Section 1.5 Equations
Linear and Rational Equations
EXAMPLES:
1. Solve the equation 7x − 4 = 3x + 8.
Solution: We have
7x − 4 = 3x + 8
7x − 4 + 4 = 3x + 8 + 4
7x − 4 = 3x + 8
7x = 3x + 12
7x − 3x = 3x + 12 − 3x
or, in short,
7x − 3x = 8 + 4
4x = 12
4x = 12
x=
12
4x
=
4
4
x=3
12
=3
4
2. Solve the equation −5(x − 4) + 2 = 2(x + 7) − 3.
Solution: We have
−5(x − 4) + 2 = 2(x + 7) − 3
−5x + 20 + 2 = 2x + 14 − 3
−5x + 22 = 2x + 11
−5(x − 4) + 2 = 2(x + 7) − 3
−5x + 22 − 2x = 2x + 11 − 2x
−7x + 22 = 11
−5x + 20 + 2 = 2x + 14 − 3
or, in short,
−7x + 22 − 22 = 11 − 22
−7x = −11
−7x = −11
x=
−11
−7x
=
−7
−7
x=
−5x − 2x = 14 − 3 − 20 − 2
11
7
1
11
7
3. Solve the equation
y+7
y+5 y−2
−
=
+ 1.
2
4
3
Solution: We have
y+5 y−2
y+7
−
=
+1
2
4
3
y+5 y−2
y+7
12 ·
−
+1
= 12 ·
2
4
3
12 ·
y−2
y+7
y+5
− 12 ·
= 12 ·
+ 12 · 1
2
4
3
6(y + 5) − 3(y − 2) = 4(y + 7) + 12
6y + 30 − 3y + 6 = 4y + 28 + 12
3y + 36 = 4y + 40
3y + 36 − 4y = 4y + 40 − 4y
−y + 36 = 40
−y + 36 − 36 = 40 − 36
−y = 4
−y(−1) = 4(−1)
y = −4
In short,
y+5 y−2
y+7
−
=
+1
2
4
3
y+5 y−2
y+7
12 ·
−
+1
= 12 ·
2
4
3
6(y + 5) − 3(y − 2) = 4(y + 7) + 12
6y + 30 − 3y + 6 = 4y + 28 + 12
30 + 6 − 28 − 12 = 4y − 6y + 3y
y = −4
4. Solve the equation
21
15
=
+ 2.
y
4y
2
4. Solve the equation
21
15
=
+ 2.
y
4y
Solution: We have
15
y
15
4y ·
y
15
4y ·
y
4 · 15
21
+2
4y
21
= 4y ·
+2
4y
21
= 4y ·
+ 4y · 2
4y
= 21 + 8y
=
or, in short,
60 = 21 + 8y
15
21
=
+2
y
4y
15
21
4y ·
= 4y ·
+2
y
4y
60 = 21 + 8y
60 − 21 = 21 + 8y − 21
39 = 8y
39 = 8y
y=
8y
39
=
8
8
39
y=
8
5. Solve the equation
39
8
y
−5
5
=
+ .
y+5
y+5 4
Solution: We have
−5
5
y
=
+
y+5
y+5 4
y
5
−5
4(y + 5) ·
= 4(y + 5) ·
+
y+5
y+5 4
4(y + 5) ·
−5
5
y
= 4(y + 5) ·
+ 4(y + 5) ·
y+5
y+5
4
4y = 4 · (−5) + (y + 5) · 5
4y = −20 + 5y + 25
4y = 5 + 5y
4y − 5y = 5 + 5y − 5y
−y = 5
(−1)(−y) = (−1)5
y = −5
3
In short,
−5
5
y
=
+
y+5
y+5 4
y
5
−5
4(y + 5) ·
= 4(y + 5) ·
+
y+5
y+5 4
4y = 4 · (−5) + (y + 5) · 5
4y = −20 + 5y + 25
20 − 25 = 5y − 4y
y = −5
6. Solve the equation
x2
3
1
11
−
=
.
