Download math hands

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

page
1
page
2
page
3
page
4
page
5
page
6
page
7
page
8
page
9
page
10
Document related concepts
no text concepts found
Transcript 
INTERMEDIATE ALGEBRA
Solving
equations w/radicals
math
hands
CH 03 SEC 3 SOLUTIONS
1. Solve
q
6x2 + 4x = 4
Solution:
q
6x2 + 4x = 4
2
q
2
2
6x + 4x = 4
6x2 + 4x = 16
2
(given)
(SBST)
(def of Rad, BI)
6x + 4x − 16 = 0
(BI)
(2x + 4)(3x − 4) = 0
2x + 4 = 0
OR
3x − 4 = 0
4
x= −2
OR
x=
3
(BI)
(ZFT)
2. Solve
(ZFT)
q
2x2 + 5x = 5
Solution:
q
2x2 + 5x = 5
q
2
2
2
2x + 5x = 5
2x2 + 5x = 25
2
2x + 5x − 25 = 0
(x + 5)(2x − 5) = 0
x+5=0
OR
x= −5
3. Solve
pg. 1
2x − 5 = 0
5
OR
x=
2
(given)
(SBST)
(def of Rad, BI)
(BI)
(BI)
(ZFT)
(ZFT)
q
3x2 + 2x = 1
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
equations w/radicals
math
hands
CH 03 SEC 3 SOLUTIONS
Solution:
q
3x2 + 2x = 1
2
q
2
2
3x + 2x = 1
3x2 + 2x = 1
2
(given)
(SBST)
(def of Rad, BI)
3x + 2x − 1 = 0
(BI)
(x + 1)(3x − 1) = 0
x+1=0
OR
3x − 1 = 0
1
x= −1
OR
x=
3
(BI)
(ZFT)
4. Solve
(ZFT)
q
6x2 + 5x = 5
Solution:
q
6x2 + 5x = 5
2
q
2
2
6x + 5x = 5
6x2 + 5x = 25
2
pg. 2
(SBST)
(def of Rad, BI)
6x + 5x − 25 = 0
(2x + 5)(3x − 5) = 0
(BI)
(BI)
OR
3x − 5 = 0
5
5
x= −
OR
x=
2
3
(ZFT)
2x + 5 = 0
5. Solve
(given)
(ZFT)
q
6x2 + 5x = − 5
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
equations w/radicals
math
hands
CH 03 SEC 3 SOLUTIONS
Solution:
q
6x2 + 5x = − 5
2
q
2
2
6x + 5x = − 5
6x2 + 5x = 25
2
(given)
(SBST)
(def of Rad, BI)
6x + 5x − 25 = 0
(BI)
(2x + 5)(3x − 5) = 0
2x + 5 = 0
OR
3x − 5 = 0
5
5
x= −
OR
x=
2
3
(BI)
(ZFT)
(ZFT)
On a careless day, this would be the end of it.. but on a careful day, we would, as suggested, check the solutions
to note neither one of these works. This is an example of a an equation where squaring both sides introduces
two extraneous solutions. In conclusion, this equation has no real solutions.
6. Solve
q
2x2 − 5x = − 5
Solution:
q
2x2 − 5x = − 5
2
q
2
2x2 − 5x = − 5
2x2 − 5x = 25
2
2x − 5x − 25 = 0
(2x + 5)(x − 5) = 0
2x + 5 = 0
x= −
OR
5
2
OR
(given)
(SBST)
(def of Rad, BI)
(BI)
(BI)
x−5=0
(ZFT)
x=5
(ZFT)
On a careless day, this would be the end of it.. but on a careful day, we would, as suggested, check the solutions
to note neither one of these works. This is an example of a an equation where squaring both sides introduces
two extraneous solutions. In conclusion, this equation has no real solutions.
