Chapter 9: Quadratic Functions
SECTION 9.5: COMPLETING
THE SQUARE TO SOLVE
QUADRATIC EQUATIONS
6/page to print
Recall how to square binomial
ƒ (x
( + 4)2=
ƒ x2 +8x +16
ƒ (x — 3)2=
ƒ x2 — 6x + 9
ƒ And these can be factored to binomial sq.
What term do we need to add to
binomial to make it factor to square
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
2
+25
=
(x
+
5)
+ 10x
x2 + 6x +9 = (x + 3)2
x2 — 14x + 49 = (x — 7)2
x2 — 2x + 1 = (x — 1)2
These are not the same value as original
expression, because we added something
But
u if you add and
a d subtract
ub a the same
a
thing,
g, it
will still be the same,
or add the same thing to both sides of
equation,
i
it
i will
ill still
ill be
b equall
x2
Use this fact to ‘complete the
square’, and apply the square
square
root property to solve quad.eq.
ƒ x2 + 10x
10 = — 25
ƒ x2 + 6x = 16
ƒ x2 — 14x = — 40
ƒ x2 — 2x = 15
ƒ On left, add what you need to make it
have a square
q
for factors
ƒ Add same thing on the right
ƒ Apply the square root property
Add the term to both sides
ƒ x2 + 10x
10 + 25=
25 — 25 + 25
ƒ x2 + 6x + 9 = 16 + 9
ƒ x2 — 14x + 49 = — 40 + 49
ƒ x2 — 2x + 1= 15 + 1
ƒ Factor on left, simplify on right
Factor on left, simplify on right
ƒ (x
( + 5)2 = 0
ƒ (x + 3)2 = 25
ƒ (x — 7)2 = 9
ƒ ((x — 1))2 = 16
ƒ And apply square root property
Apply square root property
ƒ x + 5 = ±0
ƒ x + 3 = ±5
ƒ x — 7 = ±3
ƒ x — 1 = ±4
ƒ Add or subtract constant term from x and
from right
g side
ƒ Find value of each solution when rational
g
on right
Add or subtract constant term
from x and from right side
ƒ x + 5 — 5=
5 — 5 ±0 = — 5
ƒ x + 3 — 3 = — 3 ±5 = — 8, 2
ƒ x — 7 + 7= 7 ±3 = 4, 10
ƒ x — 1 + 1 = 1 ±4 = 5, — 3
ƒ Find value of each solution because it is
± a rational number
x2 — 14x +4 = 0
ƒ x2 — 14x
14 +4
4—4=0—4
ƒ x2 — 14x = — 4
ƒ Complete the square
ƒ x2 — 14x + 49 = — 4 + 49
ƒ Factor on left, simplify on right
ƒ (x — 7)2 = 45
x2 — 14x +4 = 0
ƒ (x
( — 7)2 = 45
ƒ Apply square root property
ƒ x — 7 = ± √45 = ± 3 √5
ƒ Add 7 to both sides
x — 7 + 7 = 7 ± 3 √5
ƒ Radical remains, so write this or
ƒ x = 7 + 3 √5 , 7 — 3 √5
ƒ
2x2 — 8x = — 7
ƒ
ƒ
2
2x
8x
7
Divide by leading coefficient
=
2
2
2
x2 — 4x = — 7/2
ƒ Complete
p
the square
q
ƒ x2 —4x + 4 = — 7/2 + 4
ƒ Factor on left,
left simplify on right
ƒ (x — 2)2 = — 7/2 + 8/2 = 1/2
(x — 2)2 = 1/2
ƒ Apply square root property
1
1 2
2
x−2= ±
=±
⋅
=±
2
2
2 2
ƒ Move — 2 to other side
2
x = 2±
ƒ Can combine over one
2
denominator
Steps to complete square
ƒ Write in the Ax2+Bx=K form
ƒ Divide every term by A
ƒ Find C to add to left to make left a perfect
ƒ
ƒ
ƒ
ƒ
square that can be factored to (x+c)2
Add that value to rignt side also:
Ax2+Bx+C=K+C
+Bx+C K+C
Factor on left, simplify on right
Apply square root property
Isolate the unknown so there is a solution
5x2 — 10x + 45 = 0
ƒ Divide by leading coefficient
ƒ x2 — 2x + 9 =0
ƒ Move +9 to other side by
y subtraction
ƒ Complete the square
ƒ x2 — 2x + 1 = — 9 + 1
ƒ Factor on left, simplify on right
ƒ (x — 1)2 = — 8
(x — 1)2 = — 8
ƒ Apply square root property
x −1 = ± − 8 = ± −1 ⋅ 8
ƒ Simplify square roots on right
x − 1 = ±i 4 ⋅ 2 = ±2i 2
ƒ Move — 1 to other side
x = 1 ± 2i 2