+ 5x + 4 x + 4
x+1
Solution: We have
x2
11
3
1
−
=
+ 5x + 4 x + 4
x+1
3
1
11
−
=
(x + 4)(x + 1) x + 4
x+1
11
1
3
(x + 4)(x + 1) ·
= (x + 4)(x + 1) ·
−
(x + 4)(x + 1) x + 4
x+1
(x + 4)(x + 1) ·
3
1
11
− (x + 4)(x + 1) ·
= (x + 4)(x + 1) ·
(x + 4)(x + 1)
x+4
x+1
11 − (x + 1) · 3 = (x + 4) · 1
11 − 3x − 3 = x + 4
8 − 3x = x + 4
8 − 3x − 8 = x + 4 − 8
−3x = x − 4
−3x − x = x − 4 − x
−4x = −4
−4x
−4
=
−4
−4
x=1
4
In short,
x2
3
1
11
−
=
+ 5x + 4 x + 4
x+1
11
3
1
−
=
(x + 4)(x + 1) x + 4
x+1
1
3
11
= (x + 4)(x + 1) ·
−
(x + 4)(x + 1) ·
(x + 4)(x + 1) x + 4
x+1
11 − (x + 1) · 3 = (x + 4) · 1
11 − 3x − 3 = x + 4
−3x − x = 4 − 11 + 3
−4x = −4
x=1
7. Solve for M the equation F = G
Solution: We have
Gm
M
F =
r2
The solution is M =
=⇒
mM
.
r2
r2
Gm
F =
r2
Gm
Gm
r2
M
=⇒
r2 F
=M
Gm
r2 F
.
Gm
8. The surface area A of the closed rectangular box can be calculated from the length l, the
width w, and the height h according to the formula
A = 2lw + 2wh + 2lh
Solve for w in terms of the other variables in this equation.
5
8. The surface area A of the closed rectangular box can be calculated from the length l, the
width w, and the height h according to the formula
A = 2lw + 2wh + 2lh
Solve for w in terms of the other variables in this equation.
Solution: We have
A = 2lw + 2wh + 2lh
A = (2l + 2h)w + 2lh
A − 2lh = (2l + 2h)w
A − 2lh
=w
2l + 2h
The solution is w =
A − 2lh
.
2l + 2h
Quadratic Equations
EXAMPLES:
1. Solve each equation:
(a) x2 = 0
(b) x2 = 1
(c) x2 = 4
6
(d) x2 = 5
(e) (x − 4)2 = 7
1. Solve each equation:
(a) x2 = 0
(b) x2 = 1
(c) x2 = 4
(d) x2 = 5
Solution:
(a) We have x = 0.
(b) We have (see the Appendix) x = ±1.
√
(c) We have (see the Appendix) x = ±2 (which is ± 4).
√
(d) We have (see the Appendix) x = ± 5.
(e) We have
(x − 4)2 = 7
√
x−4=± 7
h
i
√
x−4+4=± 7+4
The solutions are x = 4 −
√
x=4±
7 and x = 4 +
√
√
7
7.
2. Solve the equation x2 + 5x = 24.
Solution: We have
x2 + 5x = 24
x2 + 5x − 24 = 0
(x − 3)(x + 8) = 0
x − 3 = 0 or x + 8 = 0
x = 3 or x = −8
The solutions are x = 3 and x = −8.
3. Solve each equation:
(a) x2 − 8x + 13 = 0
(b) 3x2 − 12x + 7 = 0
7
(e) (x − 4)2 = 7
3. Solve each equation:
(a) x2 − 8x + 13 = 0
(b) 3x2 − 12x + 7 = 0
Solution:
(a) We have
x2 − 8x + 13= 0
x2 − 8x + 13 − 13= 0 − 13
x2 − 8x= −13
x2 − 2x · 4= −13
x2 − 2x · 4 + 42 = −13 + 42
(x − 4)2 = 3
√
x − 4= ± 3
√
x − 4 + 4= ± 3 + 4
√
x= 4 ± 3
In short,
x2 − 8x + 13 = 0
x2 − 2x · 4 = −13
x2 − 2x · 4 + 42 = −13 + 42
(x − 4)2 = 3
The solutions are x = 4 −
√
√
x−4=± 3
√
x=4± 3
3 and x = 4 +
√
3.
