01 · Warm-up
x² + 10x + 7
(x + 5)² − 18Half 10 → 5. Square 5 → 25. Then 7 − 25 = −18.
Worked example · Algebra
Completing the square, from scratch
Turn x² + 6x + 2 into (x + 3)² − 7—with every move shown and every rule explained.
Step 1 of 7
Start with the pattern
What number belongs inside the square?
x² + 6x + 2
Why: In (x + h)², the middle term is 2hx. So the coefficient 6 tells us what h must be.
01
See why the pattern works
A square with side length x + h naturally splits into four pieces. Change h and watch every term change with it.
x²
3x
3x
9
The identity
(x + 3)² = x² + 6x + 9
The middle number forces the last number
The two rectangles contribute hx + hx = 2hx. The corner contributes h². That is why halving the x-coefficient and then squaring it is not a trick—it rebuilds the missing corner.
x coefficient6
halve it3
square it9
02
Complete any square
Change the coefficients. The page completes the square, identifies the vertex, and plots the same expression in its more useful form.
Vertex form
(x + 3)² − 7
- 1
- Half the effective x-coefficient: 6 ÷ 2 = 3
- 2
- Square that half: 3² = 9
- 3
- Add and subtract the needed amount: +9 − 9 = 0
- ✓
- Vertex: (−3, −7)
y = x² + 6x + 2
same curve, new form
03
Climb the practice ladder
Say the half, the square, and the useful zero aloud before revealing each answer.
02 · Negative b
x² − 8x + 5
(x − 4)² − 11Half −8 → −4. Square −4 → 16. Then 5 − 16 = −11.
03 · Fraction
x² + 5x + 1
(x + 5/2)² − 21/4Half 5 → 5/2. Square it → 25/4. Then 1 − 25/4 = −21/4.
04 · Leading coefficient
2x² + 8x + 6
2(x + 2)² − 2Factor 2 from the x-terms first: 2(x² + 4x) + 6. Then complete the square inside.
04
Catch the common slips
Open each warning and use it as a ten-second self-check.
“I added the square number, but forgot to subtract it.”
You changed the expression. The useful move is always +h² − h², because that totals zero.
“I halved b even though the x² coefficient was not 1.”
Normalize first. Factor the leading coefficient from the x² and x terms, or use h = b/(2a).
“I read the vertex of (x + 3)² − 7 as (3, −7).”
The inside sign flips. The bracket reaches zero when x = −3, so the vertex is (−3, −7).
“My answer looks right, so I skipped the check.”
Expand back. It takes seconds and proves the new form equals the original for every x.
Feynman finish
Teach it back
If you can explain these five ideas without looking, you can operate the method—not just imitate it.