Question

Ile wynosi suma kwadratów pierwiastków równania x^2+mx-1=0?

Answer 1

Answer 2

X²+mx-1=0 

Δ= m² - 4*1*(-1) = m²+4   
√Δ = m+2

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

x1 + x2 = 1 + (-m-1) = -m