8
(b) We have
3x2 − 12x + 7= 0
3x2 − 12x + 7 − 7= 0 − 7
3x2 − 12x= −7
3x2 − 12x
=
3
3x2 12x
−
=
3
3
−7
3
−7
3
7
x2 − 4x= −
3
7
x2 − 2x · 2= −
3
7
x2 − 2x · 2 + 22 = − + 22
3
5
−7
−7 4
−7 4 · 3
−7 12
−7 + 12
2
=
+4=
+ =
+
=
+
=
(x − 2) =
3
3
1
3
1·3
3
3
3
3
r
5
x − 2= ±
3
r
5
+2
x − 2 + 2= ±
3
r
5
x= 2 ±
3
In short,
3x2 − 12x + 7 = 0
3x2 − 12x = −7
7
3
7
x2 − 2x · 2 + 22 = − + 22
3
5
(x − 2)2 =
3
r
5
x−2=±
3
r
5
x=2±
3
r
r
5
5
and x = 2 +
.
The solutions are x = 2 −
3
3
x2 − 4x = −
9
Proof: We have
ax2 + bx + c= 0
ax2 + bx + c − c= 0 − c
ax2 + bx= −c
ax2 + bx −c
=
a
a
ax2 bx −c
+ =
a
a
a
x2 +
b
c
x= −
a
a
b
c
=−
2a
a
2
2
b
c
b
b
2
+
=− +
x + 2x ·
2a
2a
a
2a
x2 + 2x ·
b
x+
2a
x+
2 −4ac + b2
−c
b2
b2
b2
−c
−c
−4ac
b2
=
=
+
+
+
=
=
=
+
a
(2a)2
a
2 2 a2
a
4a2
4a2
4a2
4a2
( r
b
= ±
2a
2
√
−4ac + b
−4ac +
√
=±
2
4a
4a2
b2
)
√
√
2
b − 4ac
b2 − 4ac
=±
=± √ √
2a
4 a2
√
b
b2 − 4ac
b
b
x+
− =±
−
2a 2a
2a
2a
√
√
−b ± b2 − 4ac
b
b2 − 4ac
±
=
x= −
2a
2a
2a
10
In short,
ax2 + bx + c = 0
ax2 + bx = −c
c
b
x2 + x = −
a
a
2
2
c
b
b
b
2
=− +
+
x + 2x ·
2a
2a
a
2a
2
−4ac + b2
=
4a2
√
b
b2 − 4ac
=±
x+
2a
2a
√
−b ± b2 − 4ac
x=
2a
b
x+
2a
EXAMPLES:
1. Solve the equation 3x2 − 4x − 5 = 0.
Solution: We first rewrite the equation as 3x2 + (−4)x + (−5) = 0. Here a = 3, b = −4, and
c = −5. Therefore by the quadratic formula,
p
√
√
√
−(−4) ± (−4)2 − 4 · 3 · (−5)
4 ± 16 + 60
4 ± 76
−b ± b2 − 4ac
=
=
=
x=
2a
2·3
2·3
2·3
√
4 ± 4 · 19
=
2·3
√ √
4 ± 4 19
=
2·3
√
2 · 2 ± 2 19
=
2·3
√
2(2 ± 19)
=
2·3
√
2 ± 19
=
3
In short,
x=
−(−4) ±
p
√
√
√
√
(−4)2 − 4 · 3 · (−5)
