I’m trying to get reliable at completing the square, but I keep tripping over what I’m actually supposed to add and subtract.
Example: x^2 + 6x + 5. My instinct is: half of 6 is 3, so I wrote (x + 3)^2 + 5. But expanding gives x^2 + 6x + 14, which obviously isn’t the same as the original. I thought I was only fixing the middle term, so why did the constant term change so much? Where is that extra 9 coming from and where is it supposed to go?
Here’s a completely wrong attempt to show my confusion: for x^2 + 4x + 7, I wrote (x + 2)^2 + 7 − 2, because I figured I was “adding 2 to complete the square” and then “subtracting 2 to balance it.” I now realize that doesn’t check out, but I can’t pinpoint the exact rule I’m violating.
Could someone show the minimal, mechanically safe steps to turn x^2 + 6x + 5 into a completed square without changing its value? What exactly do I add and subtract, and at which step, so that the final expression is algebraically identical to the original?
Follow-up: if the leading coefficient isn’t 1, say 2x^2 + 7x − 3, do I have to factor out the 2 first, or is there a clean way to complete the square directly inside without introducing messy fractions? If factoring is required, what’s the neatest way to keep track of the constants so I don’t lose or mis-scale terms?
I suspect I’m mixing up adding b/2 versus (b/2)^2 somewhere in the process, but I’d like a crisp way to check myself at each step.
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3 Responses
You don’t add b/2, you add and subtract (b/2)^2: x^2 + 6x + 5 = (x + 3)^2 − 9 + 5 = (x + 3)^2 − 4.
In general ax^2 + bx + c = a(x + b/2a)^2 + (c − b^2/(4a)), so 2x^2 + 7x − 3 = 2(x + 7/4)^2 − 73/8; quick refresher: https://www.khanacademy.org/math/algebra2/polynomial-functions/completing-the-square-alg2/v/solving-quadratic-equations-by-completing-the-square
Use the identity (x + t)^2 = x^2 + 2tx + t^2. To match x^2 + bx, choose t = b/2. Then x^2 + bx = (x + b/2)^2 − (b/2)^2. This tells you exactly what to add and subtract: the square of half the x‑coefficient. For x^2 + 6x + 5, take b = 6, so add and subtract 9: x^2 + 6x + 5 = (x^2 + 6x + 9) + 5 − 9 = (x + 3)^2 − 4. The “extra 9” is t^2 with t = 3; you put it in to form the square and immediately take it back out to keep the value unchanged. Your earlier “+2 then −2” for 4x was off because you must use (b/2)^2 = 2^2 = 4, not b/2 itself.
If the leading coefficient is a, factor it from the x^2 and x terms first. Then complete the square inside. Example: 2x^2 + 7x − 3 = 2[x^2 + (7/2)x] − 3 = 2[(x + 7/4)^2 − (7/4)^2] − 3 = 2(x + 7/4)^2 − 49/8 − 3 = 2(x + 7/4)^2 − 73/8. In compact form: ax^2 + bx + c = a(x + b/(2a))^2 + c − b^2/(4a). Fractions are unavoidable unless b is a multiple of 2a. A quick self‑check is to expand your result and confirm you recover the original coefficients.
Think of completing the square like turning a lopsided rectangle into a perfect square by adding a small corner square-then, to keep the “total area” the same, you subtract that exact same little square back. The safe rule is: for x^2 + bx + c, compute m = b/2, then add and subtract m^2. That’s because (x + m)^2 = x^2 + 2mx + m^2, so 2m must equal b, and the “extra” constant you accidentally create is m^2. Example: x^2 + 6x + 5 → add and subtract (6/2)^2 = 9: x^2 + 6x + 9 + 5 − 9 = (x + 3)^2 − 4. Your extra 9 came from the m^2 in (x + 3)^2. For x^2 + 4x + 7, m = 2, so add and subtract 4: x^2 + 4x + 7 = (x + 2)^2 + 3. With a leading coefficient a ≠ 1, it’s cleanest to factor a from the x-terms, then complete the square inside: 2x^2 + 7x − 3 = 2(x^2 + (7/2)x) − 3 = 2[(x + 7/4)^2 − 49/16] − 3 = 2(x + 7/4)^2 − 73/8. A compact formula to check yourself is a x^2 + b x + c = a(x + b/(2a))^2 + c − b^2/(4a). Hope this helps! Nice walkthrough here: https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-by-completing-the-square