7. Solve
pg. 3
q
15x2 + 6x = − 3
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
math
hands
CH 03 SEC 3 SOLUTIONS
Solving
equations w/radicals
Solution:
q
15x2 + 6x = − 3
2
q
2
2
15x + 6x = − 3
15x2 + 6x = 9
2
(given)
(SBST)
(def of Rad, BI)
15x + 6x − 9 = 0
(BI)
(3x + 3)(5x − 3) = 0
3x + 3 = 0
OR
5x − 3 = 0
3
x= −1
OR
x=
5
(BI)
(ZFT)
(ZFT)
On a careless day, this would be the end of it.. but on a careful day, we would, as suggested, check the solutions
to note neither one of these works. This is an example of a an equation where squaring both sides introduces
two extraneous solutions. In conclusion, this equation has no real solutions.
8. Solve
q
q
3
15x2 − 10x = 3 25
Solution:
q
q
3
15x2 − 10x = 3 25
3 q 3
q
3
15x2 − 10x = 3 25
15x2 − 10x = 25
2
15x − 10x − 25 = 0
(5x + 5)(3x − 5) = 0
5x + 5 = 0
OR
x= −1
9. Solve
pg. 4
3x − 5 = 0
5
OR
x=
3
(given)
(CBST)
(def of Rad, BI)
(BI)
(BI)
(ZFT)
(ZFT)
q
q
3
6x2 + 7x = 3 49
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
math
hands
CH 03 SEC 3 SOLUTIONS
Solving
equations w/radicals
Solution:
q
q
3
6x2 + 7x = 3 49
3 q 3
q
3
2
6x + 7x = 3 49
6x2 + 7x = 49
2
(given)
(CBST)
(def of Rad, BI)
6x + 7x − 49 = 0
(BI)
(2x + 7)(3x − 7) = 0
2x + 7 = 0
OR
3x − 7 = 0
7
7
x= −
OR
x=
2
3
(BI)
(ZFT)
10. Solve
(ZFT)
q
q
3
7x2 + 30x = 3 25
Solution:
q
q
3
7x2 + 30x = 3 25
3 q 3
q
3
2
7x + 30x = 3 25
7x2 + 30x = 25
2
7x + 30x − 25 = 0
(x + 5)(7x − 5) = 0
x+5=0
OR
x= −5
11. Solve
pg. 5
7x − 5 = 0
5
OR
x=
7
(given)
(CBST)
(def of Rad, BI)
(BI)
(BI)
(ZFT)
(ZFT)
q
− 2 + 2 2x + 1 = 3x
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
math
hands
CH 03 SEC 3 SOLUTIONS
Solving
equations w/radicals
Solution:
q
− 2 + 2 2x + 1 = 3x
q
2 2x + 1 = 3x + 2
2
q
2
2 2x + 1 = 3x + 2
(2)2 2x + 1 = 9x2 + 12x + 4
(given)
(BI, to isolate radical term)
(SBST, to annihilate radical)
(BI)
2
(BI)
2
(CLM)
8x + 4 = 9x + 12x + 4
0 = 9x + 4x
From here, use degree two skills, such as QF, and CHECK answers since SBST may introduce extraneous
solutions
12. Solve
5+3
q
2x − 1 = 3x
Solution:
q
2x − 1 = 3x
q
3 2x − 1 = 3x − 5
q
2
2
3 2x − 1 = 3x − 5
(3)2 2x − 1 = 9x2 − 30x + 25
5+3
(given)
(BI, to isolate radical term)
(SBST, to annihilate radical)
(BI)
2
(BI)
2
(CLM)
18x − 9 = 9x − 30x + 25
0 = 9x − 48x + 34
From here, use degree two skills, such as QF, and CHECK answers since SBST may introduce extraneous
solutions
13. Solve
pg. 6
q
q
2x − 1 = 5x + 3
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
equations w/radicals
math
hands
CH 03 SEC 3 SOLUTIONS
Solution:
q
q
2x − 1 = 5x + 3
2 q
2
q
2x − 1 =
5x + 3
2x − 1 =
q 2
5x
2x − 1 = 5x
q
+ 2 · 5x · 3
q
6 · 5x + 9
(given)
(SBST)
+
+
q
2x − 1 − 5x − 9 = 6 · 5x
q
− 3x − 10 = 6 · 5x
q 2
2
− 3x − 10 = 6 · 5x
2
2 q
5x
9x2 + 60x + 100 = (36) 5x
9x2 + 60x + 100 = 6
2
9x − 120x + 100 = 0
3
2
(FOIL, or PP2, BI)
(BI, killed one radical, not both...)