4 ± 76
4 ± 4 · 19
2 · 2 ± 2 19
2 ± 19
=
=
=
=
2·3
2·3
2·3
2·3
3
11
2. Solve the equation x2 = 4.
Solution: We first rewrite the equation as
1 · x2 + 0 · x + (−4) = 0
Here a = 1, b = 0, and c = −4. Therefore by the quadratic formula,
p
√
√
√
0 ± 02 − 4 · 1 · (−4)
± 0 + 16
± 16
±4
−b ± b2 − 4ac
=
=
=
=
= ±2
x=
2a
2·1
2
2
2
3. Find all solutions of each equation.
(a) 3x2 − 5x − 1 = 0
(b) 4x2 + 12x + 9 = 0
(c) x2 + 2x + 2 = 0
12
3. Find all solutions of each equation.
(a) 3x2 − 5x − 1 = 0
(b) 4x2 + 12x + 9 = 0
(c) x2 + 2x + 2 = 0
Solution:
(a) We first rewrite the equation as 3x2 + (−5)x + (−1) = 0. Here a = 3, b = −5, and c = −1.
Therefore by the quadratic formula,
p
√
√
√
−(−5) ± (−5)2 − 4 · 3 · (−1)
5 ± 25 + 12
5 ± 37
−b ± b2 − 4ac
=
=
=
x=
2a
2·3
6
6
(b) In this quadratic equation a = 4, b = 12, and c = 9. Therefore by the quadratic formula,
√
√
√
√
−12 ± 122 − 4 · 4 · 9
−12 ± 144 − 144
−12 ± 0
−b ± b2 − 4ac
=
=
=
x=
2a
2·4
8
8
=
−12 ± 0
8
=
−12
8
=−
3
2
(c) In this quadratic equation a = 1, b = 2, and c = 2. Therefore by the quadratic formula,
√
√
√
√
−b ± b2 − 4ac
−2 ± 22 − 4 · 2
−2 ± 4 − 8
−2 ± −4
x=
=
=
=
2a
2·1
2
2
√
Since the square of any real number is nonnegative, −4 is undefined in the real number
system. The equation has no real solution.
EXAMPLES:
1. Use the discriminant to determine how many real solutions each equation has.
1
(a) x2 + 4x − 1 = 0
(b) 4x2 − 12x + 9 = 0
(c) x2 − 2x + 4 = 0
3
13
1. Use the discriminant to determine how many real solutions each equation has.
1
(a) x2 + 4x − 1 = 0
(b) 4x2 − 12x + 9 = 0
(c) x2 − 2x + 4 = 0
3
Solution:
(a) We first rewrite the equation as 1 · x2 + 4x + (−1) = 0. Here a = 1, b = 4, and c = −1.
Therefore the discriminant is
D = b2 − 4ac = 42 − 4 · 1 · (−1) = 16 + 4 = 20 > 0
so the equation has two distinct real solutions.
(b) We first rewrite the equation as 4x2 + (−12)x + 9 = 0. Here a = 4, b = −12, and c = 9.
Therefore the discriminant is
D = b2 − 4ac = (−12)2 − 4 · 4 · 9 = 144 − 144 = 0
so the equation has exactly one real solution.
1
1
(c) We first rewrite the equation as x2 + (−2)x + 4 = 0. Here a = , b = −2, and c = 4.
3
3
Therefore the discriminant is
D = b2 − 4ac
1
= (−2) − 4 · · 4 =
3
2
−4
4·1·4
4 16
4 · 3 16
12 16
12 − 16
=
4−
= −
=
−
=
−
=
<0
3
1
3
1·3
3
3
3
3
3
so the equation has no real solution.
2. An object thrown or fired straight upward at an initial speed of v0 ft/s will reach a height
of h feet after t seconds, where h and t are related by the formula
h = −16t2 + v0 t
Suppose that a bullet is shot straight upward with an initial speed of 800 ft/s. Its path is shown
in the Figure below.
(a) When does the bullet fall back to ground level?
(b) When does it reach a height of 6400 ft?
(c) When does it reach a height of 2 mi?
(d) How high is the highest point the bullet reaches?
14
2. An object thrown or fired straight upward at an initial speed of v0 ft/s will reach a height
of h feet after t seconds, where h and t are related by the formula
h = −16t2 + v0 t
Suppose that a bullet is shot straight upward with an initial speed of 800 ft/s. Its path is shown
in the Figure below.
