Question

Определите нули функции f(x)=3x^2-7x+4

Answer 1

f(x)=3x²-7x+4 3x²-7x+4=y0=0 D=(-7)²-4*3*4=49-48=1=√1=1 x1=(-(-7)+1)/4*3=8/12=2/3 x1=2/3 x2=(-(-7)-1)/4*3=6/12 x2=1/2 x0=0 y0=3*(x0)²-7*x0+4=0-0+4 x0=0 y0=4 y0=0 x1=2/3 x2=1/2