solving equations

1. Solve for x:
2x – 3 = 4
2x = 4 + 3
2x = 7
x = 7/2
Solving Equations

2. Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0 extract answers
so x = –4/3, 2

3. Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0 extract answers
so x = –4/3, 2
the roots for the equation ax2 + bx + c = 0 are
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.
For 2nd degree or quadratic equations, we may also
use the Quadratic Formula (QF):

4. Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0 extract answers
so x = –4/3, 2
the roots for the equation ax2 + bx + c = 0 are
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.
For 2nd degree or quadratic equations, we may also
use the Quadratic Formula (QF):
For 3x2 – 2x – 8 = 0, a = 3, b = –2, c = –8,
so b2 – 4ac = 100, or b2 – 4ac = 10
Hence x = (2 – 10)/6 or (2 + 10)/6
or x = –4/3, 2

5. Rational Equations
Solve rational equations by clearing all denominators
using the LCD.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
]
x – 2
2 =
x + 1
4
+ 1
x – 2
2 = x + 1
4 + 1
(x – 2)(x + 1) * [

6. Rational Equations
Solve rational equations by clearing all denominators
using the LCD.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
]
x – 2
2 =
x + 1
4
+ 1
x – 2
2 = x + 1
4 + 1
(x – 2)(x + 1) * [

7. Rational Equations
Solve rational equations by clearing all denominators
using the LCD.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
]
x – 2
2 =
x + 1
4
+ 1
(x + 1)
x – 2
2 = x + 1
4 + 1
(x – 2)(x + 1) * [

8. Rational Equations
Solve rational equations by clearing all denominators
using the LCD.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
]
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
(x – 2)(x + 1) * [

9. Rational Equations
Solve rational equations by clearing all denominators
using the LCD.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
(x – 2)(x + 1) * [

10. Rational Equations
Solve rational equations by clearing all denominators
using the LCD.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
(x – 2)(x + 1) * [
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
0 = x2 + x – 12
0 = (x + 4)(x – 3)  x = -4, 3
Both are good.

11. Power Equations
To solve a power equation, take the reciprocal power,
so if xR = c, xp/q = c
Equations of the Form xp/q = c
Example F. Solve (2x – 3)3/2 = -8

12. Power Equations
To solve a power equation, take the reciprocal power,
so if xR = c, xp/q = c
x = (±)cq/p
or
Equations of the Form xp/q = c
then x = (±)c1/R
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8  [(2x – 3)3/2]2/3 = (-8)2/3

13. Power Equations
To solve a power equation, take the reciprocal power,
so if xR = c, xp/q = c
x = (±)cq/p
or
Equations of the Form xp/q = c
then x = (±)c1/R
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8  [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
2x = 7  x = 7/2
Since x = 7/2 doesn't work because 43/2 = -8,
there is no solution.

14. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4

15. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;

16. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
0 = x2 – 10x + 9
0 = (x – 9)(x – 1)
x = 9, x = 1 Only 9 is good.

17. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”.
Absolute Value Equations

18. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
c
r
x = c + r
r
x = c – r

19. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
7
12
12
r
x = c – r

20. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
So to the right x = 7 + 12 = 19.
and to the left x = 7 – 12 = – 5.
7
12
12
x = 19
x = – 5
r
x = c – r

21. Zeroes and Domains
Example I.
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are where “D = 0”
N
D
x2 – 4.
Find the zeros and the domain f = x2 – 1

22. Zeroes and Domains
Example I.
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are where “D = 0”
N
D
x2 – 4.
Find the zeros and the domain f = x2 – 1
The zeros of f are where x2 – 1 = 0 or x = ±1.
The domain of f are
“all numbers x2 – 4 = 0, or all numbers except x = ±2”.

23. A. Solve the following equations by factoring.
5. x2 – 3x = 10
9. x3 – 2x2 = 0
6. x2 = 4
7. 2x(x – 3) + 4 = 2x – 4
10. 2x2(x – 3) = –4x
8. x(x – 3) + x + 6 = 2x2 + 3x
1. x2 – 3x – 4 = 0 2. x2 – 2x – 15 = 0 3. x2 + 7x + 12 = 0
4. –x2 – 2x + 8 = 0
11. 4x2 = x4
12. 7x2 = –4x3 – 3x 13. 5 = (x + 2)(2x + 1)
14. (x + 1)2 = x2 + (x – 1)2 15. (x + 3)2 – (x + 2)2 = (x + 1)2
B. Solve the following equations by the quadratic formula.
If the answers are not real numbers, just state so.
1. x2 – x + 1 = 0 2. x2 – x – 1 = 0
3. x2 – 3x – 2 = 0 4. x2 – 2x + 3 = 0
5. 2x2 – 3x – 1 = 0 6. 3x2 = 2x + 3
Equations

24. Rational Equations
7. 1
x
+
1
x – 1
=
5
6
8. 1
x
+
1
x + 2
=
3
4
9. 2
x
+ 1
x + 1
= 3
2
10. + 5
x + 2
= 2
2
x – 1
11. – 1
x + 1
= 3
2
12.
6
x + 2
– 4
x + 1
= 1
1
x – 2
x
6 3
1 2
3
5
2
3
–
+ = x
1. x
4 6
–3 1
8
–5
– 1
– = x
2.
x
4 5
3 2
10
7
4
3
+
– = x
3. x
8 12
–5 7
16
–5
+ 1
+ = x
(x – 20) = x – 3
100
30
100
20
5.
(x + 5) – 3 = (x – 5)
100
25
100
20
6.
C. Clear the denominators of the equations, then solve.

