Chapter 9: Quadratic Functions
9.4 USING SQUARE
Q
ROOT
PROPERTY TO SOLVE
QUADRATIC EQUATIONS
6/page to print
Simplify square root: √72
ƒ √72
ƒ
ƒ
ƒ
ƒ
≈8
8.485281374
485281374
72= 36 · 2
√72= √36 · 2 =√36 · √2
= 6 · √2
Leave irrational part of root under radical
sign
g
Simplifying a radical quotient
8 − 28 8
28 4
4⋅7 4 2 7
= −
= −
= −
10
10 10
5
10
5 10
4
7 4− 7
= −
=
5 5
5
ƒ FIRST write each term of numerator over
the denominator!!
ƒ Reduce each fraction and simplify radical
ƒ Can then rewrite over single denominator
if you choose: doesn’t
doesn t really matter
Use Square Root property
ƒ If you d
do th
the same thi
thing tto b
both
th sides
id off
equation, it is still a valid equation
ƒ Including
I l di taking
ki square root
ƒ Be sure to write ( ) around each side, so
you take the square root of the entire
side, not of separate terms on the side
Square root property to solve
equations
ƒ x2=25
25
ƒ √ x2=± √ 25
ƒ x=5, -5 : ±5
Square root property to solve
equations
ƒ x2=3
3
ƒ √ x2=± √ 3
ƒ x= ± √ 3
Using square root property
ƒ x2—24=0
24 0
ƒ Need to isolate the square root first!!
ƒ x2=24
ƒ √ x2 =± √24
ƒ x =± 2√6
Using square root property
ƒ 3
3x2—4=3
4 3
ƒ Need to isolate the square root first!!
ƒ 3x2=7
ƒ x 2=
7
3
x = ±
x
7
3
⋅
3
3
2
=
= ±
21
3
7
3
(x+4)2—36=0
36 0
ƒ Vertex form of equation
ƒ ((x+4)
4)2=36,
36 can use sq.rt.prop.
t
ƒ √(x+4)2 =± √36
ƒ x+4 = ±6: now subtract 4
ƒ x = — 4±6
ƒ Both are rational, need to evaluate!!
ƒ x = — 4+6, — 4 — 6
ƒ x= 2, — 10
(x-7)
(x
7)2=50
50
ƒ √(x— 7)2 =± √50
ƒ x — 7 = ± √50 : now simplify
i lif √50
ƒ x — 7 = ± 5√2
ƒ Add 7, but 2 stays under radical sign
ƒ x = 7 ± 5√2
ƒ Radical expression, write two of them
ƒ x = 7 + 5√2, 7 — 5√2
ƒ Or use ± sign after the 7
√49
—√49
√49
√-49
√
49
ƒ Third:
Thi d nott a reall number
b
à There is not a real number you can multiply by
itself to get a negative product
ƒ When radicand is negative, there is not a
real square root
ƒ But radicand is -1·49…
ƒ Define
f
sq. rt. off -1 as “i“
“ “ ffor imaginary
i is the square root of —1
1
ƒ Factor out — 1 from negative radicands
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
FIRST
The proceed to simplify the root
When solving equations,
equations use both roots
± sign: plus or minus
Write +
+, then underline it with the —
If results has ± radical, ok to leave ±
If result is ± a value,
value add or subtract
value from the rest, and get two answers
√ 49
√-49
ƒ √-1·49
√ 1 49
ƒ =√-1·√49
ƒ = i ·7 = 7i
— √-24
√ 24
ƒ — i √24
ƒ = — i √4·6 = — 2i √6
If there are terms with real parts
and imaginary parts
ƒ It iis called
ll d a ‘complex
‘
l number’
b ’
ƒ 2 + 5i
ƒ Any rational number could be written
ƒ 3 = 3 + 0i
ƒ But it is not complex is the coefficient of i
is zero
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 if yyou add and subtract the same 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
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
on right
g
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 = 45
ƒ Factor on left
ƒ (x — 7)2 = 45
ƒ Apply sq.rt.prop.
ƒ x — 7 = ± √45 = ± 3 √5
x2 — 14x +4 = 0
ƒ (x
( — 7)2 = 45
ƒ Apply sq.rt.prop.
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