Penjelasan dengan langkah-langkah:

It's known that:

f(x) → mx² + nx + 2n = 8x + 4

f(x) → mx² + nx - 8x + 2n - 4 = 0

f(x) → mx² + x(n-8) + (2n - 4), has roots:

x1 = n x2 = 1/m

it's also known that there is a formula for a quadratic equation f(x) → ax² + bx + c = 0 which has roots x1 and x2, so it can be applies:

x1 + x2 = -b/a

x1 × x2 = c/a

in the question:

a = m b = (n-8) c = (2n-4)

a. x1 + x2 = -b/a

n + 1/m = -((n-8)/m)

(nm+1)/m = -((n-8)/m)

nm+1 = -n+8

nm+n = 7 ..................... (1) 4m + 4 = 7 → m = ¾

x1 × x2 = c/a

n×1/m = (2n-4)/m

n = 2n-4

4 = n

b. x1 = 2m = 2.¾ = 3/2

x2 = -1/4

x1 + x2 = -b/a x1 × x2 = c/a

3/2 - ¼ = -b/a 3/2 × (-¼) = c/a

5/4 = -b/a -⅜ = c/a

10/8 = -b/a

b= -10

a = 8

c = -3

g(x) = 8x² - 10x - 3