SME
Grade 9 Math · Problem type
Complete the square
Use this for “Solve x² + 6x − 7 = 0 by completing the square.”
Diagram
x² + 6x needs 9 more unit tiles (dashed) to make a square: x² + 6x + 9 = (x + 3)².
x² + 6x − 7 = (x + 3)² − 16.
This is an example. Type your own values to start a new problem.
x²
+
b
x
+
(
c
)
=
0
k
Half of b
worked out (not typed)
m
Missing corner
worked out (not typed)
R
Right side after completing
worked out (not typed)
x₁
Smaller root
worked out (not typed)
x₂
Larger root
worked out (not typed)
Formulas
k
=
b
2
3
=
6
2
m
=
k
2
9
=
3
2
R
=
m
−
c
16 = 9 − (−7)
x₁
=
−
k
−
√
R
−7
=
−3
−
√
16
x₂
=
−
k
+
√
R
1
=
−3
+
√
16
Assumptions
•
Half of b fills each side of the x² tile; the missing corner is (b/2)² unit tiles.
•
Add the corner to both sides, so the left side is (x + b/2)².
•
Then solve by square roots.
Step-by-step
WE KNOW
b = 6, c = −7
FIND
half of b (k), missing corner (m), right side after completing (R), smaller root (x₁), larger root (x₂)
Step 1 · Find half of b (k)
Use k = b ÷ 2
Split
the
x
tiles
in
half,
one
half
on
each
side
of
the
x
2
tile.
k
=
6
2
k
=
3
Step 2 · Find missing corner (m)
Use m = k²
The
missing
corner
is
k
by
k
.
m
=
3
2
m
=
9
Step 3 · Find right side after completing (R)
Use R = m − c
Move
c
to
the
right
side
and
add
the
corner
to
both
sides:
(x +
k
)
2
=
k
2
−
c
.
R
=
9
−
(−7)
x
2
+
6x
+
9
=
7
+
9
(x + 3)
2
=
16
R
=
16
Step 4 · Find smaller root (x₁)
Use x₁ = −k − √R
Take
the
square
root
of
both
sides,
x
+
k
=
±√
R
,
then
take
k
away.
x₁
=
−3
−
√
16
x₁
=
−3
−
4
x₁
=
−7
Step 5 · Find larger root (x₂)
Use x₂ = −k + √R
The plus root.
x₂
=
−3
+
√
16
x₂
=
−3
+
4
x₂
=
1
Check
Put the numbers back in. Both sides should match.
3
=
6
2
✓
9
=
3
2
✓
16 = 9 − (−7) ✓
−7
=
−3
−
√
16
✓
1
=
−3
+
√
16
✓
Topic check
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Related lessons
Solving quadratic equations: factoring, square roots, completing the square and the quadratic formula
Main lesson
Solve by square roots
Use this for “(x − 3)² = 25” or “2(x − 1)² = 18” (a = 1 when there is no number in front).
Quadratic inequality
Use this for “Solve x² − 2x − 8 < 0” or “x² − x − 6 ≥ 0.”