(a) When does the bullet fall back to ground level?
Solution: Since the initial speed in this case is v0 = 800 ft/s, the formula is
h = −16t2 + 800t
Ground level corresponds to h = 0, so we must solve the equation
0 = −16t2 + 800t
0 = −16t(t − 50)
Thus, t = 0 or t = 50. This means the bullet starts (t = 0) at ground level and returns to
ground level after 50 s.
15
(b) When does it reach a height of 6400 ft?
Solution: Setting h = 6400 in h = −16t2 + 800t gives
6400 = −16t2 + 800t
16t2 − 800t + 6400 = 0
t2 − 50t + 400 = 0
(t − 10)(t − 40) = 0
t = 10 or t = 40
The bullet reaches 6400 ft after 10 s (on its ascent) and again after 40 s (on its descent to
earth).
(c) When does it reach a height of 2 mi?
Solution: Two miles is 2 × 5280 = 10, 560 ft. Setting h = 10, 560 in h = −16t2 + 800t gives
10, 560 = −16t2 + 800t
16t2 − 800t + 10, 560 = 0
t2 − 50t + 660 = 0
The discriminant of this equation is
D = b2 − 4ac = (−50)2 − 4 · 660 = −140
which is negative. Thus, the equation has no real solution. The bullet never reaches a height
of 2 mi.
16
(d) How high is the highest point the bullet reaches?
Solution: Each height the bullet reaches is attained twice, once on its ascent and once on its
descent. The only exception is the highest point of its path, which is reached only once. This
means that for the highest value of h, the following equation has only one solution for t:
h = −16t2 + 800t
16t2 − 800t + h = 0
This in turn means that the discriminant D of the equation is 0, and so
D = b2 − 4ac = (−800)2 − 4 · 16 · h = 0
640, 000 − 64h = 0
h = 10, 000
The maximum height reached is 10,000 ft.
Other Types of Equations
EXAMPLES:
1. Solve the equation x3 = x.
Solution 1: We have
Solution 2: If x 6= 0, then
x3 = x
x3 = x
x3 − x = 0
x
x3
=
x
x
x(x2 − 1) = 0
x2 = 1
x(x − 1)(x + 1) = 0
x=0
or
x−1 =0
or
x=0
or
x=1
or
x = ±1
x+1 =0
x = −1
2. Solve the following equations:
(a) x6 = 16x2
(b) x7 = 27x4
17
Note that x = 0 is also a solution of the equation. This
gives the same result.
2. Solve the following equations:
(a) x6 = 16x2
(b) x7 = 27x4
Solution 1(a): We have
Solution 2(a): If x 6= 0, then
x6 = 16x2
x6
x6
x2
x4
√
4
x4
x6 − 16x2 = 0
x2 (x4 − 16) = 0
x2 (x2 )2 − 42 = 0
x2 (x2 − 4)(x2 + 4) = 0
|x| = 2
x2 (x − 2)(x + 2)(x2 + 4) = 0
Since x2 + 4 6= 0, we have
x2 = 0
x=0
or
or
x−2 =0
x=2
or
or
= 16x2
16x2
= 2
x
= 16
√
= 4 16
x = ±2
Note that x = 0 is also a solution of the equation. This
gives the same result.
x+2 =0
x = −2
Solution 1(b): We have
Solution 2(b): If x 6= 0, then
x7 = 27x4
x7
x7
x4
x3
√
3
x3
x7 − 27x4 = 0
x4 (x3 − 27) = 0
x4 (x3 − 33 ) = 0
x4 (x − 3)(x2 + 3x + 9) = 0
= 27x4
27x4
= 4
x
= 27
√
= 3 27
x=3
Since the discriminant of x2 + 3x + 9 is D =
32 − 4 · 1 · 9 = −27 < 0, it follows that x2 +
3x + 9 6= 0. Therefore
x4 = 0
or
x−3 =0
x=0
or
x=3
Note that x = 0 is also a solution of the equation. This
gives the same result.
