22hharmon936
22hharmon936
12-06-2018
Mathematics
contestada
How many solutions per each equation?
Respuesta :
tonb
tonb
12-06-2018
assuming y=ax²+bx+c, calculate D = b²-4ac, D<1: no solutions; D=0: 1 solution; D>0: 2 solutions.
1-4*-3*12 = 145, so 2 solutions
36-4*2*5 = -4, no solutions
49-4*-11 = 5, so 2 solutions
64-4*-1*-16 = 0, so 1 solution
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