How many solutions per each equation?

Question

asked Dec 25, 2019 81.3k views

1 Answer

Mpjdem · Answer 1 · 2019-12-31T10:24:29+0000

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

How many solutions per each equation?

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

Please log in or register to add a comment.

No related questions found

Categories

Other Questions