3. Solve the equation 20a3 − 12a2 − 45a + 27 = 0.
Solution: We have
20a3 − 12a2 − 45a + 27 = 0
4a2 (5a − 3) − 9(5a − 3) = 0
(5a − 3)(4a2 − 9) = 0
(5a − 3) (2a)2 − 32 = 0
(5a − 3)(2a − 3)(2a + 3) = 0
5a − 3 = 0
5a = 3
3
a=
5
or
or
or
2a − 3 = 0
2a = 3
3
a=
2
18
or
or
or
2a + 3 = 0
2a = −3
3
a=−
2
4. Solve the equation
5
3
+
= 2.
x x+2
Solution: We have
3
5
+
=2
x x+2
3
5
· x(x + 2) = 2 · x(x + 2)
+
x x+2
5
3
· x(x + 2) +
· x(x + 2) = 2x2 + 4x
x
x+2
3(x + 2) + 5x = 2x2 + 4x
3x + 6 + 5x = 2x2 + 4x
0 = 2x2 + 4x − 3x − 6 − 5x
0 = 2x2 − 4x − 6
0 = x2 − 2x − 3
To solve x2 − 2x − 3 = 0 we can either factor
x2 − 2x − 3 = 0
(x − 3)(x + 1) = 0
x − 3 = 0 or x + 1 = 0
x = 3 or x = −1
or use the quadratic formula with a = 1, b = −2, and c = −3:
p
√
√
√
−(−2) ± (−2)2 − 4 · 1 · (−3)
2 ± 4 + 12
2 ± 16
2±4
−b ± b2 − 4ac
=
=
=
=
x=
2a
2·1
2
2
2
so
2+4
6
2−4
−2
x=
= =3
or
x=
=
= −1
2
2
2
2
The values x = 3 and x = −1 are only potential solutions. We must check them to see if they
satisfy the original equation.
Check: x = 3
Check: x = −1
5
3
+
=2
x x+2
5
3
+
=2
x x+2
3
5 ?
+
=2
3 3+2
3
5
?
+
=2
−1 −1 + 2
1 + 1 = 2 XTRUE
−3 + 5 = 2 XTRUE
We see that both x = 3 and x = −1 are the solutions of the equation
19
5
3
+
= 2.
x x+2
5. Solve the equation 2x = 1 −
Solution: We have
√
2 − x.
√
2x = 1 − 2 − x
√
2x − 1 = − 2 − x
√
(2x − 1)2 = (− 2 − x)2
(2x)2 − 2 · 2x · 1 + 12 = 2 − x
4x2 − 4x + 1 = 2 − x
4x2 − 4x + 1 − 2 + x = 0
4x2 − 3x − 1 = 0
To solve 4x2 − 3x − 1 = 0 we can either factor
4x2 − 3x − 1 = 0
(4x + 1)(x − 1) = 0
4x + 1 = 0 or x − 1 = 0
1
or x = 1
4
or use the quadratic formula with a = 4, b = −3, and c = −1:
p
√
√
√
−(−3) ± (−3)2 − 4 · 4 · (−1)
−b ± b2 − 4ac
3 ± 9 + 16
3 ± 25
3±5
x=
=
=
=
=
2a
2·4
8
8
8
so
−2
1
3+5
8
3−5
=
=−
or
x=
= =1
x=
8
8
4
8
8
1
The values x = − and x = 1 are only potential solutions. We must check them to see if they
4
satisfy the original equation.
x=−
Check: x = −
1
4
Check: x = 1
√
2x = 1 − 2 − x
s
√
1 ?
1
2x = 1 − 2 − x
2· −
=1− 2− −
4
4
√
?
r
2·1 =1− 2−1
1 ?
1
− =1− 2+
√
?
2
4
r
2 =1− 1
1 ?
