Question

3x²+11x-20=0

jawab dengan tiga cara
1.dengan pemfaktoran
2.dengan kuadrat sempurna
3.dengan rumus kuadratik/rumus abc
jangan ngasall ​
harus pakai jalan

Answer 1

Jawab:
x = {-5, 4/3}  ✅

Penjelasan dengan langkah-langkah:
Tentukan akar dari 3x² + 11x - 20 = 0 dengan

Pemfaktoran
3x² + 11x - 20 = 0
Ax² + Bx + C = 0
A = 3, B = 11, C = -20
↓
Ax² + Bx + C = 0
x² + Bx + AC = 0
↓
x² + 11x - 20(3) = 0
x² + 11x - 60 = 0
p+q = 11, pq = -60
p = 15, q = -4
krn 15+(-4) = 11, 15(-4) = -60
↓
Jika pemfaktoran x² + Bx + AC = 0
Bentuk pemfaktorannya
(1/A) · (Ax + p) (Ax + q) = 0
maka
(1/3) · (3x + 15) (3x - 4) = 0
maka
((3x+15)/3) · (3x - 4) = 0
((3x/3)+(15/3)) · (3x - 4) = 0
(x + 5) · (3x - 4) = 0
Untuk x+5 = 0,  x = -5
Untuk 3x-4 = 0, x = 4/3
↓
x = {-5, 4/3}  ✅

Kuadrat sempurna
3x² + 11x - 20 = 0
Ax² + Bx + C = 0
nilai A lebih dari 1, koefisien x² = A/A
(A/A)x² + (B/A)x + (C/A) = 0
(A/A)x² + (B/A)x = -(C/A)
(A/A)x² + 2((B/A)/2)x + ((B/A)/2)² = -(C/A) + ((B/A)/2)²
x² + 2((B/A)/2)x + ((B/A)/2)² = -(C/A) + ((B/A)/2)²
karena p² + 2pq + q² = (p+q)², maka
(x + ((B/A)/2))² = -(C/A) + ((B/A)/2)²
maka
(x + ((11/3)/2))² = -(-20/3) + ((11/3)/2))²
(x + (11/6))² = (20/3) + (11/6)²
(x + (11/6))² = (20(12)/(3(12))) + (121/36)
(x + (11/6))² = (240/36) + (121/36)
(x + (11/6))² = (361/36)
|x + (11/6)| = √(361/36)
|x + (11/6)| = (√361/√36)
|x + (11/6)| = (19/6)
Karena |a| = b, a = {-b, b}, maka
x = {-(19/6)-(11/6),  (19/6)-(11/6)}
x = {-30/6,  8/6}
x = {-5, 4/3}  ✅

Rumus kuadratik
3x² + 11x - 20 = 0
Ax² + Bx + C = 0
A = 3, B = 11, C = -20
Rumus kuadratik
x = {(-B+√(B²-4AC))/(2A), (-B-√(B²-4AC))/(2A)}
x = {(-11+√(11²-4·3·(-20)))/(2·3), (-11-√(11²-4·3·(-20)))/(2·3)}
x = {(-11+√(121-(-240)))/(6), (-11-√(121-(-240)))/(6)}
x = {(-11+√(121+240))/(6), (-11-√(121+240))/(6)}
x = {(-11+√(361))/(6), (-11-√(361))/(6)}
x = {(-11+19)/(6), (-11-19)/(6)}
x = {(-11-19)/(6), (-11+19)/(6)}
x = {-30/6,  8/6}
x = {-5, 4/3}  ✅

(XCVI)