1. The document describes solving several quadratic equations by factoring, completing the square, and using the quadratic formula. It provides examples of finding the roots of quadratic equations in various forms. 2. Methods for solving quadratic equations discussed include factoring, completing the square to convert it into the standard form of a quadratic equation, and using the quadratic formula. 3. Several examples are worked through step-by-step to demonstrate these different methods of solving quadratic equations algebraically.

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

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
b. 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)

3. 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
− 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
jawab :
printer A + printer B = 1,2 jam
(x – 1) + x = 1,2 jam
x – 1 + b = 1,2
2x – 1 = 1,2

4. 2x = 2,2
x = 1,1 jam
jawaban 2 :
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.

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

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
x− =−
x2 = 1
3. Selesaikanpersamaankuadratberikutinidenganrumusabc

7. 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
atau
x2 =
x2 =
x2= −3

8. 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=