9
− =1−
2 = 1 − 1 FALSE
2
4
1
3
− =1−
XTRUE
2
2
1
1
We see that x = − is a solution but x = 1 is not. So, the only solution is x = − .
4
4
√
√
6. Solve the equation 6 + 2x − x + 7 = −2.
20
6. Solve the equation
Solution 1: We have
√
6 + 2x −
√
x + 7 = −2.
√
6 + 2x − x + 7 = −2
√
√
6 + 2x = −2 + x + 7
√
√
( 6 + 2x)2 = (−2 + x + 7)2
√
√
6 + 2x = (−2)2 + 2(−2) x + 7 + ( x + 7)2
√
6 + 2x = 4 − 4 x + 7 + x + 7
√
6 + 2x − 4 − x − 7 = −4 x + 7
√
x − 5 = −4 x + 7
√
(x − 5)2 = (−4 x + 7)2
√
x2 − 2 · x · 5 + 52 = 16(x + 7)
x2 − 10x + 25 = 16x + 112
x2 − 10x + 25 − 16x − 112 = 0
x2 − 26x − 87 = 0
To solve x2 − 26x − 87 = 0 we can either factor
x2 − 26x − 87 = 0
(x − 29)(x + 3) = 0
x − 29 = 0 or x + 3 = 0
x = 29 or x = −3
or use the quadratic formula with a = 1, b = −26, and c = −87:
p
√
√
√
−(−26) ± (−26)2 − 4 · 1 · (−87)
−b ± b2 − 4ac
26 ± 676 + 348
26 ± 1024
x=
=
=
=
2a
2·1
2
2
26 ± 32
=
2
so
58
26 − 32
−6
26 + 32
=
= 29
or
x=
=
= −3
x=
2
2
2
2
The values x = 29 and x = −3 are only potential solutions. We must check them to see if they
satisfy the original equation.
Check: x = 29
√
√
6 + 2x − x + 7
√
√
6 + 2 · 29 − 29 + 7
√
√
64 − 36
8−6
Check: x = −3
√
√
6 + 2x − x + 7
p
√
6 + 2 · (−3) − −3 + 7
√
√
0− 4
0−2
= −2
?
= −2
?
= −2
= −2 FALSE
= −2
?
= −2
?
= −2
= −2 XTRUE
We see that x = −3 is a solution but x = 29 is not. So, the only solution is x = −3.
21
Solution 2: We have
√
6 + 2x − x + 7 = −2
√
√
( 6 + 2x − x + 7)2 = (−2)2
√
√
√
√
( 6 + 2x)2 − 2 6 + 2x x + 7 + ( x + 7)2 = 4
p
6 + 2x − 2 (6 + 2x)(x + 7) + x + 7 = 4
p
−2 (6 + 2x)(x + 7) = 4| − 6 − 2x
{z − x − 7}
|
{z
}
√
6x+42+2x2 +14x=2x2 +20x+42
−3x−9
√
−2 2x2 + 20x + 42 = −3x − 9
√
2 2x2 + 20x + 42 = 3x + 9
√
(2 2x2 + 20x + 42)2 = (3x + 9)2
4(2x2 + 20x + 42) = (3x)2 + 2 · 3x · 9 + 92
8x2 + 80x + 168 = 9x2 + 54x + 81
0 = 9x2 + 54x + 81 − 8x2 − 80x − 168
0 = x2 − 26x − 87
and the same result follows.
7. Solve the equation x4 − 8x2 + 8 = 0.
Solution: Setting W = x2 , we get
x4 − 8x2 + 8 = 0
(x2 )2 − 8x2 + 8 = 0
W 2 − 8W + 8 = 0
To solve this equation, we use the quadratic formula with a = 1, b = −8, and c = 8:
p
√
√
−(−8) ± (−8)2 − 4 · 8
−b ± b2 − 4ac
8 ± 64 − 32
W =
=
=
2a
2
2
√
√ √
√
√
√
√
8 ± 16 · 2
8 ± 16 2
8±4 2
8 4 2
8 ± 32
=
=
=
= ±
=4±2 2
=
2
2
2
2
2
2
so
q
√
√
2
x = 4 ± 2 2 =⇒ x = ± 4 ± 2 2
It follows that there are four solutions:
