Question

find the nature of the roots of 2x^2+3x+1=0
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Answer 1

Answer:

2x²+3x+1=0

2x²+2x+1x+1=0

2x(x+1)+1(x+1)=0

(2x+1)(x+1)=0

2x+1=0 or x+1=0

2x=-1 or x=-1

x=-1/2 or -1

Answer 2

Answer:

$d \: = b { }^{2} - 4ac \\ = > d = {3 }^{2} - 4 \times 2 \times 1 \\ = > d = 9 - 8 \\ = > d = 1$

Since d<0, 2x^2+3x+1=0 has 2 real roots.