1. MenemukanKonsepPersamaanKuadrat
1. Diketahui :
a. Ukurantanahkosong = 60 m × 30 m.
b. Luaslapangan yang direncanakan 1000 m.
c. Untukmemperolehluas
yang
diinginkanmakaukuranpanjangdanlebartanah di kurang x:
P = 60 – x
L = 30 – x
Ditanya :
a. Temukansebuahpersamaankuadrat = …?
Penyelesaian :
L=p×l
1000 = (60 – x) (30 – x)
1000 = 1800 – 60x – 30x + x
2
1000 = 1800 – 90x + x2
0 = 1800 – 1000 – 90x + x2
0 = 800 – 90x + x2
0 = x2 – 90x + 800
x2 – 90x + 800 = 0
2. Diketahui :
a. Ukuran plat seng :
P = 50 cm
l = 40 cm
b. Luas alas balok = 200 cm2
SMK NEGERI 1 TABANAN
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MATEMATIKA KLS X

2. Ditanya :
a. Persamaankuadrat = …?
b. Volume tempatair = …?
Penyelesaian :
a. p = 50 – 2x
l = 40 – 2x
L=p×l
200 = (50 − 2x) (40 – 2x)
200 = 2000 – 100x − 80x +4x2
200 = 2000 – 180x + 4x2
4x2 – 180x + 2000 – 200 = 0
4x2 – 180x + 1800 = 0
x2 – 45x + 450 = 0
x1.2 =
x1.2 =
x1.2 =
x1.2 =
x1 =
atau
x2 =
x1 =
x2 =
x1 = 30
x2 = 15
Pengujian :
P = 50 – 2x
P = 50 – 2x
atau
P = 50 – 2.30
P = 50 – 2.15
P = 50 – 60
P = 50 – 30
P = −10
P = 20
l = 40 – 2x
l = 40 – 2x
atau
l = 40 – 2.30
l = 40 – 2.15
l = 40 – 60
l = 40 – 30
l = −20
l = 10
SMK NEGERI 1 TABANAN
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MATEMATIKA KLS X

3. b. maka :
p = 50 cm – 30 cm
= 20 cm
l = 40 cm – 30 cm
= 10 cm
t = 15 cm
Volume balok = p × l× t
= 20 cm × 10 cm × 15 cm
= 3000 cm3
3. Volume mula-mula
V=⅓
V=⅓
Volume Karenapenambahanjari-jarisebesar 24 cm
V1 = ⅓
V1 =
(r + 24)2 . 3
(r + 24)2
Volume karenapenambahantinggi 24 cm
V2 = ⅓
(3 + 24)
V2 = ⅓
.27
V2 = 9
Jadi :
V1 = V2
(r + 24)2 = 9
r2 + 48r + 576 = 9
0=9
– r2 – 48r – 576
0 = 8 - 48r – 576
0=
– 6r – 72
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MATEMATIKA KLS X

4. − 6r – 72r = 0
(r – 12) (r + 6) = 0
r1 – 12 = 0
atau
r+6=0
r = −6
r1 = 12
Jari-jarikerucutsemula = 12 cm
4. Jawaban 1 :
a. Misal printer keduaperluwaktu x
b. Maka printer pertamaperluwaktu = x – 1
P1 = x−1 jam
P2 = x jam
P1 + P2 = 1,2 jam
Penyelesaian :
= 1,2
x2 – x = 2,4x – 1,2
x2 = 2,4x + x – 1,2
x2 = 3,4x – 1,2
0 = −x2 + 3,4x – 1,2
x2 – 3,4x + 1,2 = 0
10x2 – 34x + 12 = 0
5x(2x – 6) + 2(2x – 6) = 0
(5x + 2) (2x – 6) = 0
(5x + 2) = 0
atau
(2x – 6) = 0
5x = −2
x1 =
2x = 6
x2 = 3
jadiwaktu yang dibutuhkan printer jeniskeduaadalah 3 jam
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MATEMATIKA KLS X

5. MenentukanAkar-AkarPersamaanKuadrat
1. Selesaikanpersamaankuadratdibawahdengancaramemfaktorkan.
a. x2 + 5x – 50 = 0
x2 + 10x – 5x – 50 = 0
x (x + 10) – 5 (x + 10) = 0
(x – 5) (x + 10) = 0
x1= 5 atau
x2 = −10
b. x2 + 3x = 0
x (x + 3) = 0
x1 = 0 atau
x2 = −3
c. x2 – 4 = 0
(x – 2) (x + 2) = 0
x1 = 2 atau
x2 = −2
2
d. 2x + 3x + 1 = 0
2x2 + 2x + 1x + 1 = 0
2x (x + 1) + 1 (x + 1) = 0
(2x + 1) (x + 1) = 0
2x + 1 = 0
atau
x+1=0
2x = −1
x2 = 0 − 1
x1 = −
x2 = −1
e. 3x2 + 5x – 2 = 0
3x2 + 6x – 1x – 2 = 0
3x (x + 2) – 1 (x + 2) = 0
(3x – 1) (x + 2) = 0
3x – 1 = 0
atau
x+2=0
3x = 1
x2 = 0 – 2
x1 = 1
x2 = −2
2. Selesaikanpersamaankuadratberikutinidenganmelengkapkankuadratsempurna.
a. x2 + 5x + 4 = 0
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MATEMATIKA KLS X

6. x2 + 5x = −4
x2 + 5x +( )2 = −4 + ( )2
(x+ )2= −4 +
(x + )2=
+
(x + ) =
x+
=±√
x+ =±
x+ =
atau
x1= −
atau
x1=−
x2 = −
x1= −1
x+ =−
x2 = − −
x2 = −4
b. 2x2 – 14x + 12 = 0
x2 – 7x + 6 = 0
x2 – 7x = −6
x2 – 7x + (−
2
= −6 + (− )2
(x −
(x −
(x −
2
2
= −6 +
=−
2
+
=
x − = ±√
x− =±
x− =
atau
x1=
x2 = −
x1 =
x2 =
x1 = 6
x2 = 1
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x− =−
MATEMATIKA KLS X

7. 3. Selesaikanpersamaankuadratberikutinidenganrumusabc
a. x2 – 15x + 30 = 0
x1,2
x1,2
x1,2
x1,2
x1=
atau x2 =
b. x2 + 8x – 20 = 0
x1,2=
x1.2 =
x1,2=
x1,2 =
x1,2=
x1 =
atau
x2 =
x1 =
x2 =
x1= 2
x2 = −10
c. x2 + 3x = 0
x1,2=
x1,2=
x1,2 =
x1,2 =
x1,2=
x1=
x1= 0
SMK NEGERI 1 TABANAN
atau
x2 =
x2 =
7
MATEMATIKA KLS X

8. x2= −3
4. (p + 3)x2 + 3x – 4 = 0
SyaratD = 0
D = b2 – 4ac
0 = 32− 4(p+3) – 4
0 = 9 + 16 (p+3)
0 = 9 + 16p + 48
0 = 16p + 57
0 −57 = 16p
−57 = 16p
=p
p=
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MATEMATIKA KLS X