q
q
√
√
4+2 2
4−2 2
q
√
− 4+2 2
8. Solve the equation x1/3 + x1/6 − 2 = 0.
22
q
√
− 4−2 2
8. Solve the equation x1/3 + x1/6 − 2 = 0.
Solution: Setting W = x1/6 , we get
x1/3 + x1/6 − 2 = 0
(x1/6 )2 + x1/6 − 2 = 0
W2 + W − 2 = 0
(W − 1)(W + 2) = 0
W −1=0
or
W +2=0
W =1
W = −2
x1/6 = 1
x1/6 = −2
(x1/6 )6 = 16
(x1/6 )6 = (−2)6
x=1
x = 64
The values x = 1 and x = 64 are only potential solutions. We must check them to see if they
satisfy the original equation.
Check: x = 1
Check: x = 64
x1/3 + x1/6 − 2 = 0
x1/3 + x1/6 − 2 = 0
?
?
11/3 + 11/6 − 2 = 0
1 + 1 − 2 = 0 XTRUE
641/3 + 641/6 − 2 = 0
4 + 2 − 2 = 0 FALSE
We see x = 1 is a solution but x = 64 is not. So, the only solution is x = 1.
23
Absolute Value Equations
EXAMPLES:
1. Solve the equation |2x − 5| = 3.
Solution: By the definition of absolute value, |2x − 5| = 3 is equivalent to
2x − 5 = 3
or
2x − 5 = −3
2x = 8
2x = 2
x=4
x=1
The solutions are x = 4 and x = 1.
2. Solve the equation 5|2 − 4x| + 3 = 53.
Solution: We have
5|2 − 4x| + 3 = 53
5|2 − 4x| = |53{z
− 3}
50
|2 − 4x| =
50
= 10
5
By the definition of absolute value, |2 − 4x| = 10 is equivalent to
2 − 4x = 10
or
−4x = 10
− 2}
| {z
−4x = −10
| {z− 2}
8
x=
2 − 4x = −10
−12
8
= −2
−4
x=
−12
=3
−4
The solutions are x = −2 and x = 3.
3. Solve the equation | − x + 7| = |2x − 4|.
Solution: By the definition of absolute value, | − x + 7| = |2x − 4| is equivalent to
−x + 7 = 2x − 4
or
− x + 7 = −(2x − 4)
−x − 2x = −4 − 7
−x + 7 = −2x + 4
−3x = −11
x=
The solutions are x =
−x + 2x = 4 − 7
11
−11
=
−3
3
x = −3
11
and x = −3.
3
24
Appendix
1. Solve the equation x2 = 1.
Solution 1: We have
Solution 2: We have
x2 = 1
√
√
x2 = 1
x2 = 1
x2 − 1 = 0
x2 − 12 = 0
|x| = 1
(x + 1)(x − 1) = 0
x = ±1
x + 1 = 0 or x − 1 = 0
x = −1 or x = 1
The solutions are x = −1 and x = 1.
2. Solve the equation x2 = 4.
Solution 1: We have
Solution 2: We have
x2 = 4
√
√
x2 = 4
x2 = 4
x2 − 4 = 0
x2 − 22 = 0
|x| = 2
(x + 2)(x − 2) = 0
x = ±2
x + 2 = 0 or x − 2 = 0
x = −2 or x = 2
The solutions are x = −2 and x = 2.
3. Solve the equation x2 = 5.
Solution 1: We have
Solution 2: We have
x2 = 5
√
√
x2 = 5
√
|x| = 5
√
x=± 5
x2 = 5
x2 − 5 = 0
√
x2 − ( 5)2 = 0
√
√
(x + 5)(x − 5) = 0
√
√
x + 5 = 0 or x − 5 = 0
√
√
x = − 5 or x = 5
√
√
The solutions are x = − 5 and x = 5.
25