persamaan kuadrat 2x²-px+1=0 dengan p>0 mempunyai akar alfa dan beta jika x²-5x+q=0 mempunyai akar akar 1/a² dan 1/beta² maka q-p=

Rumus: ax² + bx + c = 0
A + B = -b/a
A . B = c/a

Persamaan pertama:
A + B = p/2
A . B = 1/2

Persamaan kedua:
1/A² + 1/B² = 5
5 = [ ( B² + A² ) ] / [ A²B² ]
5 = [ ( B + A )² - 2AB ] / [ (AB)² ]
5 = [ ( p/2 )² - 2(1/2) ] / [ (1/2)² ]
5 = [ p²/4 - 1 ] / [ 1/4 ]
5/4 = ( p² - 4 )/4
Penyebut sama, coret aja
5 = p² - 4
p² = 9
p = 3

1/A² . 1/B² = q
q = 1 / A²B²
q = 1 / ( AB )²
q = 1 / [(1/2)²]
q = 1/1/4 = 4

Maka, q - p = 4 - 3 = 1
A = alfa, B = beta
Kalo bingung bisa ditanggapin ya.

2x^2 - px + 1 = 0
a + b = p/2
a . b = 1/2

Misal 1/a^2 = m dan 1/b^2 = n
m + n = 1/a^2 + 1/b^2
= (b^2 + a^2)/a^2.b^2
= ((b + a)^2 - 2ab) / (ab)^2
= ((p/2)^2 - 2(1/2)) / (1/2)^2
= ((p^2)/4 - 1) / (1/4)
= p^2 - 4
m.n = 1/a^2. 1/b^2 = 1/(ab)^2 = 1/(1/2)^2 = 1/(1/4) = 4

Persamaan kuadrat yg akar-akarnya 1/a^2 dan 1/b^2 adalah
x^2 - (m + n)x + mn = 0
x^2 - (p^2 - 4)x + 4 = 0
x^2 - 5x + q = 0 ===> (diketahui disoal)
Jadi
p^2 - 4 = 5 => p^2 = 9 => p = 3
Dan
q = 4

q - p = 4 - 3 = 1