(CLA, isolate and annihilate, again...)
(BI, clean up)
(SBST, isolate, annihilate)
(BI)
(BI)
(CLA, BI)
From here, use degree two skills, such as QF, and CHECK answers since SBST may introduce extraneous
solutions
14. Solve
pg. 7
q
q
x − 2 = 3x + 1
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
equations w/radicals
math
hands
CH 03 SEC 3 SOLUTIONS
Solution:
q
q
x − 2 = 3x + 1
2 q
2
q
x−2 =
3x + 1
x−2=
q 2
3x
x − 2 = 3x
q
+ 2 · 3x · 1
q
2 · 3x + 1
(given)
(SBST)
+
+
q
x − 2 − 3x − 1 = 2 · 3x
q
− 2x − 3 = 2 · 3x
q 2
2
− 2x − 3 = 2 · 3x
2
2 q
3x
4x2 + 12x + 9 = (4) 3x
4x2 + 12x + 9 = 2
2
4x + 9 = 0
1
2
(FOIL, or PP2, BI)
(BI, killed one radical, not both...)
(CLA, isolate and annihilate, again...)
(BI, clean up)
(SBST, isolate, annihilate)
(BI)
(BI)
(CLA, BI)
From here, use degree two skills, such as QF, and CHECK answers since SBST may introduce extraneous
solutions
15. Solve
pg. 8
q
q
5x − 2 = 2x + 7
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
equations w/radicals
math
hands
CH 03 SEC 3 SOLUTIONS
Solution:
q
q
5x − 2 = 2x + 7
2 q
2
q
5x − 2 =
2x + 7
q 2
2x
(given)
(SBST)
+
2·
+ 14 ·
q
5x − 2 − 2x − 49 = 14 · 2x
q
3x − 51 = 14 · 2x
q 2
2
3x − 51 = 14 · 2x
q
2x
5x − 2 =
5x − 2 = 2x
q
2x · 7
+
+
49
2
2 q
2x
9x2 − 306x + 2601 = (196) 2x
9x2 − 306x + 2601 = 14
2
9x − 698x + 2601 = 0
7
2
(FOIL, or PP2, BI)
(BI, killed one radical, not both...)
(CLA, isolate and annihilate, again...)
(BI, clean up)
(SBST, isolate, annihilate)
(BI)
(BI)
(CLA, BI)
From here, use degree two skills, such as QF, and CHECK answers since SBST may introduce extraneous
solutions
16. Solve
pg. 9
q
q
5x − 2 = 3x + 1
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
equations w/radicals
math
hands
CH 03 SEC 3 SOLUTIONS
Solution:
q
q
5x − 2 = 3x + 1
2 q
2
q
5x − 2 =
3x + 1
5x − 2 =
q 2
3x
5x − 2 = 3x
q
+ 2 · 3x · 1
q
2 · 3x + 1
+
q
5x − 2 − 3x − 1 = 2 · 3x
q
2x − 3 = 2 · 3x
q 2
2
2x − 3 = 2 · 3x
2
2 q
3x
4x2 − 12x + 9 = (4) 3x
4x2 − 12x + 9 = 2
2
4x − 24x + 9 = 0
(given)
(SBST)
+
1
2
(FOIL, or PP2, BI)
(BI, killed one radical, not both...)
(CLA, isolate and annihilate, again...)
(BI, clean up)
(SBST, isolate, annihilate)
(BI)
(BI)
(CLA, BI)
From here, use degree two skills, such as QF, and CHECK answers since SBST may introduce extraneous
solutions
pg. 10
c
2007-2011
MathHands.com v.1010
Related documents