25. Radical Equations and Power Equations
D. Isolate one radical if needed, square both sides, do it again
if necessary, to solve for x. Check your answers.
1. x – 2 = x – 4 2. x + 3 = x + 1
3. 2x – 1 = x + 5 4. 4x + 1 – x + 2 = 1
5. x – 2 = x + 3 – 1 6. 3x + 4 = 3 – x – 1
7. 2x + 5 = x + 4 8. 5 – 4x – 3 – x = 1
E. Solve by raising both sides to an appropriate power.
No calculator.
1. x –2 = 1/4 2. x –1/2 = 1/4
3. x –3 = –8 4. x –1/3 = –8
5. x –2/3 = 4 6. x –3/2 = 8
7. x –2/3 = 1/4 8. x –3/2 = – 1/8
9. x 1.5 = 1/27 10. x 1.25 = 32
11. x –1.5 = 27 12. x –1. 25 = 1/32

26. F. Solve for x.
1. Is it always true that I+x| = x? Give reason for your answer.
2. Is it always true that |–x| = x? Give reason for your answer.
Absolute Value Equations
3. |4 – 5x| = 3 4. |3 + 2x| = 7 5. |–2x + 3| = 5
6. |4 – 5x| = –3 7. |2x + 1| – 1= 5 8. 3|2x + 1| – 1= 5
9. |4 – 5x| = |3 + 2x|
11. |4 – 5x| = |2x + 1| 12. |3x + 1| = |5 – x|
10. |–2x + 3|= |3 – 2x|
Solve geometrically for x. Draw the solution.
13. |x – 2| = 1 14. |3 + x| = 5 15. | –9 + x| = –7
x – 4
G. Find the zeros and the domain of the following rational
formulas. (See 2.1)
2x – 1
1. x2 – 1.
x2 – 4
3.
5x + 7
2.
3x + 5
x2 – x
x2 – x – 2
4.
x2 – 4x
5.
x2 + x – 2
x2 + 2x
6.
x2 + x + 2
2x2 – x – 1
7.
x3 + 2x
x4 – 4x
8.
x3 – 8
16. |2 + x| = 1 17. |3 – x| = –5 18. | –9 – x| = 8

27. (Answers to odd problems) Exercise A.
Exercise B.
1. No real solution 3. 𝑥 =
3
2
±
17
2
5. 𝑥 =
1
4
3 ± 17
1. 𝑥 = −1, 𝑥 = 4 3. 𝑥 = −4, 𝑥 = −3 5. 𝑥 = −2, 𝑥 = 5
7. 𝑥 = 2 9. 𝑥 = 0, 𝑥 = 2 11. 𝑥 = ±2, 𝑥 = 0
13. 𝑥 = −3, 𝑥 =
1
2
15. 𝑥 = ±2
1. 𝑥 =
13
9
3. 𝑥 = 23 5. 𝑥 = 30 7. 𝑥 =
2
5
, 𝑥 = 3
9. 𝑥 =
1
6
3 ± 57 11. 𝑥 =
1
6
5 ± 145
Exercise C.
Equations

28. Exercise D.
1. 𝑥 = 4 3. 𝑥 = 4 5. 𝑥 = 6 7. 𝑥 = 4, 𝑥 =
4
9
Exercise E.
1. 𝑥 = ±2 3. 𝑥 = 2
2
3 5. 𝑥 =
1
8
7. 𝑥 = 8
9. 𝑥 =
1
9
11. 𝑥 =
1
9
1. If 𝑥 < 0, the statement is not true 3. 𝑥 =
1
5
, 𝑥 = 7/5
5. 𝑥 = −1, 𝑥 = 4 7. 𝑥 = 5/2, 𝑥 = −7/2
9. 𝑥 = 1/7, 𝑥 = 7/3 11. 𝑥 =
3
7
, 𝑥 =
5
3
13. |𝑥 − 2| = 1
𝑥 = 1 or 𝑥 = 3
Exercise F.
2
1 3
left 1 right 1
Radical Equations and Power Equations

29. 15. No solution 17. No solution
Exercise G.
1. Zeros: 𝑥 = 1/2, domain: 𝑥 ≠ 4
3. Zeros: 𝑥 = ±2, domain: 𝑥 ≠ ±4
5. Zeros: 𝑥 = 0, 𝑥 = 4, domain: 𝑥 ≠ −2, 𝑥 ≠ 1
7. Zeros: 𝑥 = −
1
2
, 𝑥 = 1, domain: 𝑥 ≠ 0
Absolute Value Equations