MISSION · Algebra
Completing the Square
Forcing a quadratic into a shape you can solve by taking a square root
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The idea in one sentence
Any quadratic can be rewritten as a perfect square plus a constant, and once it’s in that form you can solve it by square-rooting and read its turning point straight off.
The problem it solves
\[x^2 + 6x + 1 = 0\]You can’t undo this directly: \(x\) appears twice, once squared and once not, so there’s no single operation to reverse. If it looked like \((x + 3)^2 = 8\) instead, you could square-root both sides and be finished. Completing the square is the manoeuvre that gets you there.
The method
\[x^2 + 6x + 1\]- Take half the coefficient of \(x\). Half of 6 is 3.
- Write \((x + 3)^2\). Expanding gives \(x^2 + 6x + 9\) — the right first two terms, but 9 has appeared from nowhere.
- Subtract what you added. \((x+3)^2 - 9\) is now exactly \(x^2 + 6x\).
- Bring back the constant. \((x+3)^2 - 9 + 1 = (x+3)^2 - 8\).
Now solving is straightforward:
\[(x+3)^2 = 8 \Rightarrow x + 3 = \pm\sqrt{8} \Rightarrow x = -3 \pm 2\sqrt{2}\]Note the \(\pm\). Forgetting it loses a root, and it’s the most common slip in the whole method.
Why “half of \(b\)”
Because \((x + p)^2 = x^2 + 2px + p^2\). The middle term carries a factor of 2, so to match \(6x\) you need \(2p = 6\), giving \(p = 3\). The constant that turns up is then \(p^2 = 9\), which is why it’s always the square of the half.
Reading the graph for free
\[y = (x + 3)^2 - 8\]A square is never negative, so the smallest \((x+3)^2\) can be is 0, which happens at \(x = -3\). There, \(y = -8\). The vertex is \((-3, -8)\), and it took no calculus and no formula to find it.
In general \(y = a(x - h)^2 + k\) has its turning point at \((h, k)\) — a minimum if \(a > 0\), a maximum if \(a < 0\). This is the form to use whenever a question asks for a maximum or minimum value.
When the leading coefficient isn’t 1
\[2x^2 + 12x + 5\]Factor the 2 out of the first two terms only, then complete the square inside:
\[2(x^2 + 6x) + 5 = 2[(x+3)^2 - 9] + 5 = 2(x+3)^2 - 18 + 5 = 2(x+3)^2 - 13\]The bracket’s \(-9\) gets multiplied by the 2 on the way out. That’s where the marks go.
It’s also where the quadratic formula comes from
Complete the square on \(ax^2 + bx + c = 0\) with letters instead of numbers and you get
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]The formula isn’t a separate fact to memorise. It’s this method, done once, in general.
Common mix-ups
- Forgetting to subtract \(p^2\), which changes the expression instead of rewriting it.
- Losing the ± when square-rooting.
- Not multiplying the subtracted constant by the factored-out leading coefficient.
- Reading the vertex as \((+h, k)\) from \((x + 3)^2\) — the \(x\)-value is \(-3\), because that’s what makes the bracket zero.
Check yourself
Write \(x^2 + 6x + 1\) in the form \((x+3)^2 - k\), then state the minimum value.
Half of 6 is 3, and \((x+3)^2\) overshoots by 9, so the expression is \((x+3)^2 - 8\) — giving k = 8. The square can’t go below 0, so the minimum value is −8, at x = −3.
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