Akar akar persamaan kuadrat (x+1)=36-x² adalah α dan β. Persamaan kuadrat baru yang akar akarnya (α+β-3) dan 2/5(αβ)) adalah.. Tolong sama caranya yaa hehehe

(x + 1) = 36 - x²
x + 1 - 36 + x² = 0
x² + x - 35 = 0

alpha + beta = -b/a
= -1/1
= -1

alpha . beta = c/a
= -35/1
= -35

cari akar akar yg baru , misal x1 dan x2
x1 = ( alpha + beta - 3)
= (-1-3)
= -4

x2 = 2/5 (alpha.beta)
= 2/5 (-35)
= -14

PK :
x² - (x1+x2)x + (x1.x2) = 0
x² - (-4+(-14))x + (-4.-14) = 0
x² + 18x + 56 = 0

Matematika VIII SMP
→→ Persamaan Kuadrat ←←

Pembahasan :
x + 1 = 36 - x²
x² + x - 35 = 0

Berdasarkan Teorema Vietta
aβ = c/a
aβ = -35

a + β = -b/a
a + β = -1

Maka...
x² - (a + β - 3 + 2/5aβ)x + (a + β - 3 . 2/5aβ) = 0
x² - (-4 - 14)x + (-4 . (-14)) = 0
x² + 18x + 56 = 0