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 number do I add to complete the square?
Half of the middle coefficient, squared: \((\tfrac{b}{2})^2\). Add it to both sides so the left side becomes a perfect square.
Why use completing the square instead of factoring?
It works even when the quadratic doesn’t factor with nice numbers, and it directly reveals the vertex (vertex form). Factoring is faster only when the equation factors cleanly.
What if the leading coefficient isn’t 1?
Divide every term by \(a\) first so the \(x^2\) coefficient is 1, then complete the square as usual.
How does it relate to the quadratic formula?
The quadratic formula is the result of completing the square on the general equation \(ax^2 + bx + c = 0\) — same method, done once for all.
Related Topics
Continue Your Study
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