jika diketahui jumlah kuadrat akar persamaan 2x² + (m+1)x-2m=0 sama dengan 10,maka jumlah akar persamaan kuadrat tersebut adalah... Terima kasih ☺

Mapel : Matematika
Kelas : X SMA
Bab : Persamaan Kuadrat

Pembahasan :
2x² + (m + 1)x - 2m = 0

X1² + X2² = 10
(X1 + X2)² - 2X1.X2 = 0
(X1 + X2)² - 2(2m/2) = 0
(X1 + X2)² - 2m = 0
(-m - 1)²/4 - 2m = 0
m² + 2m + 1 = 8m
m² - 6m + 1 = 0
m = 3 + 2√2
m = 3 - 2√2

Jika m = 3 + 2√2
m = 6

Persamaan Kuadratnya menjadi...
2x² + 7x - 12 = 0

X1 + X2 = -b/a
X1 + X2 = -7/2

Jika m = 3 - 2√2
m = 0

Persamaan Kuadratnya menjadi...
2x² + x = 0

X1 + X2 = -b/a
X1 + X2 = -1/2

2x^2 + (m + 1)x - 2m = 0
a = 2
b = m +1
c = - 2m

x1 + x2
= -b/a
= -(m+1)/2
= (-m-1)/2

x1.x2
= c/a
= -2m/2
= -m

(x1)^2 + (x2)^2 = 10
(x1 + x2)^2 - 2(x1.x2) = 10
( (-m-1)/2)^2 - 2(-m) = 10
( (m^2 + 2m + 1)/4) + 2m = 10
Kedua ruas × 4
m^2 + 2m + 1 + 8m = 40
m^2 + 10m + 1 - 40 = 0
m^2 + 10m - 39 = 0
(m + 13)(m - 3) = 0
m = - 13
Atau
m = 3

Mengikuti x1 + x2
= (-m-1)/2
Jika m = - 13 , maka jmlh akar :
= -(-13)-1/2
= 12/2
= 6

Jika m = 3 , maka jmlh akar
= -3-1/2
= -4/2
= - 2