Question

Determine the number and type of complex solutions and possible real solutions for each of the following equations.

2x^2+5x+3=0

Answer 1

are you familiar with Descartes' rule of signs ?

Answer 2

THe solution of this equation -1 and -3/2.

Answer 3

use b^2-4ac i.e. discriminent to find the nature of roots since the coefficients are real therefore complex roots will be conjugate pairs.....

Answer 4

@yunus could you help me see how you got that becuase im not seein it

Answer 5

Ok i will try to explain.

Answer 6

see if D i.e disriminent is negative then its root has to be a complex quantity .....
since the roots are of the form..\[(-b \pm \sqrt{D})/2a\]
therefore u can see that if one is "-b+sqrtD" other will be "-b-sqrtD" where sqrtD is itself a complex number

Answer 7

Here i made a drawing. i hope u understand.

Answer 8

I want to explain something may help you.
if delta is greather than 0, then the equation has two real solution.
if delta is equal to zero then the equation has a real solution and solutions are same. that is x1=x2
if delta is less than zero then the equation has 2 complex solution.

Answer 9

In this case delta is 1. so the equation has 2 real different root (solutions).

Answer 10

You got it?

Answer 11

2x^2+5x+3=0

Answer 12

http://answers.yahoo.com/question/index?qid=20130121073646AA5xc4v