Solve for
x
x
size 12{x} {}
.
x
2
=
18
x
2
=
18
size 12{x rSup { size 8{2} } ="18"} {}
x
2
=
0
x
2
=
0
size 12{x rSup { size 8{2} } =0} {}
x
2
=
−
60
x
2
=
−
60
size 12{x rSup { size 8{2} } = - "60"} {}
x
2
+
8x
+
12
=
0
x
2
+
8x
+
12
=
0
size 12{x rSup { size 8{2} } +8x+"12"=0} {}
- a. Solve by factoring.
- b. Now, we’re going to solve it a different way. Start by adding four to both sides.
- c. Now, the left side can be written as
(x+something)2(x+something)2 size 12{ \( x+"something" \) rSup { size 8{2} } } {}. Rewrite it that way, and then solve from there.
- d. Did you get the same answers this way that you got by factoring?
Fill in the blanks.
(
x
−
3
)
2
=
x
2
−
6x
+
___
(
x
−
3
)
2
=
x
2
−
6x
+
___
size 12{ \( x - 3 \) rSup { size 8{2} } =x rSup { size 8{2} } - 6x+"___"} {}
(
x
−
3
2
)
2
=
x
2
−
___
x
+
___
(
x
−
3
2
)
2
=
x
2
−
___
x
+
___
size 12{ \( x - { {3} over {2} } \) rSup { size 8{2} } =x rSup { size 8{2} } - "___"x+"___"} {}
(
x
+
___
)
2
=
x
2
+
10
x
+
___
(
x
+
___
)
2
=
x
2
+
10
x
+
___
size 12{ \( x+"___" \) rSup { size 8{2} } =x rSup { size 8{2} } +"10"x+"___"} {}
(
x
−
___
)
2
=
x
2
−
18
x
+
___
(
x
−
___
)
2
=
x
2
−
18
x
+
___
size 12{ \( x - "___" \) rSup { size 8{2} } =x rSup { size 8{2} } - "18"x+"___"} {}
Solve for
x
x
size 12{x} {}
by completing the square.
x
2
−
20
x
+
90
=
26
x
2
−
20
x
+
90
=
26
size 12{x rSup { size 8{2} } - "20"x+"90"="26"} {}
3x2+2x−4=03x2+2x−4=0 size 12{3x rSup { size 8{2} } +2x - 4=0} {}
Start by dividing by 3. The x2x2 size 12{x rSup { size 8{2} } } {} term should never have a coefficient when you are completing the square.
"This is the "main" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing about them in […]"