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Algebra · Quadratic equations

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Completing the square

Why x² + bx is a square with a corner missing, why the bracket takes HALF of b, and how the completed form solves equations exactly and hands you the turning point for free.

Algebra · Quadratic equations

Completing the square

Why x² + bx is a square with a corner missing, why the bracket takes HALF of b, and how the completed form solves equations exactly and hands you the turning point for free.

Why it works

Expand (x+3)2(x + 3)^2 and watch where the pieces go: (x+3)2=x2+6x+9(x + 3)^2 = x^2 + 6x + 9. The bracket's 33 appears doubled in the middle term. So to run this backwards for x2+6xx^2 + 6x, the bracket must take half the xx-coefficient — and then the expansion drags in an unwanted +9+9 (=32= 3^2) that has to be subtracted off:

x2+6x+2=(x+3)2−9+2=(x+3)2−7.x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7.

That's the whole method: halve bb into the bracket, subtract the square of the half, tidy the constants. Check by expanding back — ten seconds.

Why bother? The form (x+p)2+q(x + p)^2 + q makes the quadratic transparent.
  • Solving, exactly. (x+5)2−7=0(x + 5)^2 - 7 = 0 unwinds like an onion:
(x+5)2=7(x + 5)^2 = 7, so x+5=±7x + 5 = \pm\sqrt{7} — both square roots, or half the answers vanish — and x=−5±7x = -5 \pm \sqrt{7}. Surd-form answers with no formula needed.
  • The minimum, at sight. A square is never negative: (x+p)2≥0(x + p)^2 \ge 0,
hitting 00 only when x=−px = -p. So (x+3)2−7(x + 3)^2 - 7 has minimum value −7-7, at x=−3x = -3. On the graph of y=(x+p)2+qy = (x + p)^2 + q the turning point is (−p,q)(-p, q) — mind the sign: the bracket (x−4)2(x - 4)^2 bottoms out at x=+4x = +4.
  • "Show it's always positive." x2−4x+9=(x−2)2+5≥5>0x^2 - 4x + 9 = (x - 2)^2 + 5 \ge 5 > 0
for every xx — the square does the arguing for you.

A leading coefficient gets factored out of the xx-terms first. For 2x2+12x+72x^2 + 12x + 7: take the 2 out of the terms that contain xx — 2(x2+6x)+72(x^2 + 6x) + 7 — complete the square inside, then let the 2 back in:

2[(x+3)2−9]+7=2(x+3)2−18+7=2(x+3)2−11.2\big[(x + 3)^2 - 9\big] + 7 = 2(x + 3)^2 - 18 + 7 = 2(x + 3)^2 - 11.

The subtracted 99 gets doubled on the way out — forgetting that is the classic slip.