Jika x1 dan x2 merupakan akar akar persamaan 2x^2+5x =3 , maka persamaan kuadrat baru yg akar akarnya x1/x2 dan x2/x1 adalah

misal persmaan kuadrad :
ax²+bx+c=0
-b/a = jumlaj akar2nya
c/a = perkalian akar-akarnya

2x²+5x=3
2x²+5x-3=0
-5/2 = x1 + x2
-3/2 = x1.x2

dibuat persmaan kuadrat bru yg akar-akarnya x1/x2 dan x2/x1
misal persmaan kuadrat baru
ax²+bx+c=0 dengan a = 1
x²+ bx + c = 0
-b/1 = x1/x2 + x2/x1
-b = (x1²+x2²)/(x1.x2)
-b = ((x1+x2)²-2x1.x2)/(x1.x2)
-b = ((-5/2)²-2(-3/2))/(-3/2)
-b = (37/4)/(-3/2)
-b = -37/6
b = 37/6

c/1 = (x1/x2)(x2/x1)
c = 1

jdi prsmaan kuadrat baru adalah
x²+(37/6)x + 1 = 0
6x² + 37x + 6 = 0

Pers. Kuadrat

2x²+5x-3 = 0
• x1+x2 = -5/2
• x1.x2 = -3/2
...
Misal,
p → x1/x2 + x2/x1
p = x1²/x1.x2 + x2²/x1.x2
p = x1²+x2²/x1.x2
p = (x1+x2)²-2x1.x2 / x1.x2
p = (-5/2)²-2(-3/2) / (-3/2)
p = 25/4 + 12/4 / (-3/2)
p = 37/4 / -3/2
p = -37/6
...
q → x1/x2 . x2/x1
q = 1

Maka, PK barunya :
x² - px + q = 0
x² - (-37/6)x + 1 = 0
6x²+37x+1 = 0