Section 1.5 Equations
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− = − = − = − =
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− = − = − = − =
