = 0\).
- Isolate: \((x-2)^2 = 5\), so \(x – 2 = \pm\sqrt5\).
- Solve: \(x = 2 \pm \sqrt5\), the curve still crosses, just at irrational points.
Answer: \(x = 2 \pm \sqrt5\)
Example D, Read the vertex
Find the vertex of \(y = x^2 + 6x + 5\).
- Completed form: \((x+3)^2 – 4\).
- Vertex form is \((x-h)^2 + k\) with vertex \((h,k)\).
- So the vertex is \((-3, -4)\). It falls right out.
Answer: \((-3, -4)\)
Where You’ll Use It
Completing the square is the bridge to vertex form, so it’s how you find a parabola’s max or min exactly. It also solves quadratics that won’t factor, and it’s the derivation behind the quadratic formula itself, understanding it makes that formula far less mysterious.
Slip-Ups That Cost Easy Points
- Adding to only one side. Whatever you add to complete the square, add to both sides to keep the equation balanced.
- Wrong number. It’s \((\tfrac{b}{2})^2\), half, then square. For \(x^2 + 6x\) that’s \(9\), not \(36\).
- Dropping the \(\pm\). Taking a square root gives two cases: \(x + 3 = 2\) and \(x + 3 = -2\).
- Forgetting a leading coefficient. If \(a \ne 1\), divide through by \(a\) first.
Your Turn
Complete the square or solve, then reveal the answers.
- Complete the square: \(x^2 + 8x + 3\)
- Solve: \(x^2 – 2x – 8 = 0\)
- Solve: \(x^2 + 4x + 1 = 0\)
- Vertex of \(y = x^2 – 4x + 1\)?
Show answers
- \(\color{blue}{(x+4)^2 – 13}\)
- \(\color{blue}{x = 4, -2}\)
- \(\color{blue}{x = -2 \pm \sqrt3}\)
- \(\color{blue}{(2, -3)}\)
Make Your Own Worksheet
Generate fresh completing-the-square problems with a full answer key, print or save as a PDF.
Frequently Asked Questions
What does it mean to solve by completing the square?
It means rewriting a quadratic so one side is a perfect square, like (x + 3)², and then taking the square root of both sides. Instead of hunting for factors, you build a square on purpose by adding the right constant. The same rewriting also gives vertex form, so it shows the parabola’s turning point.
What are the steps to solve a quadratic equation by completing the square?
Take x² + 6x − 7 = 0. Move the constant: x² + 6x = 7. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 16. Write the left side as a square: (x + 3)² = 16. Take square roots: x + 3 = ±4. So x = 1 or x = −7. Don’t forget the ± sign, or you lose a solution.
What is the formula for completing the square?
The key identity is x² + bx + (b/2)² = (x + b/2)². In words, take half the coefficient of x, square it, and add it to make a perfect square trinomial. For x² − 10x, half of −10 is −5 and (−5)² = 25, so x² − 10x + 25 = (x − 5)². Applying this to ax² + bx + c = 0 in general produces the quadratic formula.
How do you solve x² + x − 20 = 0 by completing the square?
Move the constant: x² + x = 20. Half of 1 is ½ and (½)² = ¼, so add ¼ to both sides: (x + ½)² = 81/4. Take square roots: x + ½ = ±9/2. That gives x = 4 or x = −5. Fractions appear whenever b is odd, so keep them as fractions rather than switching to decimals.
How do you solve 9y² − 12y + 2 = 0 by completing the square?
First divide every term by 9 so the leading coefficient is 1: y² − (4/3)y + 2/9 = 0. Move the constant: y² − (4/3)y = −2/9. Add (2/3)² = 4/9 to both sides: (y − 2/3)² = 2/9. Take square roots: y − 2/3 = ±√2/3, so y = (2 ± √2)/3.
Can you solve any quadratic equation by completing the square?
Yes. Completing the square works for every quadratic, even ones that do not factor. If the leading coefficient is not 1, divide by it first. If you end up with a negative number on the right side, such as (x − 1)² = −4, the equation has no real solutions, only complex ones, which matches a negative discriminant.
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